The figure figure shows shows the extern external al forces forces actactProblem Problem 3.1 The ing on an object in equilibrium. The forces F 1 = 32 N and F 3 = 50 N. Determine F 2 and the angle α.
y F 1
30°
12° F 3
Solution:
x
α
Write the the forces in component component form.
F 2
F1 = 32sin30◦ i + 32cos30◦ j
27 .7 j N F1 = 16 i + 27. y
−50cos12◦ i − 50sin12◦ j = −48. 48.9i − 10. 10.4 j (N) = F2 cos αi − F2 sin α j
F2 = F2 F2
F 1
30°
Sum components in x and y directions
F x = 16
− 48. 48.9 + F 2 cos α = 0 27.7 − 10. 10.4 − F2 sin α = 0 Fy = 27.
Solving, we get
x
12°
α
F 3 F 2
F z = 37. 37 .2 N α = 27. 27.73◦
and the the angl anglee α = Problem Problem 3.2 Theforce F 1 = 100 N and negligible. Determine the the 60 . The weight of the ring is negligible. forces F 2 and F 3 .
y
◦
F 2
30° x F 1
α
F 3
Solution:
Write the the forces in component component form.
F1 = F 1 i + 0 j
F 2
−F 2 cos30◦i + F 2 sin30◦ j = −F 3 cos αi − F 3 sin α j
F2 = F3
F x = 0 and F = 0, thus We know equilibrium equations, we have
− F 2
F y = 0. Writing the
30° x F 1
cos30◦
F x
= F 1
F y
= F 2 sin30◦
F 1 = 100 N, α = 60
y
− F 3 cos α = 0 − F 3 sin α = 0
α
◦
Solving, we get
F 2 = 86. 86.6 N, F 3 = 50 N
F 3
Consider Consider the force forcess shown shown in ProbProblem Problem 3.3 lem 3.2. 3.2. Suppose Suppose that that F 2 = 100 100 N and you want to choose the angle α so that the magnitude of F 3 is a minimum. What is the resulting magnitude magnitude of F 3 ? Strategy: Draw a vector vector diagram diagram of the sum sum of the three forces. Solution:
|F2| = 100◦ N, F1 is horizontal, and α = 90 and |F3 | = 50 N From the diagram, α =
F = 0. F1
α F2
F3 min 60° 30° x
The beam is in in equili equilibrium brium.. If A Problem 3.4 A x = beam’ss weightis weightis negli negligib gible, le, A x 77 kN, B = 400 kN, andthe beam’ what are the forces Ay and C ?
A y
B
30°
2m
C
4m
Solution:
→ ↑ +
+ Ax
F x F y
sin30◦ = 0 − C sin30 = A y − B + C cos30 cos30◦ = 0 = A x
A x B
A y
= 77 kN, B = 400 kN
2m
Solving, we get
30° 4m
Ay = 267 kN C = = 154 kN
Suppose se that that themassof thebeamshown thebeamshown Problem Problem 3.5 Suppo inProble inProblem m 3.4 3.4 is20 kgand itis in equi equili libr briu ium. m. The The forc forcee Ay points upward. If A y = 258 kN and B = 240 kN, what are the forces Ax and C ? Solution:
→ ↑ +
+ Ay
F x F y
Solving, we get
sin30◦ = 0 − C sin30 = 0 = A y − B − (20)(9. (20)(9.81) + C cos30 cos30◦ = 0
A x
C
= 0 = A x
= 258 kN, B = 240 kN
A y
30°
B
(20 kg) (9.81)m/s 2
Ax = 103 kN C = = 206 kN
C
Problem 3.6 A zoologist estimates that the jaw of a predator, Martes, is subjected to a force P as large as 800 N. What forces T and M must be exerted by the temporalis and masseter muscles to support this value of P ? 22° T P
M
36°
Solution:
Resolve the forces forces into scalar components, components, and solve solve the equilibrium equations . . .Express the forces in terms of horizontal and vertical unit vectors:
cos 22◦ + j sin sin 22◦ ) = T (0. (0.927i + 0. 0 .375 j) T = T (i cos
| |
| |
P = 800(i cos270◦ + j sin270◦ ) = 0 i ◦
M = M (i cos144 + j sin144
| |
◦
− 800 j ) = |M|(−0.809i + 0. 0 .588 j)
Apply the equilibrium conditions,
F =0 = T+M+P =0
Collect like terms:
(0.927 T Fx = (0.
| | − 0.809|M|)i = 0
Fy = ( 0.375 T
| | − 0.588|M| − 800) j = 0 .809 |M| = 0. Solve the first equation, |T| = 0 0 .873|M| 0.927
Substitute this value into the second equation, reduce algebraically, and solve:
|M| = 874 N, |T| = 763. 763.3 N 22° T P
M
36°
are identical, identical, with unProblem 3.7 The two springs are stretched lengths 250 mm and spring constants k = 1200 N/m. A. (a) Draw the free-body diagram of block A B . (b) Draw the free-body diagram of block B (c) What are the masses masses of the two blocks? blocks?
300 mm
A
280 mm
B
Solution: The tension in the upper upper spring acts on block block A in the positive Y direction, Solve the spring force-deflection equation for the tension in the upper upper spring. Apply the equilibrium equilibrium conditions conditions A. Repeat the steps for block B B . to block A 300 mm TUA = 0i +
1200
N m
(0. (0.3 m
− 0.25 m) j = 0 i + 60 j N A
Similarly Similarly,, the tensionin tensionin the lower lower spring spring actson block block A in the negativ negativee Y direction direction 280 mm
−
TLA = 0i
1200
N m
The weight is WA = 0i
(0. (0.28 m
− 0.25 m) j = 0 i − 36 j N B
− |WA | j j
The equilibrium conditions are
F=
Fx +
0, Fy = 0,
F = WA + TUA + TLA = 0
Tension, upper spring
Collect and combine like terms in i , j
Solve
Fy = (
−|WA| + 60 − 36) j = 0 |WA | = (60 − 36) = 24 N
A
Tension, lower spring
The mass of A A is
mA =
Weight, mass A
|WL| = 24 N = 2. |g| 9.81 m/s2 2 .45 kg
B is shown. The free body diagram for block B y
The tension in the lower spring TLB = 0i + 36 j WB j j The weight: W B = 0 i Apply the equilibrium conditions to block B .
−|
|
B x
F = WB + TLB = 0
Collect and combine like terms in i , j :
Solve:
Fy = (
−|WB | + 36) j = 0
|WB | = 36 N
B is given by mB = The mass of B
|WB | |g|
=
Tension, lower spring
36 N 9.81 m/s2
= 3. 3 .67 kg
Weight, mass B
The two two spri spring ngss in Prob Proble lem m 3.7are iden iden-Problem Problem 3.8 The tical, with unstretched lengths 250 mm and spring constants k . The sum of the masses masses of blocks blocks A and B is 10 kg. Determine the value of k and the masses of the two blocks.
Solution: All of the forces forces are in the vertical vertical direction direction so we will use scalar equation equations. s. First, First, consider consider the upper spring supporti supporting ng both masses (10 kg total mass). The equation of equilibrium equilibrium for block the entire assem (m ( mA + m B )g = 0, bly supported by the upper spring is A is T UA UA where T UA N. The equation equati on of equilibrium equilibrium for block block = k ( 0. 0 . 25) UA U , where wher e N. The Th e equaequaB is T UB m g = 0 T = k ( 0. 0 . 25) UB B UB UB L + T m g = 0 tion of equilibrium for block A alone is T UA where UA LA LA A T LA T UB Using g = 9.81 m/s2 , and solving simultaneously, we get LA = UB . Using
− −
−
−
−
k = 1962 N/m, mA = 4 kg, and m B = 6 kg .
Problem 3.9 The 200-kg horizontal steel bar is suspended by the three springs. The stretch of each spring is 0.1 m. The constant of spring B is k B = 8000 N/m. Determine the constants kA = kC of springs A and C .
Solution: δ = 0.1 m K A = K C C +
↑
F y = K A δ + K B δ + K C C δ
(200)(9.81) = 0 − (200)(9.
2K A (0. (0.1) + (8000)(0. (8000)(0.1) = 1962 N Solving
K A = 5810 N/m = K = K C C
A
K A δ
B
C
K B δ
K C δ
(200) (9.81) N
A
B
C
−
Problem 3.10 The mass of the crane is 20 Mg (megagrams), and the tension in its cable is 1 kN. The crane’s crane’s cable is attached to a caisson whose mass is 400 kg. Determi termine ne the magnit magnitude udess of the norma normall and fricti friction on forces forces exerted on the crane by the level ground.
Draw thefree-bo thefree-body dy diagra diagram m ofthe crane crane and Strategy: Draw the part of its cable within the dashed line.
45°
Solution:
Resolve the forces forces into scalar components, components, and solve solve the equilibrium equations. The external forces are the weight, the friction force, the normal force, and the tension in the cable. The weight vector is
W = 0i
)(9.81 m/s2 ) j W = 0 i − 196, 196,200 j − mc |g| j j = 0 i − (20000 kg)(9.
45°
The normal force vector is N = (0i + Ny + j). The friction force by definition acts at right angles to the normal force, in a direction that holds the crane in place. Fx =
y
45°
−|Fx|i + 0 j
F
The angle angle between between thetensionvectorand thepositive thepositive x axisis 315◦ , hence the tension vector projection is T = T (i cos315◦ + j sin315◦ ) = 707i
| |
T
−45◦ =
W
N x
− 707 j N Solve:
The equilibrium conditions are,
Fx = (
−|Fx| + 707)i = 0,
Fy = ( Ny
| | − 707 − 196200) j = 0
196,907 N. |Fx| = 707 N |Ny | = 196200 + 707 = 196, Thus the friction force is directed toward the left, and the normal force acts upward.
What is the tension tension in the horizontal horizontal Problem 3.11 cable AB in Example 3.1 if the 20 angle is increased to 25 ? ◦
◦
Solution: A
m = 1440 kg mg = mg = (1440)9. (1440)9.81
or
− − N sin25 sin25◦ = 0 F y = N cos25 cos25◦ − mg = mg = 0 F x = T
− −
T N sin25 sin25◦ = 0 N cos25 cos25◦ 14. 14.126 kN = 0
−
25°
y
Solving, we get
T
T = 6. 6 .59 kN, N = 15. 15 .59 kN
x
25°
mg
N
B
2400-lb car will remain remain in equilibequilibProblem 3.12 The 2400-lb rium rium on the slopin sloping g road road only only if the fricti friction on force force exert exerted ed on the car by the road is not greater than 0.6 times the normal force. What is the largest angle angle α for which the car will remain in equilibrium? equilibrium?
α
Solution:
F x = W sin W sin α
y
0, − f = 0,
F y = N
cos α = 0. − − W cos
0 .6 N and write the equilibrium equations as Set f = 0.
α
W sin sin α = 0. 0 .6 N, (1) W cos cos α = N.
W
(2)
f
Divide Eq. (1) by Eq. (2):
sin α = tan α = 0.6. cos α Solving,
α = 31. 31 .0
N
◦
x
Problem Problem 3.13 The crate crate is in equili equilibri brium um on the smooth surface. (Remember that “smooth” “smooth” means that friction friction is negligi negligible) ble).. The spring spring constant constant is k = 2500 N/m and the stretch of the spring is 0.055 m. What is the mass of the crate? 20°
Solution: K = = 2500 N/m δ = 0.055 m
+ +
−
F x =
−Kδ + m(9. (9.81) 81) sin sin 20◦ = 0
F y = N-m(9. (9.81) 81) cos cos 20◦ = 0
20°
(2500)(0. (2500)(0.055) + 3. 3.355 m = 0 N 9.218 m = 0
− −
y
Solving, m = m = 41. 41.0 kg, N = = 378 (N)
δ K δ
x N
20°
mg = 9.81 m
A 600-lb 600-lb box is held held in place place on the Problem 3.14 smooth bed of the dump truck by the rope AB . α = 25 , what is the tension in the rope? (a) (a) If α (b) If the rope will will safely support a tension of 400 400 lb, what is the maximum allowable value of α? ◦
B A
α
Solution: Isolate the box. Resolve the forces into scalar components, and solve the equilibrium equations. equations. The external forces are the weight, the tension in the rope, and the normal force exerted by the surface. The angle between the x axis and the weight vector is (90 α) (or 270 + α). The weight vector is is
− −
W = W (i sin α
| |
− j cos α) = (600)(i sin α − j cos α)
The projections of the rope tension and the normal force are T=
−|Tx |i + 0 j
N = 0 i + Ny j j
| |
The equilibrium conditions are
F = W+N+T =0
Substitute, and collect like terms
Fx = (600sin α
− |Tx|)i = 0 Fy = (−600cos α + |Ny |) j = 0
Solve for the unknown tension when
α = 25◦ Tx = 600sin α = 253. 253.6 lb.
| |
For For a tens tensio ion n of400lb,( 600sin α angle
sin α =
− 400) = 0. Solvefor Solvefor the unknown unknown
400 = 0. 0 .667 or α = α = 41. 41.84◦ 600
A B
α
y T
x N W
α
forces act on the free-body diaProblem 3.15 Three forces gram of a joint of a structure. If the structure is in equilibrium and F A = 4.20 kN, what are F B and F C C ?
F C
F B
15° 40° F A
Solution:
F x F y
F A
= F A cos40◦
− F B cos15◦ = 0 = F A sin40◦ − F B sin15◦ − F C C = 0
= 4. 4 .20 kN
Substitute in the value for F A and solve the resulting two equations in two unknowns. We get
F B = 3.33 kN, F C 1 .84 kN C = 1.
F C
F B
15°
40°
F A
y F A
40° x
15° F B
FC
The weight weightss of thetwo blocksare blocksare W 1 = Problem Problem 3.16 The 200 lb and W 2 = 50 lb. Neglecting friction, determine the force the man must exert to hold the blocks in place?
30° W 1
30°
W 2
Solution: Isolate block W 2 and apply equilibrium conditions. W 1 . Repeat for block W For W 2 : The weight vector: vector: W 2 = 0i
T2
y
− 50 j
j The rope tension: T 2 = 0i + T2 j
| |
The equilibrium conditions are x
Fx = 0,
W2
Fy = ( 50 + T2 ) j = 0 , or T2 = 50
−
| |
|
For W 1 : The magnitude of of the rope tension tension T2 is unchanged by passage over the frictionless lower pulley, hence,
| |
y T1 x
T2 = T2 i + 0 j = 50 i + 0 j.
| |
T1 i + 0 j. j. The normal The rope tension T 1 : T1 = normal force force is is N = 0 i + N j. j. The angle between the x axis and the weight vector is (90 α) (or 270 + α). The projection of the weight vector is
− −
−| |
| |
W = W (i sin α
| |
173.2 j. − j cos α) = 100i − 173.
The equilibrium conditions are
F = T1 + T2 + N + W1 = 0
Substitute and collect like terms,
Fx = (
0, −|T1 + 50 + 100)i = 0, 173.2) j = 0 Fy = (|N − 173.
Solve: T1 = 150 lb. Since the frictionless pulley pulley does not change themagnitu themagnitudeof deof theropetension theropetension,, then then the the tensio tension n at the the man’shand man’shandss is T1 = 150 lb.
| |
| |
30°
W 1
30°
W 2
N
α W1
T2
have the the same unstretProblem 3.17 The two springs have ched length, and the inclined inclined surface is smooth. smooth. Show that the magnitudes of the forces exerted by the two springs are
F 1
=
W α , sin
F 2
k
=
1 + k2 1
W α
W
k
1 + k1 2
α
Isolate Isolate the block block.. Apply Apply the linear linear spring spring forceforcedeflection relations to find the ratios of the spring spring forces. The spring forces are,
−|F1 |i + 0 j,
F2 =
−|F2 |i + 0 j.
j. The angle between the x axis The normal force is, N = 0i + N j. and the weight vector is (90 α) (or 270 + α). The weight vector is
− −
W = W (i sin α
| |
− j cos α).
| |
The equilibrium conditions are
F = W + N + F1 + F2 = 0.
Fx = ( W sin α
Substitute and collect like terms,
Fy
0, | | − |F1| − |F2|)i = 0, = ( |N| − W cos cos α) j = 0. 0.
For equal extensions, ∆L ∆ L1 = ∆L2 = ∆L, the forces are F1 = k1 ∆L, and F2 = k 2 ∆L.
| |
| |
|F |
The ratio is, |F1 | = 2
k1 k2
.
| |
| |
Substitute to eliminate the unknowns F2 , and F1 . Solve,
|F2| = |W| sink α , |F1| = |W| sink α
1+
1
k2
k 2
sin
Solution:
F1 =
k 1
1+
2
k1
k 1 k 2 W
α
y F1 F2 x N W
α
A 10-kg 10-kg painting painting is suspen suspended ded by a Problem 3.18 wire. If α α = 25 , what is the tension in the wire? ◦
α
α
Solution:
Isolate support pin pin fixed to the wall or other other support. The angle of the right hand wire with the positive x axis is α, hence the tension is
−
F2 = F2 (i cos α
| |
α
α
− j sin α)
The angle of the left hand wire is (180◦ + α) hence F1 = F1 ( i cos α
| |−
− j sin α).
j The weight is W = 0 i + W j
| |
The equilibrium conditions are
y
W
F = W + F1 + F2 = 0
Fx = ( F2 cos α
x
Substitute the vector forces, and collect like terms,
Fy
0, | | − |F1 | cos α)i = 0, = ( |W| − |F2 | sin α − |F1 | sin α) j = 0. 0.
| | | |
|F1| = |F2 |
α
α
F2
α = 25◦ and W = (10 kg) 9.81 m With α = s2
Thus F1 = F2 , and
1 = 2
F1
| |
| | W
sin α
|F1 | = |F2| =
.
1 2
98. 98.1 0.423
= 116. 116.06 N
If the wire supporting supporting the the suspended suspended Problem 3.19 painting in Problem 3.18 breaks when the tension exceeds 150 N and you want a 100 percent safety factor (that is, you want the wire to be able to support twice the actual weight of the painting), what is the smallest value α you can use? of α Solution:
From Problem 3.18
|F1| = |F2 |
and W = (10 kg)
| |
1 = 2
W
9.81
m
s2
y
W
sin α
= 98 .1 N.
x
Thus
sin α = sin α =
| | 1 2
98. 98.1
1 2
98. 98.1 75
F
For a tension F =
| |
150 = 75 , 2
= 0. 0 .654 or α = α = 40. 40.8◦
F1
α
α
F2
= 98 .1 N
Assume that that the 150-lb climber climber is is in Problem 3.20 equilibri equilibrium. um. What are the tensions tensions in the rope on the left and right sides?
14°
Solution:
F x = T R cos(15◦ )
− T L cos(14◦) = 0 F y = T R sin(15◦ ) + T L sin(14◦ ) − 150 = 0
Solving, we get T L = 299 lb, T R = 300 lb
14°
15°
y
14°
T R
T L
15° x
150 lb
15°
If the mass mass of the climb climber er shown shown in Problem 3.21 Problem 3.20 is 80 kg, what are the tensions in the rope on the left and right sides? Solution:
F x F y
y
= T R cos(15◦ )
− T R cos(14◦ ) = 0 = T R sin(15◦ ) + T R sin(14◦ ) − mg = mg = 0
Solving, we get
T R
T L
T L = 1.56 kN,
T R = 1.57 kN
14°
15° x mg = (80) (9.81) N
construction worker worker holds a 180-kg Problem 3.22 A construction crate in the position shown. What force must she exert on the cable? 5°
30°
Solution:
mg
F x F y
Eqns. of Equilibrium Equilibrium :
= T 2 cos30◦
sin 5◦ = 0 − T 1 sin = T 1 cos cos 5◦ − T 2 sin30◦ − mg = mg = 0
= (180)(9. (180)(9.81) N
Solving, we get
T 1 = 1867 N T 2 = 188 N 5° 5°
y
T 1
30°
x
30°
T 2
mg = (180) (9.81) N
A construct construction ion worker worker on on the moon moon Problem 3.23 (acceleration due to gravity 1.62 m/s2 ) holds the same crate described in Problem 3.22 in the position shown. What force must she exert on the cable?
5°
30°
Solution Eqns. of Equilibrium Equilibrium
F x F y
mg
= T 2 cos30◦ cos cos 5◦
sin 5◦ = 0 − T 1 sin mg = 0 − T 2 sin30◦ − mg =
= T 1 = (180)(1. (180)(1.62) N
Solving, we get
T 1 = 308 N T 2 = 31. 31.0 N
5°
30°
5°
y
T 1
x
30°
T 2
mg = (180) (1.62) N
needs to Problem 3.24 A student on his summer job needs pull pull a crate crate acrossthe acrossthe floor floor. Pullin Pulling g as shown shown in Fig. Fig. a, he canexert canexert a tensio tension n of 60 lb. lb. He finds finds that that thecrate thecrate doesn’ doesn’tt move, so he tries the arrangement in Fig. b, exerting a verti verticalforceof calforceof 60 lb on therope. What What is themagnitud themagnitudee of thehorizont thehorizontal al force force he exert exertss on thecratein each each case? case?
20°
(a)
10°
(b)
Solution: (a)
The force diagram for for part (a) is as shown. shown. The horizontal component of the 60 lb is
20°
◦ F horiz 56.4 lb . horiz = (60) cos(20 ) = 56.
(b)
(a)
The free body body diagram for the point where where the student’ student’ss hands grasp the rope is shown to the right. right. The equations of equilibequilibrium are
and
◦ F x = F floor floor cos(10 )
F y = 60 lb
10°
− F box box = 0, ◦
− F floor floor sin(10
(b)
) = 0. 0.
Solving these two equations simultaneously, we find that y
F floor 345.5 lb, and F box 340.3 lb . floor = 345. box = 340.
60 lb
Note: We should keep this problem in mind when we try to exert a large force on an object. Here, the floor did most of the pulling and the arrangement amplified the student’s effort by a factor of almost six. Note that the angles are critical in this Problem. Small changes can make big differences.
20°
x
y
60 lb 10° F box
x F floor
The 140-kg 140-kg traffic traffic light is suspended suspended Problem 3.25 above the street street by two cables. What is the tension in the cables?
20 m
20 m
B
C
12 m
A
Solution:
Isolate the traffic light. light. From symmetry, symmetry, the angles α formed by the suspension cables are equal.
tan α =
12 m 20 m
∼
α = 30. 30.964◦ = 31 ◦
= 0. 0 .6,
+x axis is α. The tension is The angle formed by cable C and the +x C = C (i cos α + j sin α).
| |
+ x axis is (180◦ The angle formed by cable B and the +x tension is B = B (i cos(180
| |
− α) + j sin(180 − α)). )).
The weight is W = 0 i
− α). The
− j|W|. The equilibrium conditions conditions are
0. F = W + B + C = 0.
Substitute, and collect like terms. From the first equation, B = C . Substitute this into the second equation
| | | |
|B| = |C| =
| |
.
1 2
W
sin α
For values of
|W| = (140 kg) 9.81 sm2 30 .96 ∼ and α = 30. = 31 ◦ , |B| = |C| =
1 2
1373. 1373.4 sin α
= 1334. 1334.7 N.
C
B A
y
B
20 m
20 m 12 m
= 1373. 1373.4 N
α
α
W
x C
the suspended suspended traffic traffic light light in Problem 3.26 Consider the Problem 3.25. To raise the light temporarily temporarily during a parade, an engineer wants to connect the 17-m length of cable DE to to the midpoints of cables AB and AC as as shown. However, However, for safety considerations, considerations, he doesn’t want to subject any of the cables to a tension larger than 4 kN. Can he do it?
17 m D
E
B
C
A
Solution: Determine Determine the the length of AC and AB from Problem 3.26: The distance between support support poles is 40 m. The vertical drop distance of the light is 12 m. Each triangle is a right triangle so that the length of the cables is
(20)2 + (12)2 = 23. 23 .32 m.
DAC = D AB =
17 m D
Since the cable DE is is attached to the midpoint of the cables AC and and are eachhalfof this this dista distance nce,, or or 11.66 11.66 m. The cable cable AB , AD and AE are DE is is given to be 17 m. From this, the angle α is found:
cos α =
8.5 m 11. 11.66 m
E
B
C A
◦
= 0. 0 .729 α = 43. 43.2 .
Isolate Isolate the traffic light light as shown. shown. The angle formed formed by cable AD α ). The tensio and the positive x axis is (180◦ α) tension n is: is: TAD = TAD (i cos(180 α) + j sin(180 α)).
|
|
−
−
−
The angle formed by cable AE and and the positive x axis is α, hence the tension is TAE = TAE (i cos α + j sin α).
|
y
|
The weight is W = 0 i
x
− j|W|. The equilibrium conditions conditions are
F = W + TAD + TAE = 0.
Fx = (
α
D
α
y x
E
C
A
θ
D
α
Substitute and collect like terms
Fy
Solve,
W
0, −|TAD | cos α + |TAE | cos α)i = 0, = ( |TAD | sin α + |TAE | sin α − |W|) j = 0. 0.
A
Thus the tension in the cable CE exceeds the allowable limit of 4 kN. between AE and the positive x axis is Check: Use components: The angle between
|TAD | = |TAE |. . , |TAD | = |TAE | =
For
|W| = (140 kg)
9.81
|
| |
m
|
| | 1 2
W
sin α
= 1373. 1373.4 N
s2
and α = α = 43. 43.2◦ , TAD = TAE =
1 2
1373. 1373.4 sin43. sin43.2
= 1003 N.
Isolate Isolate the cablejuncture E as as shown. shown. The angle angle θ is foundas foundas follo follows: ws: The cable EC is11.66 is11.66 m. Thedistanc Thedistancee betwee between n poles40 poles40 m. Thecable Thecable DE is is 17 m, and cable DE is is horizontal. Thus EC projects projects onto the x-axis
H EC EC =
1 2
(40
11.5 m. − 17) = 11.
11. 11.5 11. 11.66
= 0. 0 .9863, or θ = 9.5◦ .
From the law of sines:
|TAD | sin θ
|TCE | = sin(180 − α)
4159.6 N. from which TCE = 4159.
|
.
is (180◦ + α). The tension in AE is TAE = TAE (i cos(180 + α) + j sin(180 + α)) TAE
| | = |TAE |(−i cos α − j sin α).
The tension in ED is: T ED =
−|TED |i + 0 j. j.
The angle between CE and the positive x axis is θ . The tension in CE is: T CE = TCE (i cos θ + j sin θ). The equilibrium conditions are
|
|
F = TAE + TED + TCE = 0.
Fx = (
Substitute and collect like terms:
The ratio is the cosine of the angle,
cos θ =
E
|
Fy
0, −|TAE | cos α − |TED | + |TCE | cos θ)i = 0, = ( −|TAE | sin α + |TCE | sin θ ) j = 0. 0.
From the second equation, TCE =
|
|
sin α sin θ
|
|
TAE .
α = 43. 43.2◦ , θ = θ = 9.5◦ , and TAE = 1003. 1003.2 N, For α =
|
|TCE | =
0.6845 0.1651
|
(1003. (1003.2) = 4159. 4159.6 N
check .
The massof massof the the susp suspen ende ded d crat cratee is5 kg. kg. Problem Problem 3.27 The What are the tensions in the cables AB and AC ?
10 m
B
C
5m
Solution:
7m
Find the interior interior angles in in the figure, then apply the equilibrium conditions conditions to the isolated crate. Given the triangle shown, shown, with known sides A, B , and C , find the unknown interior angles α, β , and γ using using the. law of cosines
B 2 = A 2 + C 2 Solve: cos β =
cos γ =
A2
C 2
cos β − 2AC cos 2 − B . Similarly,
+ 2AC
A2 + B 2 C 2 . 2AB
−
10 m
27. 27.66◦
For A = 10, B = 7, C = 5, γ = third angle is
α = (180
A
and β =
40. 40.54◦ .
The B
C
27.66 − 40. 40.64) = 111. 111.8◦ − 27.
5m
7m A
Isolate the cable juncture at A. The angle between the positive positive x axis and the tension TAC is γ . The tension is
+ j sin γ ). TAC = TAC (i cos γ +
|
|
The angle angle between between the positive positive x axis and the tension TAB is (180◦ β ),
−
A
TAB = TAB (i cos(180◦
β
TAB
C
| | − β ) + j sin(180◦ − β )))) = |TAB |(−i cos β + + j sin β ).
The weight is W = 0 i
0. F = W + TAB + TAC = 0.
Fx = ( TAC cos γ
|TAB | = |TAC | = For and
| | | | | | TAB ,
W cos γ sin(β sin(β + + γ )
,
W cos β sin(β sin(β + + γ )
.
|W| = (5 kg)
9.81
β = 40. 40 .54◦ , 46 .79 N, |TAB | = 46. |TAC | = 40. 40 .15 N
m
s2
= 49. 49 .05 N,
γ = = 27. 27.66◦
C
γ
β
| | − |TAB | cos β )i = 0 = ( |TAC | sin γ + + |TAB | sin β − |W|) = 0 cos β cos γ
x
B
Substitute and collect like terms,
|TAC | =
B
y
Solve:
α
− |W| j. j.
The equilibrium conditions are
Fy
γ
A W
tensions in the upper and Problem 3.28 What are the tensions lower lower cables? cables? (Your (Your answers answers will be in terms of W . Neglect the weight of the pulley.) 45°
30°
W
Solution:
Isolate the weight. The frictionless pulley changes changes the direction but not the magnitude magnitude of the tension. The angle between the right hand upper cable and the x axis is α , hence
TUR = TU (i cos α + j sin α).
| |
45°
30°
The angle between the positive x and the left hand upper pulley is (180◦ β ), hence
−
TUL = TU (i cos(180
| | − β ) + j sin(180 − β )))) = |TU |(−i cos β + + j sin β ). TL =
The lower cable exerts a force: The weight:
W = 0i
W
−|TL|i + 0 j
− |W| j j
TU
The equilibrium conditions are
TU α
β
F = W + TUL + TUR + TL = 0
y T L
Substitute and collect like terms,
W
Fx = ( Fy
−|TU | cos β + + |TU | cos α − |TL |)i = 0 = ( |TU | sin α + |TU | sin β − |W|) j = 0. 0.
Solve:
From which
For
|TU | =
|W|
(sin α + sin β )
,
|TL| = |TU |(cos α − cos β ). |T | = |W| cos α − cos β L
α = 30 ◦ and β = β = 45◦
0 .828|W|, |TU | = 0. 0 .132|W| |TL| = 0.
sin α + sin β
.
x
Two tow tow truckslifta truckslifta motorc motorcycl yclee outof a Problem Problem 3.29 Two ravine following an accident. If the 100-kg motorcycle is in equilibrium in the position shown, what are the tensions in cables AB and AC ?
y
(10, 9) m (3, 8) m
C
B
A (6, 4.5) m
x
Solution:
We need need tofind unit unit vector vectorss eAB and eAC . Then write write TAB = T AB AB eAB and TAC = T AC AC eAC . Finally, write and solve the equations of equilibrium. equilibrium. For the ring at A.
(10, 9) m (3, 8) m y
C
B
From the known locations of points A, B , and C , rAB
eAB =
eAC =
rAC
|rAB | |rAC | 3 .5 j m |rAB | = 4. 4 .61 m rAB = −3i + 3. 4.5 j m |rAC | = 6. 6 .02 m rAC = 4 i + 4. 0 .759 j eAB = −0.651i + 0.
A
(6, 4.5) m
0 .664i + 0. 0.747 j eAC = 0. TAB =
x
651T AB 0 .759T 759T AB −0.651T AB i + 0. AB j
0 .664T 664T AC 0 .747T 747T AC TAC = 0. AC i + 0. AC j W =
(100)(9.81) j N −mg j = −(100)(9.
y
T AB
For equilibrium, T AC
TAB + TAC + W = 0 In component form, we have
F x
=
F y
= +0 .759T 759T AB 0 .747T 747T AC AB + 0. AC
−0.651T 651T AB 0 .664T 664T AC AB + 0. AC = 0 − 981 = 0
Solving, we get
T AB AB = 658 N, T AC AC = 645 N
A (6, 4, 5)
x
A
B (3, 8) C (10, (10, 9) mg = (100) (9.81) N
astronautt candidate candidate conducts conducts experexperProblem Problem 3.30 An astronau iments on an airbearing platform. platform. While he carries out calibrations, the platform is held in place by the horizontal tethers AB , AC , and AD . The forces exerted exerted by the tethers are the only horizontal forces acting on the platform. If the tension in tether AC is is 2 N, what are the tensions in the other two tethers?
TOP VIEW D
4.0 m
A
3.5 m
B
C
3.0 m
Solution:
1.5 m
Isolate the platform. platform. The angles α and β are are B
tan α = tan β =
Also,
1.5 3.5
= 0. 0 .429, 429,
3.0 3.5
= 0. 0 .857, 857,
α = 23. 23.2◦ .
3.0 m
β = = 40. 40.6◦ .
hence
B x
The angle angle between between thetether AC andthe positiv positivee x axis axis is (180◦ + α). The tension is TAC = TAC (i cos(180◦ + α) + j sin(180◦ + α))
| | = |TAC |(−i cos α − j sin α).
|
The equilibrium condition:
Fx = (
Substitute and collect like terms,
Fy
β α
D
A
C
The tether tether AD is aligned aligned withthe positiv positivee x axis, TAD = TAD i + j. 0 j.
4.0 m
y
| | − β ) + j sin(180◦ − β )))) = |TAB |(−i cos β + + j sin β ).
0. F = TAD + TAB + TAC = 0.
C 3.5 m
− β ),
TAB = TAB (i cos(180◦
D
1.5 m
The angle angle between between the tether tether AB and the positiv positivee x axis axis is (180◦
TAB
A
0, −|TAB | cos β − |TAC | cos α + |TAD |)i = 0, = ( |TAB | sin β − |TAC | sin α) j = 0. 0.
|
Solve:
|TAB | = |TAD | =
| | | sin α sin β
TAC ,
|
sin(α + β ) TAC sin(α sin β
.
23.2◦ and β = = 40. 40.6◦ , For TAC = 2 N, α = 23.
| | 1 .21 N, | TAD | = 2 .76 N |TAB | = 1.
The forces forces exerte exerted d on the shoes shoes and and Problem 3.31 back of the 72-kg climber by the walls of the “chimney” are perpendic perpendicular ular to the walls walls exertin exerting g them. them. The tension tension in the rope is 640 N. What is the magnitude of the force exerted on his back?
10°
4°
Solution:
Draw a free body body diagram of the the climber-treating climber-treating all forces as if they act at a point. Write the forces in components components and then apply the conditions for particle equilibrium.
mg
F x F y
= F FEET cos 4◦ FEET cos
◦ cos 3◦ − T ROPE − F BACK BACK cos ROPE sin10 = 0 ◦ = F FEET sin 4◦ + F BACK sin 3◦ + T ROPE FEET sin BACK sin ROPE cos10 − mg = 0
= (72)9. (72)9.81 N, T ROPE ROPE = 640 N
Solving, we get
F BACK BACK = 559 N, F FEET FEET = 671 N
10°
T ROPE
10° y
F FEET
F BACK
3°
x
4°
mg = (72) (9.81) N
3°
Problem 3.32 The slider A is in equilibrium and the bar is smooth. What is the mass of the slider?
20°
200 N
A
45°
Solution: The pulley does not change change the tension in the rope that passes over it. There is no friction between the slider and the bar. Eqns. of Equilibrium:
F x
= T sin20 T sin20◦ + N cos45 cos45◦ = 0
F y
= N sin45 sin45◦
+ T cos20 cos20◦
(T = T = 200 N)
mg = 0 − mg =
g = 9.81 m/s2
Substituting Substituting for T and g , we have two eqns in two unkno unknowns wns (N and and m). Solving, we get N =
96.7 N, m = 12. 12.2 kg. −96.
20°
200 Ν
A
45°
y
T 200 N
20°
x
N
45° mg = (9.81) g
The unstretched unstretched length length of the spring spring Problem 3.33 AB is 660 mm, and the spring constant k = 1000 N/m. What is the mass of the suspended object?
400 mm
600 mm
B
350 mm
A
Solution: Use the linear spring spring force-extension force-extension relation to find the magnitude of the tension in spring AB . Isolate juncture A . The forces are the weight and the tensions in the cables. The angles are
tan α = tan β =
350 600
= 0. 0 .5833, 5833, α = 30. 30.26◦ .
350 400
= 0. 0 .875, 875, β = = 41. 41.2◦ .
400 mm
600 mm
B
350 mm
A
The angle between the x axis and the spring is α. The tension is
TAB = TAB (i cos α + j sin α).
|
|
C
The angle between the x axis and AC is is (180
B
β
− β ). The tension is
α
y
T AC AC = TAC (i cos(180
| | − β ) + j sin(180 − β )))) + j sin β ). TAC = |TAC |(−i cos β + The weight is: W = 0 i
A
− |W| j. j.
The equilibrium conditions:
W
0. F = W + TAB + TAC = 0. x
Substitute and collect like terms
The tension is TAB = k∆ k ∆L = (1000)(0. (1000)(0.03462) = 34. 34.6 N.
|
Fx = ( TAB cos α Fy
| | − |TAC | cos β )i = 0, 0, = ( |TAC | sin α + |TAB | sin β − |W|) j = 0. 0.
Solve: TAC =
|
|
|
| cos α cos β
TAB and W =
|
| |
sin(α sin(α+β ) cos β
α = 30. 30.26◦ and β = β = 41. 41.2◦ ; the weight is For α =
|
|W| = TAB |.
|
The tension TAB is found from the linear spring force-deflection relation. The spring extension is
∆L =
(350)2 + (600)2
|
694.62 − 660 = 34. 34.62 mm − 660 = 694.
0.948 0.752
The mass is m =
(34. (34.6) = 43. 43.62 N;
|W| |g|
=
43. 43.62 9.81
= 4. 4 .447 kg
unstretched length of the spring in Problem 3.34 The unstretched Problem 3.33 is 660 mm. If the mass of the suspended object is 10 kg and the system is in equilibrium in the position shown, what is the spring constant? Solution: First, find the distance AB to determine the stretched length of the spring. Write unit vectors from A toward B and from A toward C . Write the forces, in terms of these unit unit vectors. Then write the equations of equilibrium and solve for the unknowns. From the diagram, A is at ( 0, 0), B is at (0.6, 0.35) m, and C is at ( 0.4, 0.35) m.
−
400 mm
600 mm
B
C
350 mm
eAB
=
eAC
=
TAB TAC W
rAB |rAB | rAC |rAC |
= 0. 0 .864i + 0. 0 .504 j =
−0.753i + 0.0.659 j
A
= 0. 0 .864T 864T AB 0.504T 504T AB AB i + 0. AB j = 0.753T 753T AC + 0. 0 . 659T 659 T AC i AC AC j = 0 i 98. 98.1 j (N)
−
−
From our calculations
0 .695 m |rAB | = 0.
y
C
B
••• the stretched length of the spring. Thus, the stretch in the spring is given by
δ = rAB
| − unstretched = 0. 0 .6946 − 0.6600 = 0. 0.0346 (m) |
δ
T AB T AC A
(0,0)
mg = (10) (9.81) N
We know that
T AB AB = K δ = 0.0346 The equilibrium equations are
F x F y
= 0 .864T 864T AB AB
753T AC − 0.753T AC = 0 = 0 .504T 504T AB 659T AC 98.1 = 0 AB + 0 .659T AC − 98.
Solving, we get
T AB 77 .88 N T AC 89 .38 N AB = 77. AC = 89. Finally solving for K , we get
K = = 2250 N/m
x
Problem 3.35 The collar A slides on the smooth vertical bar. bar. The masses masses m A = 20 kg and m B = 10 kg. When h = 0.1 m, the spring is unstretched. unstretched. When the system is in equilibrium, h = 0.3 m. m. Determine Determine the spring constant k .
0.25 m
h A B
k
Solution: The triang trianglesforme lesformed d by therope segmen segments ts andthe horhorizontal line level with A can be used to determine the lengths L u and Ls . The equations are
Lu
0.1 m
0.25 m
Lu =
(0. (0.25)2 + (0. (0.1)2 and Ls =
(0. (0.25)2 + (0. (0.3)2 . Ls
Thestretchin Thestretchin thespringwhenin thespringwhenin equili equilibri brium um isgiven isgiven by δ = L s Lu . Carrying out the calculations, we get Lu = 0.269 m, Ls = 0.391 m, and δ = 0.121 m. The angl angle, e, θ, between the rope at A and the horizontal when the system is in equilibrium is given by ta n θ = diagram for mass A, we 0.3/0.25, or θ = 50. 50.2◦ . From the free body diagram get two equilibrium equations. They are
Lu
−
and
0.25 m
F x = F y
cos θ = 0 −N A + T cos = T sin T sin θ − mA g = 0.
0.3 m
T N A
A
We have have two two equati equationsin onsin twounknow twounknowns ns andcan solve. solve. We get N A = 163. 163.5 N and T = T = 255. 255.4 N. Now we go to the free body diagram for B , where the equation of equilibrium equilibrium is T mB g kδ = 0. This equation has only one unknown. Solving, we get k = 1297 N/m
− −
m A g
−
0.25 m
T
h B A B
k
m Bg K δ δ
You are designing designing a cable system system to Problem 3.36 support a suspended object of weight W . The two wires must be identical identical,, and the dimensio dimension n b is fixed. fixed. The ratio ratio of the tension T in each wire to its cross-sectional area A must equal a specified value T /A = σ . The “cost” “cost” of your design is the total volume of material in the two wires, V = 2A b2 + h2 . Determine the value value of h that minimizes the cost.
√
b
b
h
W
Solution:
From the equation
F y = 2T sin sin θ
0, − W = 0,
W = we obtain T = 2 sin sin θ
√ b +h 2
W
2h
2
.
√
2 2 Since T /A = /A = σ σ , A = A = T = W 2bσh+h σ
√
2 A b2 + h2 = and the “cost” is V = 2A
W ( W (b2 +h2 ) . σh
h that minimizes V , we set To determine the value of h dV W = dh σ
−
(b2 + h2 ) +2 =0 h2
and solve for h , obtaining h = h = b b .
T
T
θ
θ
W
system of cables suspends suspends a 1000Problem 3.37 The system lb bank of lights above a movie set. Determine the ten C D, and CE C E . sions in cables AB , CD
20 ft
18 ft
B
D C E
45°
Solution: Isolate Isolate juncture A, andsolve theequilibriumequatio theequilibriumequations. ns.
30°
Repeat for the cable juncture C .
A
The angle between the cable AC and the positive x axis is α . The The tension in AC is is TAC = TAC (i cos α + j sin α)
|
|
The angle between the x axis and AB is (180◦ TAB = TAB (i cos(180
|
TAB = (
|
− β ). The tension is
− β ) + j sin(180 − β ))))
+ j sin β ). −i cos β +
The weight is W = 0 i
− |W| j. j.
The equilibrium conditions are
| | = |TCA | cos α, |TCD | = |TCA | sin α;
Solve: TCE
0. F = 0 = W + TAB + TAC = 0.
α = 30◦ , for TCA = 732 lb and α =
Substitute and collect like terms,
| | 896.6 lb, |TAB | = 896. |TCE | = 634 lb, |TCD | = 366 lb
Fx = ( TAC cos α Fy
| | − |TAB | cos β )i = 0 = ( |TAB | sin β + + |TAC | sin α − |W|) j = 0. 0.
Solving, we get
|TAB | =
cos α cos β
|
|
TAC
and
W| cos β |TAC | = |sin(α sin(α + β )
20 ft
B
α = 30◦ , β = = 45◦ |W| = 1000 lb, and α = = 732. 732.05 lb |TAC | = (1000) 00..7071 9659
|TAB | = (732)
D C
0.866 0.7071
18 ft
,
E
45°
30° A
= 896. 896.5 lb
Isolate juncture C . The The angle angle between between the positiv positivee x axis axis andthe cable cable CA is (180◦ α). The tension is
−
)), TCA = TCA (i cos(180◦ + α) + j sin(180◦ + α)), or TCA
| | = |TCA |(−i cos α − j sin α).
B
β
The tension in the cable CE is
C A
α y
TCE = i TCE + 0 j.
|
|
x
W
The tension in the cable CD is TCD = 0 i + j TCD .
|
|
The equilibrium conditions are
F = 0 = TCA + TCE + TCD = 0
Fx = ( TCE
D
Substitute t and collect like terms,
Fy
0, | | − |TCA | cos α)i = 0, = ( |TCD | − |TCA | sin α) j = 0. 0.
C
α
90°
E y
A x
1000-lb bank of lights lights in Problem 3.38 Consider the 1000-lb Problem 3.37. A technician changes the position of the lights by removing the cable CE . What is the tension tension in cable AB after the change? Solution:
The original configuration configuration in Problem 3.35 is is used to solve for the dimensions and the angles. Isolate the juncture A , and solve the equilibrium conditions.
18 ft
20 ft
The lengths are calculated as follows: The vertical interior distance in the triangle is 20 ft, since the angle is 45 deg. and the base and altitude of a 45 deg triangle are equal. The length AB is given by
D
B
C
α
β
20 ft AB = = 28 .284 ft. cos cos 45◦
A
The length AC is is given by
AC =
18 ft = 20. 20 .785 ft. cos cos 30◦
The altitude altitude of the triangle triangle for which AC is the hypotenuse hypotenuse is distance CD isgiven isgiven by 20 10. 18tan30◦ = 10. 10 .392 ft. The distance 10.392 = 9.608 ft.
38
B
β
−
α
28.284
The distance AD is given by
D
20.784 + 9.608 = 30.392
β
AD = AD = AC + + CD = 20. 20.784 + 9. 9.608 = 30. 30.392
α A
The new angles are given by the cosine law
AB 2 = 38 2 + AD 2
B
2(38)(AD)cos )cos α. − 2(38)(AD
D
cos α = cos β =
A
β
Reduce and solve:
382 + (30. (30.392)2 (28. (28.284)2 2(38)(30. 2(38)(30.392)
−
(28. (28.284)2 + (38)2 (30. (30.392)2 2(28. 2(28.284)(38)
−
α
= 0. 0 .6787, 6787, α = 47. 47.23◦ .
y x
W
= 0. 0 .6142, 6142, β = = 52. 52.1◦ .
Isolate the juncture A . The angle angle between between the the cable AD and the positive x axis is α. The tension is: TAD = TAD (i cos α + j sin α).
|
|
The angle between x and the cable AB is (180◦
− β ). The tension isis
+ j sin β ). TAB = TAB ( i cos β +
|
|−
The weight is W = 0 i
− |W| j j
The equilibrium conditions are
F = 0 = W + TAB + TAD = 0.
Fx = ( TAD cos α
Substitute and collect like terms,
Fy
0, | | − |TAB | cos β )i = 0, = ( |TAB | sin β + + |TAD | sin α − |W|) j = 0. 0.
Solve:
|TAB | =
and
|TAD| =
| | | | cos α cos β
TAD ,
W cos β
sin(α sin(α + β )
.
α = 51. 51.2◦ , β = β = 47. 47.2◦ For W = 1000 lb, and α =
| |
|TAD | = (1000)
0.6142 0.989
= 621. 621.03 lb,
0.6787 0.6142
= 687. 687.9 lb
|TAB | = (622. (622.3)
While workin working g on anothe anotherr exhib exhibit it,, a cuProblem Problem 3.39 While rator at the Smithsonian Institution pulls the suspended Voyager aircraft aircraft to one side by attaching three horizontal cables as shown. The mass of the aircraft is 1250 kg. Determine the tensions in the cable segments AB , BC , C D. and CD
D
50°
B
Solution: Isolate Isolate each cablejuncture, cablejuncture, beginnin beginning g with A and solve solve the equilibrium equilibrium equations at each juncture. juncture. The angle between the cable AB and the positive x axis is α = 70◦ ; the tension in cable AB is T AB = TAB (i cos α + j sin α). The weig weight ht is is W = 0i W j. j. The tension T i + 0 j. j. The tension in in cable AT is T = The equilibrium conditions are
|
−| |
|
30°
C
A
70°
−| |
F = W + T + TAB = 0. 0.
Substitute and collect like terms
Fx ( TAB cos α
0, − |T|)i = 0, 0. Fy = (|TAB | sin α − |W|) j = 0.
|
|
Solve: the tension in cable AB is TAB =
|
For W = (1250 kg)
| |
|TAB | =
9.81
12262. 12262.5 0.94
m s2
|
D C
|W| sin α
.
= 12262. 12262.5 N and α = α = 70◦
70°
A
T
= 13049. 13049.5 N
= 70◦ , andthe tensio Isolate juncture B . The anglesare anglesare α = 50◦ , β = tension n cable BC is is TBC = TBC (i cos α + j sin α). The angle between the cable BA and the positive x axis is (180 + β ); the tension is
|
|
)) TBA = TBA (i cos(180 + β ) + j sin(180 + β ))
| | = |TBA |(−i cos β − j sin β )
The tension in the left horizontal cable is T = equilibrium conditions are
y B x
−| T|i + 0 j. j.
The The
α
A
T
0. F = TBA + TBC + T = 0.
Substitute and collect like terms
W
Fx = ( TBC cos α Fy
| | − |TBA | cos β − |T|)i = 0 = ( |TBC | sin α − |TBA | sin β ) j = 0. 0.
Solve: TBC =
|
|
| sin β sin α
|
|
|TBC | = (13049. (13049.5)
y
0.9397 0.7660
|TCD | =
sin β sin α
β
= 16007. 16007.6 N A
|
y
D
x
TCB .
|
(16007.6) 0.07660 = 24525. 24525.0 N. Substitute: TCD = (16007. .5
|
α
B
T
Isolate the cable juncture C . The angles are α = α = 30◦ , β = β = 50◦ . By symmetry with the cable juncture B above, the tension in cable CD is
|
C x
TBA .
13049.5 N, and α = α = 50◦ , β = β = 70◦ , For TBA = 13049.
|
50°
B
T
α
C
β
This completes the problem solution.
B
30°
A truck dealer wants wants to suspend a 4Problem 3.40 Mg (megagram (megagram)) truck as shown for advertisi advertising. ng. The distance b = 15 m, and the sum of the lengths of the cables AB and B C is is 42 m. What are are the tensions tensions in the cables?
40 m b A
C
B
Solution:
Determine the dimensions dimensions and angles angles of the cables. Isolate the cable juncture B , and solve the equilibrium conditions. The dimensions of the triangles formed by the cables:
40 m b C
A
b = 15 m,
L = 25 m,
AB + AB + BC = BC = S = S = 42 m. B
Subdivide into two right triangles with a common side of unknown length. Let the unknown length of this this common side be d, then by the Pythagorean Theorem b2 + d2 = AB 2 , L 2 + d2 = BC 2 . Subtract Subtract the first equation equation fromthe second second to eliminate eliminate the unknown unknown d, L2 b2 = B C 2 AB 2 .
−
−
Note that BC 2
( BC − AB)( AB )(BC BC + + AB) AB ). − AB2 = (BC
15 m
Substitute and reduce to the pair of simultaneous equations in the unknowns
BC
− AB
Solve:
=
BC = =
L2
− b2
S
,
b
L
A
C
β
α
BC + BC + AB = AB = S B
1 2
L2
1 2
252
and AB = S
25 m
− b2 + S
S
y
− 152 + 42
42
A
= 25. 25 .762 m
B
β
BC = 42 − 25. 25.762 = 16. 16.238 m. − BC =
α
C
W
The interior angles are found from the cosine law:
cos α = cos β =
(L + b)2 + BC 2 AB 2 2(L 2(L + b)(BC )(BC ))
−
(L + b)2 + AB 2 BC 2 2(L 2(L + b)(AB )(AB))
−
x ◦
= 0. 0 .9704 α = 13. 13.97
Substitute and collect like terms
= 0. 0 .9238 β = = 22. 22.52◦
Isolate cable juncture B . The angle between BC and the positive x axis is α; the tension is TBC = TBC (i cos α + j sin α)
|
|
The angle between B A and the positive x axis is (180◦ tension is TBA = TBA (i cos(180
| | − β ) + j sin(180 − β )))) = |TBA |(−i cos β + + j sin β ).
The weight is W = 0 i
− |W| j. j.
The equilibrium conditions are
0. F = W + TBA + TBC = 0.
Fx = ( TBC cos α Fy
Solve:
− β ); the
0, | | − |TBA | cos β )i = 0, = ( |TBC | sin α + |TBA | sin β − |W|) j = 0
|TBC | =
and TBA =
|
|
| | | | cos β cos α
TBA ,
W cos α
sin(α sin(α + β )
.
(4000)(9.81) = 39240 N, For W = (4000)(9.
| |
α = 13. 13.97◦ , β = β = 22. 22.52◦ , and α =
|TBA | = 64033 = 64 kN, |TBC | = 60953 = 61 kN
distance h = 12 in., in., andthe tensio tension n Problem Problem 3.41 The distance in cable AD is 200 lb. What are the tensions tensions in cables AB and AC ?
B
12 in.
A D C
12 in.
h
8 in.
12 in.
Solution:
Isolated Isolated the cablejuncture. cablejuncture. Fromthe sketch, sketch, the angles angles
are found from
tan α = tan β =
8 in.
B
8 12
4 12
= 0. 0 .667 α = 33. 33.7◦
A
12 in.
D
C
= 0. 0 .333 β = = 18. 18.4◦
12 in.
h
8 in.
Theangle Theangle betwee between n thecable thecable AB and the positiv positivee x axis axis is (180◦ the tension in AB is:
− α),
TAB = TAB (i cos(180 TAB
| | − α) + j sin(180 − α)) = |TAB |(−i cos α + j sin α).
8 in.
12 in.
+ β )). The The angle between AC and the positive x axis is (180 + β The tension is
y B
12 in
α
8 in A
D
)) TAC = TAC (i cos(180 + β ) + j sin(180 + β ))
| | TAC = |TAC |(−i cos β − j sin β ).
4 in
β C
The tension in the cable AD is
x
TAD = TAD i + 0 j.
|
|
The equilibrium conditions are
0. F = TAC + TAB + TAD = 0.
Fx = (
Substitute and collect like terms,
Fy
Solve:
−|TAB | cos α − |TAC | cos β + + |TAD |)i = 0 = ( |TAB | sin α − |TAC | sin β ) j = 0. 0.
|TAB | =
and TAC =
|
|
sin β sin α
| | |
TAC ,
sin α sin(α sin(α + β )
TAD .
|
For TAD = 200 lb, α = α = 33. 33.7◦ , β = = 18. 18.4◦
|
|
|TAC | = 140. 140.6 lb, |TAB | = 80. 80 .1 lb
You are designing designing a cable system system to Problem 3.42 support a suspended object of weight W . Because your design requires points A and B to be placed as shown, you have have no contro controll over over the angle angle α, but but you you can can choo choose se the angle β by placin placing g point point C whereve whereverr you wish. Show that to minimize the tensions in cables AB and BC , you must choose β = α if the angle α 45 . diagram of the sum of the forces forces Strategy: Draw a diagram exerted by the three cables at A.
≥
◦
Solution:
Draw the free body body diagram of the the knot at point point A . Then draw the force triangle involving involving the three forces. Remember that α is fixed fixed andthe force force W has bothfixed magnitude magnitude and directio direction. n. From the force triangle, we see that the force T AC AC can be smaller than T AB AB for a large range of values for β . By inspection, we see that the minimum simultaneous values for T AC AC and T AB AB occur when the two = β . Note: forces forces are equal. equal. This occurs occurs when α = β Note: this this does nothappen when α < 45 ◦ .
In this case, we solved the problem without writing the equations of equilibrium. equilibrium. For reference, these equations are:
and
F x =
−T AB = 0 AB cos α + T AC AC cos β = = T AB 0. AB sin α + T AC AC sin β − W = 0.
F y
α
B
β
C
A W
y T AC T AB
B
α A
x
W
Possible locations for C lie on line C ? C ? B
α T AB
Candidate β
W
Candidate values for T AC Fixed direction for line AB
B
β
α
A
W
C
Problem m 3.42, 3.42, suppos supposee that that you have have Problem Problem 3.43 In Proble no control over the angle α and you want to design the cable cable systemso systemso that that thetensionin thetensionin cable cable AC is minimum. minimum. What is the required angle β ? Solution:
From Problem 3.32 above, above, the angle required to mini = α . However, mize mize the the tensio tension n incable AC , forlargevalu forlargevalues es of α is β = for small values of , the situation is different. different. In this situation, the force triangle is as shown in the figure. It is obvious from the figure that the minimum value for tension in cable AC is is obtained when the T AC AC is perpendicular to T AB AB .
Possible locations for C lie on line
∝
C ?
B
C ?
α T AB
Candidate β
W
Fixed direction for line AB Candidate values for T AC
Problem 3.44 The masses of the boxes on the left and right are 25 kg and 40 kg, respectively. respectively. The surfaces are smooth and the boxes are in equilibrium. equilibrium. Determine the tension in the cable and the angle α.
Solution:
α
30°
α
30°
We now now need to write the equilibrium equilibrium equations for
each box. For the left box,
F x F y
= T
− − mLg sin α = 0 = N L − mL g cos α = 0
For the right box,
F x
F y
m Lg = (25) (9.81) N
−T + mR g sin sin 30◦ = 0 = N R − mR g cos cos 30◦ = 0 =
We have four equations in the four unknowns T , N L , N R , and α . (mL = 25 kg, m R = 40 kg). Solving, we get
T y
α
m Rg = 40 (9.81) N
x
30°
T
y′
N L = 147 N, N R = 340 N T = T = 196. 196.2 N, α = 53. 53.1◦ N L
α
N R
30°
x ′
Consider the system system shown shown in Prob- Solution: Use the free body diagrams of Problem 3.44. Problem 3.45 3.44. The equations equations of lem 3.44. The angle α = 45 , the surfaces are smooth, equilibrium are the same as for Problem 3.44. and the boxes are in equilibrium. Determine the ratio of T − − mLg sin α = 0 the mass of the right box to the mass of the left box. ◦
N L mL g cos α = 0 T + + mR g sin sin 30◦ = 0 N R mR g cos cos 30◦ = 0
−
− −
g = 9.81 m/s 81 m/s 2 , m L = 1, α = α = 45◦ . where g = Solving, we get mR = 1.41.
∴
The 3000-lb 3000-lb car and the 4600-lb tow tow Problem 3.46 truck truck arestationa arestationary ry.. The muddy muddy surfac surfacee on which which thecar rests exerts a negligible negligible friction force on the car. What is the tension in the tow cable?
mR/mL = 1. 1 .41/ 41/1 = 1.41
10°
18°
26°
Solution:
From the geometry, geometry, the the angle between the cable and the x axis is 8 ◦ . From the free body diagram, the equations of equilibrium are
and
cos(8◦ ) + 3000sin(26◦ ) = 0 −T cos(8 = N − − 3000 cos(26◦ ) = 0.
F x = F y
The firstequation firstequation can be solved solved forthe tensionin tensionin thecable. Thetension T = 3000sin(26◦ )/ cos(8◦ ) = 1328 lb. is T
18°
10°
26°
N
18° y
T
3000 lb 25°
x
hydraulic cylinder cylinder is subjected subjected to to Problem 3.47 The hydraulic three forces. An 8-kN force is exerted exerted on the cylinder cylinder at B that is parallel to the cylinder and points from B toward C . The link link AC exe exertsa rtsa force force at C that that is parall parallel el C D exerts a force at to the line from A to C . The link C C that that is parallel to the line from C to to D . (a) Draw the free-body diagram of the cylinder. cylinder. (The cylinder’s weight is negligible). (b) Determine the magnitudes magnitudes of the forces forces exerted by by the links AC and and CD C D.
1m D C
Hydraulic cylinder
1m 0.6 m B
A
0.15 m
0.6 m
Scoop
Solution:
From the figure, if C is is at the origin, then points A, B , and D are located at
1m D
A(0. (0.15, 15, 0.6) B (0. (0.75, 75, 0.6) D(1. (1.00, 00, 0.4)
− −
C
Hydraulic cylinder
1m 0.6 m
and forces FCA , FBC , and FCD are parallel to CA , BC , and CD , respectively. A
We need to write unit vectors in the three force directions and express the forces in terms of magnitudes and unit vectors. The unit vectors are given by
0.15 m
0.6 m
B
Scoop
y
rCA
eCA = eCB
0 .243i − 0.970 j |rCA | = 0. rCB = 0 .781i − 0.625 j |r | = 0.
D
CB
eCD =
FCD
rCD
0 .928i + 0. 0.371 j |rCD | = 0.
C
Now we write the forces in terms of magnitudes and unit vectors. We can write FBC a s FCB = 8eCB kN or as FCB = 8( eCB ) kN (because we were told it was directed from B toward C and and had a magnitude of 8 kN. Either way, we must end up with
−
FCB =
−
Similarly,
0 .243F 243F CA FCA = 0. CA i
− 0.970F 970F CA CA j
0 .928F 928F CD 0 .371F 371F CD FCD = 0. CD i + 0. CD j For equilibrium, FCA + FCB + FCD = 0 In component form, this gives
F x
= 0 .243F 243F CA 0 .928F 928F CD CA + 0. CD
F y
=
− 6.25 (kN) = 0
970F CA 0 .371F 371F CD 5 .00 (kN) = 0 −0.970F CA + 0. CD + 5.
Solving, we get
F CA CA = 7.02 kN, F CD CD = 4.89 kN
FCA
A
−6.25i + 5.5.00 j kN
x
F BC
B
on two smooth Problem 3.48 The 50-lb cylinder rests on surfaces. (a) Draw the free-body free-body diagram of the cylinder cylinder.. α = 30 , what are the magnitudes of the forces (b) (b) If α exert exerted ed on thecylinde thecylinderr by theleftand rightsurf rightsurface aces? s? ◦
α
45°
Solution:
Isolate the cylinder. cylinder. (a) The free body diagram diagram of the isolat isolated ed cylind cylinder er is shown. shown. (b)The forces forces actingare actingare theweight theweight andthe normal forces exerted by the surfaces. The angle between the normal force on the right and the x axis is (90 + β ). The normal force is
NR
| | = |NR |(−i sin β + + j cos β ).
The angle between the positive positive x axis and the left hand force is normal α); the normal force is NL = NL (i sin α + j cos α). The W j. j. The equilibrium conditions weight is W = 0 i conditions are
(90
−
y
| |
−| |
F = W + NR + NL = 0.
Fx = (
N L
−|NR| sin β + + |NL | sin α)i = 0, 0, = ( |NR | cos β + + |NL | cos α − |W|) j = 0. 0.
β
α
Substitute and collect like terms,
Fy
45°
α
)) NR = NR (i cos(90 + β ) + j sin(90 + β ))
Solve:
N R
W x
|NR| =
and NL =
| |
| | | | sin α sin β
N L ,
W sin β
sin(α sin(α + β )
.
α = 30◦ , β = β = 45◦ , the normal forces are For W = 50 lb, and α =
| |
36 .6 lb, |NR | = 25. 25 .9 lb |NL| = 36.
Problem 3.49 For the the 50-lb cylinder in Problem Problem 3.48, obtain an equation for the force exerted on the cylinder by the left surface in terms of the angle α in two ways: (a) using a coordinate system with the y axis vertical, (b) using a coordinate system with the y axis parallel to the right surface. Solution: The solution for Part (a) is given in Problem 3.48 (See free body diagram).
|NR| =
| sin α sin β
N L
| |NL| =
|
W sin β
|
sin(α sin(α + β )
y
N R N L
Part (b): The isolated cylinder with the coordinate system is shown. The angle between the right hand normal force and the positive x axis N R = NR i + 0 j. j. The angle between is 180◦ . The normal force: force: N between the left hand normal force and the positive x is 180 (α + β + β ). The cos(α + β ) + j sin(α sin(α + β )) )). normal force is NL = NL ( i cos(α
−|
|
| |−
| |
The equilibrium conditions are
F = W + NR + NL = 0.
−
−β .
W x
−
The angle between the weight vector and the positive x axis is The weight vector is W = W (i cos β j sin β ).
β
α
.
Substitute and collect like terms,
Fx = ( Fy
Solve:
cos(α + β ) + |W| cos β )i = 0, 0, −|NR | − |NL| cos(α = ( |NL | sin(α sin(α + β ) − |W| sin β ) j = 0. 0.
W| sin β |NL| = |sin(α sin(α + β )
The 50-kg 50-kg spher spheree is at rest rest on the Problem Problem 3.50 smooth horizontal horizontal surface. The horizontal force F = exerted on the sphere 500 N. What is the normal force exerted by the surface?
30°
F
Solution:
Isolate the sphere sphere and solve the the equilibrium equilibrium equations. α). The The angle between the cable and the positive x is (180 The tension:
−
30°
T = T ( i cos α + j sin α).
| |−
The other forces are F = F i + 0 j,
| |
F
N = 0 i + N j j,
| |
W = 0i
− |W | j j.
The equilibrium conditions are
0. F = T + F + N + W = 0.
Fx = (
T
α
Substitute and collect like terms,
Fy
0, −|T| cos α + |F|)i = 0, = ( |N| − |W| + |T| sin α) j = 0. 0.
F
| | | | cos , and |N| = |W| − |F| tan α. α For |W| = (50)(9. (50)(9.81) = 490. 490.5 N, |F| = 500 N, and α = α = 30◦
Solve: T =
F
W
N
490.5 − (500)(0. (500)(0.577) = 201. 201.8 N |N| = 490.
Problem Problem 3.51 Consider Consider the stationa stationary ry sphere sphere in Problem 3.50. (a) Draw a graph of the normal normal force force exerted exerted on the sphere by the surface as a function of the force F from F = 0 to F = 1 kN. (b) In the result of (a), (a), notice that the normal force decrease creasess to zero zero andbecomes andbecomes negat negativ ivee as F increases. increases. What does that mean? Solution:
From the solution of Problem Problem 3.50,
|N| = |W| − |F| tan α. (a)
The commercia commerciall package package TK Solver Plus was used to produce the graph of the normal force vs. the applied force, for W = (50)(9. (50)(9.81) = 490. 490.5 N and α = α = 30◦ , as shown.
| |
(b)
The normal force becomes negative negative when the the cylinder cylinder is lifted lifted from the surface (it would take a negative force to keep it in contact with the surface).
N o r m a l
Normal Force vs Force
500 400 300 200
F 100 o r 0 c e −100 0
200
400 600 Force
800
10 1000
1440-kg car is moving moving at constant Problem 3.52 The 1440-kg speed on a road with the slope shown. The aerodynamic forces on the car are the drag D = 530 530 N, which is parallel to the road, and the lift L = 360 N, which is perpendicular to the road. Determine the magnitudes magnitudes of the total normal and friction forces exerted on the car by the road.
Solution:
From the free free body diagram, diagram, the equations equations of equilib-
rium are
sin15◦ = 0 − − D − W sin15 = L + N − cos15◦ = 0 − W cos15
F x = f F y
W = mg = (1440)(9. (1440)(9.81) N L = 360 N, D = 530 N m = 1440 kg, g = 9.81 m/s2
Solving, we get
f = f = 4.19 kN, N = 13. 13 .29 kN
L D
15°
y
L
x
F D
15°
W
N
15°
L D
15°
device shown shown is towed beneath a Problem 3.53 The device shipto measure measure water water temperat temperature ure and salinity salinity.. Themass of the device is 130 kg. The angle α = 20 . The motion of the water relative to the device causes a horizontal drag force D . The hydrostatic pressure distribution distribution in the water exerts a vertical “buoyancy” force B . The magnitude of the buoyancy force is equal to the product of the volume of the device, V = 0.075 m 3 , and the 3 weight weight densit density y of the water water,, γ = = 9500 N/m . Determine the drag force D and the tension in the cable. ◦
α B
D
Solution: Calculate the the magnitude of the buoyancy buoyancy force. Draw a free body diagram diagram of the device. device. The drag, buoyancy buoyancy and drag forces are
y
α T
D = D i + 0 j,
| |
B = 0 i + B j j,
W = 0i
| |
− j|W|.
The angle between the tow cable and the positive x axis is (90◦ + α); the cable tension is
T = T (i cos(90 + α) + j sin(90 + α))
| | T = |T|(−i sin α + j cos α). The equilibrium conditions are
0. F = W + B + T + D = 0.
Substitute and collect terms
Fx = ( D Fy
| | − |T| sin α)i = 0 = ( |T| cos α + |B| − |W|) j = 0. 0.
The magnitude of the buoyancy force is
B = ρV ρV = (970)(0. (970)(0.15) = 145. 145.5 N.
Solve: D = T sin α, and T =
| | | |
| |
|W|−|B| cos α
.
(130)(9.81) = 1275. 1275.3 N, and α = 20◦ , the tension in For W = (130)(9. 411.2 N the cable and the drag are T = 1202 N, D = 411.
| |
| |
α B
| |
D B
W x
Problem 3.54 The mass of each pulley of the system is m and the mass of the suspended object A is mA . Determine the force T necessary for the system to be in equilibrium.
A T
Solution:
Draw free body body diagrams diagrams of each pulley pulley and the object
A. Each pulley and the object A must be in equilibrium. The weights of the pulleys and object A are W = mg and W A = mA g. The The equilibrium equations for the lower pulley, pulley, middle pulley, pulley, and upper 2T W = 0, B 2A W = 0, pulley are, respectively, respectively, A and C 2B equation for the weight is W = 0. The equilibrium equation T + + A + B W A = 0. Solving the first equation for A in terms of T and W , substituting for A in the second equation and solving for B in terms of T T and W , we get A = A = 2T + + W and + 3W . and B = 4T + Substituting Substituting for A and B in the equilibrium equilibrium equation for the weight, 7 T = W A 4W = m A g 4mg. Thus, the tension, T , in we get 7T g terms of masses and g is T = 7 (mA 4m)
− −
− − −
− −
−
−
T A
C B
B
B W A
A
T
A
T W
W T
W A
B
C
−
−
−
Problem 3.55 The mass of each pulley of the system is m and the mass of the suspended object A is mA . Determine the force T necessary for the system to be in equilibrium.
Solution:
Draw free body body diagrams diagrams of each pulley pulley and the object
A. Each pulley and the object A must be in equilibrium. The weights of the pulleys and object A are W = mg and W A = mA g. The The equilibr equilibriumequatio iumequations ns forthe weight weight A, thelowerpulley, thelowerpulley, second second pulley pulley,, third pulley, and the top pulley are, respectively, B W A = 0, 2C B W = 0, 2D 2 D C W = 0, 2T 2 T D W = 0, and solve for B , F S 2T W = 0. Begin with the first equation and solve substitute for B in the second equation and solve for C , substitute for C in in the third equation and solve for D , and substitute for D in the fourth equation and solve for T , to get T in in terms of W and and W A . The
− − − − − −
− − −
result is
B = W A , D =
C = =
W A W + , 2 2
W A 3 W W A 7W 7 W + + , , and T = 4 4 8 8
or in terms of the masses,
g (mA + 7m 7 m). 8
T =
T
A
F s W T W D D
D
C
W
C C
B B W A
T T
T W
− − − −
T A
system is in equili equilibri brium um.. What What are Problem Problem 3.56 The system the coordinates of A A?
y b
Solution: Determine Determine from geometry geometry the coordinat coordinates es x, y . Isolate the cable juncture A. Since the frictionless pulleys pulleys do not change the magnitude of cable tension, and since each cable is loaded with the same weight, arbitrarily set this weight to unity, W = 1 . The angle between between the cable AB and the positiv positivee x axis axis is α; the the tensio tension n in AB is
| |
h W
x
|TAB | = i cos α + j sin α. The angle between AC and the positive x axis is (180◦ tension is
A
− β ); the
W W
+ j sin β ). TAC = TAC ( i cos β +
|
|−
The weight is W = 0 i
| |
x , y) ( x
− j1. The equilibrium conditions conditions are
0. F = TAB + TAC + W = 0.
Fx = (cos α
y b
Substitute and collect like terms,
− cos β )i = 0, 0, 0. Fy = (sin α + sin β − 1) j = 0.
h W
From the first equation cos α = cos β . On the realistic realistic assumption that both angles are in the same quadrant, then α = β . From From the the ◦ . With the angles known, second equation sin α = 1 or α = 30 2 geometry can be used to determine the coordinates x , y . The origin origin of the x, y coordinate system is at the pulley B , so that the coordinate x of the point A is positive. Define the positive distance ε as shown, so that
A W W
y
B
ε = tan α. x
Similarly,
h+ε b x
−
C A
= tan α.
β
α
Reduce to obtain
x = b = b
− h cot α − ε cot α.
W
Substitute into the first equation to obtain
x =
1 2
(b
x
− h cot α).
y
ε = x x tan α to obtain Multiply this equation by tan α and use ε = ε =
tan α 2
b
(b
− h cot α). −
−
1 2
(b
− h cot α),
y =
α
h
The sign of the coordinate y is determined as follows: Since the co ( b h cot α) > 0 is required; ordinate x is positive, the condition (b with this inequality satisfied (as it must be, or the problem is invalid), ε is also positive, as required. But the angle α is in the first quadrant, so that the point A is below the pulley B . Thus Thus y = ε and the coordinates of the point A are:
x =
C
− 12 (b tan α − h),
α = 30◦
x B
0
0
x
α α
α A
x
light fixtureof fixtureof weight weight W is is suspensuspenProblem Problem 3.57 The light ded from a circular arch by a large number N of of equally spaced cables. The tension T in each cable is the same. same. Show that
T =
πW 2 N
d θ θ
θ
.
Strategy: Consider an element element of the arch defined defined by an angle dθ measured from the point where the cables join: Since the total angle described by the arch is π radians, the the numb number er of cabl cables es atta attach ched ed to the the elem elemen entt is (N/ (N/ pi)dθ. You can use this this result result to write write the equil equilibr ibrium ium equati equations ons for the part of the cable system where the cables join. Solution:
The angle between between any cable cable and the positiv positivee x axis is π is the number of δθ intervals, one less than the number of cables. The tension in the k th kδθ ). The weight cable is T is T k = Tk (i cos kδθ + j sin kδθ) weight is W is W = 0i W j. j.
kδθ , where k = 0, 1, 2, 3 . . . K , where K =
| |
−| |
Substitute into the solution for the tension
The equilibrium conditions are
|T| =
|
W
K
K
F=W+
Tk = 0
∼
π π δθ = K If δθ = N
k=0
|π |T| = | W 2 N
where N is is the number of cables. Substitute and collect like terms:
K
Fx =
| | | |
( Tk cos kδθ) kδθ )i = 0
k=0
K
Fy =
( Tk sin kδθ) kδθ )
k=0
− |W|
j = 0
Since the tension in each cable is the same, Tk = T , the tension can be removed from the sum, and the second equation solved for the tension:
| | | |
|T| =
|
|
W
K
sin kδθ
k=0
.
The trigonometric sum can be found in handbooks1 :
K
sin kδθ =
k=0
sin
1 (N + + 2
1)δθ 1)δθ sin
sin
1 δθ 2
1 N δθ 2
.
π K = δθ The angle is divided into K intervals intervals over the arc, K = .
Substitute into the sum to obtain K
k=0
sin kδθ =
1 δθ 2 1 δθ 2
cos sin
.
=
δθ W sin 1 2 cos
sin kδθ
k=0
| | | ∼ 1, tan
δθ 2
=
1 δθ 2
δθ , 2
= W tan
therefore:
| |
1 δθ 2
.
The solutio solution n to Problem Problem 3.57 3.57 is an an Problem 3.58 “asymptotic” result whose accuracy increases as N increases. Determine the exact tension tension T exact exact for N = 3, 5, 9, and 17, and confirm the numbers in the following table. (For example, for N = 3, the cables are attached at θ = 0, θ = 90 , and θ = 180 ). ◦
◦
N
3
5
9
17 17
T exact exact πW/ 2N
1.91
1.32
1.14
1.07
Solution:
|T| =
1 is the number of intervals in an arc of π radians. where N radians. If the angle angle increment δθ is sufficiently sufficiently small,
− −
From Problem 3.57, the tension is
| K
W|
sin kδθ
k=01
tan
K
sin(kδθ sin(kδθ)) =
k=0
sin
1 (K + + 1)δθ 1) δθ sin 2 sin 12 δθ
1 Kδθ 2
π and N = δθ
π δθ
K
sin(kδθ sin(kδθ)) =
+ 1, 1,
k=0
1 δθ 2 1 δθ 2
cos sin
.
Substitute this into the expression for the tension:
|T| =
|
|
W
N
Since
δθ =
| | δθ W sin 1 2
=
sin kδθ
k=0
cos
1 δθ 2
= W tan
| |
π , N −1
the exact solution for the tension in a cable is given by
|T| = |W| tan
π 2(N 2(N
δθ Wπ Wπ , and T = = 2 2(N 2(N 1) 2 N
| |
3 5 7 9 17
since the number of cables is one more than the number of intervals Substitute this into the sum to obtain
=
N
where N isthe numberof numberof cables cables.. Theangle Theangle isdividedinto isdividedinto K segments over the interval, thus
K = =
δθ 2
| | ∼| | − −
is the asymptotic solution. The asymptotic solution, the exact solution, and the ratio of the exact solution to the asymptotic solution for the two configurations are given in the table below for 3, 5, 9, and 17 cables for the two configurations.
where the denominator is
∼
− − 1
)
1 δθ 2
.
π 2N
0.5235 0. 0.3142 0. 0.2244 0. 0.1745 0.0924
tan
π 2(N 2(N −1)
1 0.4142 0.2679 0.1989 0.0985
Ratio 1.909 1.318 1.194 1.140 1.066
Problem 3.59 If the coordinates of point A in Example ple 3.5 3.5 are are chan change ged d to (0, (0, 2, 0) m, what what are are the the tens tensio ions ns in cables AB , AC , and AD ?
−
Solution:
We need to write unit vectors eAB , eAC , and eAD .
0 .816i + 0. 0.408 j + 0. 0 .408k eAB = 0. 0 .577 j 0.577k eAC = 0.577i + 0. 0.426 j + 0. 0 .640k eAD = 0.640i + 0.
− −
−
We now need to write the four forces acting at point A.
TAB TAC TAD W
= 0. 0 .816T 816T AB 0.408T 408T AB 0.408T 408T AB AB i + 0. AB j + 0. AB k i j = 0.577T 577T AC + 0. 0 . 577T 577 T 0 . 577T 577 T AC AC AC AC AC k = 0.640T 640T AD + 0. 0 . 426T 426 T + 0. 0 . 640T 640 T i j AD AD AD AD AD k = 981 j (N)
− − −
−
Equilibrium: T AB + TAC + TAD + W = 0
F x
= 0. 0 .816T 816T AB AB
F y
= 0. 0 .408T 408T AB 577T AC 426T AD AB + 0.577T AC + 0.426T AD
F z
577T AC 640T AD − 0.577T AC − 0.640T AD = 0 − 981 = 0 = +0. +0 .408T 408T AB 577T AC 0 .640T 640T AD AB − 0.577T AC + 0. AD = 0
Solving, we get
T AB AB = 848 N T AC AC = 900 N T AD AD = 271 N
C
B T AC T AB
D
T AD
A W = mg j mg j = (100) (9.81) N j
y C (Ð2, 0, Ð2) m B
(Ð3, 0, 0 , 3) m D
x
(4, 0, 2) m z
A
(0, Ð2, Ð2, 0) m
100 kg
Problem 3.60 The force F = 5i (kN) acts on point A where the cables AB , AC , and AD are joined. What are the tensions in the three cables? Strategy: Isolate part of the cable cable system near near point A. See Example 3.5.
y D (0, 6, 0) m A
F
(12, 4, 2) m C B
(6, 0, 0) m
x
(0, 4, 6) m
z
Solution:
Isolate the cable juncture A. Get the unit vectors parallel to the cables using the the coordinates of the end points. points. Express the tensions in terms of these unit vectors, and solve the equilibrium conditions. The coordinates of points A, B , C , D are:
y D
(0, 6, 0) m A
A(12, (12, 4, 2), 2),
B (6, (6, 0, 0), 0),
C (0, (0, 4, 6), 6),
D(0, (0, 6, 0). 0).
(12, 4, 2) m C (0, 4, 6) m
The unit vector eAB is, by definition,
z
eAB
− rA (6 − 12)i + (0 − 4) j + (0 − 2)k = |rB − rA | = (6 − 12)2 + (4)2 + (2)2 −6 i − 4 j − 2 k = rB
7.483
7.483
0 .8018i eAB = 0.
7.483
− 0.5345 j − 0.267k.
Similarly, the other unit vectors are
eAB = eAD
0 .3163k, −0.9487i + 0 j + 0. = −0.9733i + 0. 0 .1622 j − 0.1622k.
The tensions in the cables are expressed in terms of the unit vectors,
TAB = TAB eAB ,
|
|
TAC = TAC eAC ,
|
|
TAD = TAD eAD .
|
|
The external force acting on the juncture is, F = 5 i + 0 j + 0 k. The equilibrium conditions are
0. F = 0 = TAB + TAC + TAD + F = 0.
Substitute and collect like terms,
Fx = ( 0.8018 TAB Fy Fz
F
− | | − 0.9487|TAC | − 0.9733|TAD | + 5)i = 0 = ( −0.5345|TAB | + 0|TAC | − 0.1622|TAD |) j = 0 = ( −0.2673|TAB | − 0.3163|TAC | − 0.1622|TAD |)k = 0. 0.
A handheld calculato calculatorr was usedto solve solve thesesimultaneou thesesimultaneouss equations equations.. The results are:
0 .7795 kN, |TAC | = 1. 1 .9765 kN, |TAD | = 2. 2 .5688 kN. |TAB | = 0.
B
(6, 0, 0) m
x
cables in Problem 3.60 will safely Problem 3.61 The cables support a tension of 25 kN. Based on this criterion, what is the largest safe magnitude of the force F = F i? Solution:
This problem problem offers a new challenge. challenge. We need to be able to solve the problem with one of the forces FAB , FAC , or FAD equal to 25 kN and the other two forces must be smaller. Note that in all of our e arlier work, forces have appeared linearly in our equations of equilibrium. equilibrium. This means that if we increase F by some factor, all other forces increase by the same factor.
y D (0, 6, 0) m A
Plan of Attack : Assume F has a value of 1 kN and solve for all forces. Find the largest force in the three cables and scale it up to 25 kN— increasing all forces by the same scale factor.
eAB =
rAB
|rAB | ,
eAC =
rAC
|rAC | ,
eAD =
rAD
|rAD |
where the points A, B , C , and D are
A: (12, (12, 4, 2) m, B : (6, (6 , 0, 0) m C : (0, (0 , 4, 6) m, D : (0, (0 , 6, 0) m The unit vectors are
eAB eAC eAD
= = =
−0.802i − 0.535 j − 0.267k −0.949i + 0 j + 0.0.316k 0 .162 j − 0.162k −0.973i + 0.
The forces are
TAB TAC TAD F
= 0.802T 802T AB 0.535T 535T AB 0.267T 267T AB AB i AB j AB k = 0.949T 949T AC 0.316T 316T AC AC i + 0 j + 0. AC k = 0.973T 973T AD 0.162T 162T AD 0.162T 162T AD AD i + 0. AD j AD k = F i
− − −
−
− −
Summing forces in the three coord. directions, we get
F x
=
F y
=
F z
802T AB 949T AC 973T AD −0.802T AB − 0.949T AC − 0.973T AD + F = 0 535T AB + 0. 0 . 162T 162 T = 0 −0.535T AB AD AD = −0.267T 267T AB + 0. 0 . 316T 316 T 162T AD AB AC AC − 0.162T AD = 0
We set F = 1 and solve the three eqns in 3 unknowns. Solving, we get
F AB 0 .155 kN, F AC AB = 0. AC = 0.395 kN 0 .513 for F = 1 kN and F AD AD = 0. Scaling, we want F AD AD FAD = 25 kN
|
|
→ 25 kN we get F
= 48. 48.7 kNi kN i. When When
(12, 4, 2) m
C B
(0, 4, 6) m
We must write our forces in terms of unit vectors.
z
F
(6, 0, 0) m x
tension in the Problem 3.62 To support the tent, the tension rope AB must be 40 lb. What are are the tensions tensions in the ropes AC , AD , and AE ?
y
(0, 5, 0) ft
Solution:
Get the unit vectors vectors parallel parallel to the cables using using the coordinates of the end points. points. Express the tensions in terms of these unit vectors, and solve the equilibrium conditions. conditions. The coordinates of points A, B , C , D , E are: are:
C
(0, 6, 6) ft
(5, 4, 3) ft (8, 4, 3) ft
D
A
B x
A(5, (5, 4, 3), 3),
B (8, (8, 4, 3), 3),
C (0, (0, 5, 0), 0),
D(0, (0, 6, 6), 6),
E (3, (3, 0, 3). 3).
E (3, 0, 3) ft
The vector locations of these points are, z
rA = 5 i + 4 j + 3 k,
rB = 8i + 4 j + 3 k,
rC = 0 i + 5 j + 0 k,
rD = 0 i + 6 j + 6 k,
y
rE = 3 i + 0 j + 3 k. The unit vector parallel to the tension acting between the points A, B in the direction of B is by definition
(0, 5, 0) ft C (0, 6, 6) ft
eAB
A
D
− rA = |rB − rA | . rB
E (3, 0, 3) ft
z
eAB = 1 i + 0 j + 0k
C
eAC =
eAE
0 .1690 j − 0.5071k −0.8452i + 0. = −0.8111i + 0. 0 .3244 j + 0. 0.4867k = −0.4472i − 0.8944 j + 0k
A B D
E
The tensions in the cables are,
TAB = TAB eAB = 40 eAB , TAD
| | = |TAD |eAD ,
TAC = TAC eAC ,
|
|
TAE = TAE eAE .
|
|
The equilibrium conditions are
0. F = 0 = TAB + TAC + TAD + TAE = 0.
Substitute the tensions,
(8, 4, 3) 3) ft ft B x
Perform this operation for each unit vector. We get
eAD
(5, 4, 3) ft
Fx = (40
− 0.8452|TAC | − 0.8111|TAD | − 0.4472|TAE |)i = 0 (+0.1690|TAC | − 0.3244|TAD | − 0.8944|TAE |) j = 0 Fy = (+0. Fz = ( −0.5071|TAC | − 0.4867|TAD |)k = 0 .
This set of simultaneous equations in the unknown forces may be solved using any of several standard algorithms.: The results are:
11 .7 lb, |TAC | = 20. 20 .6 lb, |TAD | = 21. 21 .4 lb. |TAE | = 11.
bulldozer er exerts exerts a force force F = 2i (kip) Problem Problem 3.63 The bulldoz at A. What What are the tensio tensions ns in cables cables AB , AC , and and AD?
y
6 ft C
8 ft
2 ft
B A
3 ft
Solution:
Isolate Isolate thecable juncture juncture.. Express Express the tensionsin tensionsin terms terms of unit vectors. Solve the equilibrium equilibrium equations. The coordinates of points A, B , C , D are:
D
z
4 ft
8 ft x
A(8, (8, 0, 0), 0),
B (0, (0, 3, 8), 8),
C (0, (0, 2, 6), 6),
−
D(0, (0, 4, 0). 0).
−
The radius vectors for these points are
rA = 8 i + 0 j + 0 k,
rB = 0i + 3 j + 8 k,
rC = 0 i + 2 j + 6 k,
rD = 0 i + 4 j + 0 k. y
By definition, the unit vector parallel to the tension in cable AB is 6 ft C
8 ft
− rA . eAB = |rB − rA | rB
B
Carrying out the operations for each of the cables, the results are:
z
2 ft A
3 ft D
4 ft 8 ft
eAB =
0 .2563 j − 0.6835k, −0.6835i + 0. 0 .1961 j − 0.5883k, eAC = −0.7845i + 0. 0 .4472 j + 0k. eAD = −0.8944i + 0. The tensions in the cables are expressed in terms of the unit vectors,
TAB = TAB eAB ,
|
|
TAC = TAC eAC ,
|
|
TAD = TAD eAD .
|
|
The external force acting on the juncture is F = 2000i + 0 j + 0 k. The equilibrium conditions are
0. F = 0 = TAB + TAC + TAD + F = 0.
Substitute the vectors into the equilibrium conditions:
Fx = ( 0.6835 TAB
|− 0.7845|TAC |− 0.8944|TAD | +2000)i = 0 = ( 0.2563|TAB | + 0. 0.1961|TAC | − 0.4472|TAD |) j = 0 = ( 0.6835|TAB | − 0.5883|TAC | + 0 |TAD |)k = 0 −
Fy Fz
|
The commercial program TK Solver Plus was used to solve these equations. The results are
780.31 lb , |TAC | = 906. 906.9 lb , |TAD | = 844. 844.74 lb . |TAB | = 780.
x
Problem 3.64 Prior to its launch, a balloon carrying a set of experiments to high altitude is held in place by groups of student volunteers holding the tethers at B , C , and D . The mass of the balloon, experiments package, and the gas it contains is 90 kg, and the buoyancy force on the balloo balloon n is 1000 1000 N. The superv supervisi ising ng profes professor sor conconservatively estimates that each student can exert at least a 40-N tension on the tether for the necessary length of time. Based on this estimate, estimate, what minimum numbers of students are needed at B , C , and D ?
y
A (0, 8,0) m
(10,0, –12) m C (10,0,
Solution:
F y = 1000
D (–16,0,4) (–16,0,4) m
− (90)(9. (90)(9.81) − T = 0
x
(16,0, 16) m B (16,0,
z
T T = 117. 117.1 N y
A(0, (0, 8, 0) B (16, (16, 0, 16) C (10, (10, 0, 12)
−
D( 16, 16, 0, 4)
−
We need to write unit vectors eAB , eAC , and eAD . eAB = 0. 0 .667i
− 0.333 j + 0.0.667k 0 .570i − 0.456 j − 0.684k eAC = 0. 0 .218k eAD = −0.873i − 0.436 j + 0.
A (0, 8,0 ) m
C (10,0, Ð12) m
We now write the forces in terms of magnitudes and unit vectors
FAB FAC FAD T
D (Ð16, 0, 4) m z
x (16,0, 16) m B (16,0,
= 0. 0 .667F 667F AB 0.333F 333F AB 0 .667F 667F AB AB i AB j + 0. AB k = 0. 0 .570F 570F AC 0 . 456F 456 F 0 . 684F 684 F i j AC AC AC AC AC k = 0.873F 873F AD 0 . 436F 436 F + 0. 0 . 218F 218 F AC i j AD AC AC AC k = 117. 117.1 j (N)
−
− −
−
1000 N
−
(90) g
The equations of equilibrium equilibrium are
F x = 0. 0 .667F 667F AB 0 .570F 570F AC AB + 0. AC F y F z
− 0.873F 873F AD AD = 0 = −0.333F 333F AB 456F AC 436F AC 117.1 = 0 AB − 0.456F AC − 0.436F AC + 117. = 0. 0 .667F 667F AB 684F AC 218F AC AB − 0.684F AC + 0 .218F AC = 0
T
y
Solving, we get
T (0, 8, 0)
F AB 64 .8 N AB = 64.
∼ 2 students F AC 99 .8 N ∼ 3 students AC = 99. F AD 114.6 N ∼ 3 students AD = 114.
A
F AC
F AD C (10, 0, −12) m D x
(−16, 0, 4) z
B (16, 0, 16) m
The 20-k 20-kg g mass mass is susp suspen ende ded d by cabl cables es Problem Problem 3.65 The attached attached to three vertica verticall 2-m posts. Point Point A is at (0, 1.2, 0) m. Determin Determinee the tensions tensions in cables AB , AC , and AD .
y C B D A
1m 1m 2m
0.3 m
x
z
Solution: Points A, B , C , and D are located at A(0, (0, 1.2, 0), 0), C (0, (0, 2, 1), 1),
−
y
B ( 0.3, 2, 1), 1), D(2, (2, 2, 0)
C
−
F AC B
F AB
F AD
Write the unit vectors eAB , eAC , eAD
D A
eAB =
0 .760k −0.228i + 0.0.608 j + 0. 0 .625 j − 0.781k eAC = 0 i + 0.
W
eAD = 0. 0 .928i + 0. 0 .371 j + 0k
x
(20) (9.81) N
z
The forces are y C
FAB =
228F AB 0.608F 608F AB 0 .760F 760F AB −0.228F AB i + 0. AB j + 0. AB k 0 F AC 0 .625F 625F AC 781F AC FAC = 0F AC i + 0. AC j − 0.781F AC k
B D
0 .928F 928F AD 0.371F 371F AD FAD = 0. AD i + 0. AD j + 0 k W =
A
(20)(9.81) j −(20)(9.
The equations of equilibrium equilibrium are
F x
=
228F AB 0 .928F 928F AD −0.228F AB + 0 + 0. AD = 0
F y
= 0. 0 .608F 608F AB 0 .625F 625F AC 371F AD AB + 0. AC + 0.371F AD
F z
= 0. 0 .760F 760F AB AB
781F AC − 0.781F AC + 0 = 0
We have 3 eqns in 3 unknowns solving, we get
F AB 150.0 N AB = 150. F AC 146.1 N AC = 146. F AD 36.9 N AD = 36.
1m 1m
20(9.81) = 0 − 20(9.
0.3 m z
2m x
secProblem 3.66 The weight of the horizontal wall section is W = 20,000 lb. lb. Determine the tensions tensions in the cables AB , AC , and AD .
Solution: Set the the coordinate coordinate origin origin at A with axes as shown. shown. The upward force, T , at point A will be equal to the weight, W , since the cable at A supports the entire wall. wall. The upward force at A is T = W k. From the figure, the coordinates of the points in feet are A(4, (4, 6, 10), 10),
B (0, (0, 0, 0), 0),
C (12, (12, 0, 0), 0),
and
D (4, (4, 14, 14, 0). 0).
The three unit vectors are of the form eAI =
− Z A)k) , − Z A)2
4 ft
−0.324i − 0.487 j − 0.811k, 0 .566i − 0.424 j − 0.707k, eAC = 0. 0.625 j − 0.781k. and eAD = 0 i + 0. The forces are TAB = T AB AB eAB ,
TAC = T AC AC eAC ,
and
TAD = T AD AD eAD .
The equilibrium equation for the knot at point A is T + TAB + TAC + TAD = 0. From the vector equilibrium equation, write the scalar equilibrium equations in the x , y , and z directions. We get three linear equations equations in three unknowns. Solving these equations simultaneously, simultaneously, we get
T AB AB = 9393 lb, T AC AC = 5387 lb, and
T AD 10,977 lb AD = 10,
A
14 ft
C
B
4 ft
7 ft
D
10 ft
8 ft
W
T z A
10 ft 6 ft
8 ft W
eAB =
T B
T D
y D
7 ft
T C
14 ft C
B
4 ft
X
8 ft W
7 ft C
B
(xI xA )2 + (y ( yI yA )2 + (z ( zI
where I takes takes onthe value valuess B , C , and and D. The denominat denominators ors of the unit vectors vectors are the distances distances AB , AC , and and AD , respectively. respectively. Substitution of the coordinates of the points yields the following unit vectors:
6 ft
D
10 ft
6 ft
((x ((xI xA )i + (y ( yI yA ) j + (z ( zI
A
14 ft
Problem 3.66, 3.66, each cable will safely safely Problem 3.67 In Problem support a tension of 40,000 lb. Based on this criterion, what is the largest safe value of the weight W ? Solution: There There aretwo possiblesoluti possiblesolutionsto onsto thisproblem, thisproblem, dependependingon howwe interpre interprett theproblem. Onesolution Onesolution considersthe considersthe cable extending upward from A as one of the cables subject to the 40,000 lb limit and the other does not. (a)
Assumethatthecabl Assumethatthecablee upwardfro upwardfrom m A is subjectto subjectto thelimit. From From the solution to Problem 3.66, we see that the largest tension in the cables is the tension in the cable extending upward from A. If we double the weight, we increase the tension in this cable to
40,000 lb. 40,000 lb. For this case, W MAX MAX = 40, (b)
Assume that the cable upward from A is not subject to the limit. From the solution to Problem 3.66, the largest force in the three supporting cables is T AD lb. The scale factor must must AD = 10977 lb. increase this force to 40,000 lb. The scale factor, f , is given by
f = f = 40, 40,000/ 000/10, 10,977 = 3. 3.644. 644. The maximum allowable weight is
W MAX 20,000f 000f = = (20, (20,000)(3. 000)(3.644) = 72, 72,880 lb. MAX = 20,
suspended from from the the Solution: Problem 3.68 The 680-kg load suspended helicopte helicopterr is in equilibr equilibrium. ium. The aerodynam aerodynamic ic drag force force on the load load is horizo horizonta ntal. l. The y axis axis is vertic vertical, al, and cable cable ◦ F x = T OA OA sin10 − D = 0 , the magnitude of OA lies in the x -y plane. Determine the ◦ F y = T OA (680)(9.81) = 0. 0. OA cos10 − (680)(9. the drag force and the tension in cable OA .
D = 1176 N, T OA Solving, we obtain D = OA = 6774 N. y
B
(2, 2, 0) m y
C
(5, 2, Ð1) m
A
x
10°
A z
O
(3, 0, 4) m
x
y T OA B C
D
10° D
x
(680) (9.81) N
Problem 3.68, 3.68, the coordinates of the the Problem 3.69 In Problem three cable attachment points B , C , and D are ( 3.3, 4.5, 0) m, (1.1, 5.3, 1) m, and (1.6, 5.4, 1) m, respectively respectively.. What are the tensions in cables OB , OC , and OD ?
−
− − −
−
Solution:
The position position vectors from O to pts B , C , and D are
rOB =
−3.3i − 4.5 j (m), 1 .1i − 5.3 j + k (m), rOC = 1. 1 .6i − 5.4 j − k (m). rOD = 1. Dividing by the magnitudes, we obtain the unit vectors
eOB =
−0.591i − 0.806 j, 0 .200i − 0.963 j + 0. 0 .182k, eOC = 0. 0 .280i − 0.944 j − 0.175k. eOD = 0. Using these unit vectors, we obtain the equilibrium equations
◦ F x = T OA OA sin10
F y F z
591T OB 0 .200T 200T OC 280T OD − 0.591T OB + 0. OC + 0.280T OD = 0 , ◦ = T OA 806T OB 963T OC 944T OD OA cos10 − 0.806T OB − 0.963T OC − 0.944T OD = 0, = 0. 0 .182T 182T OC 175T OD OC − 0.175T OD = 0.
From the solution of Problem 3.68, T OA OA = 6774 N. Solving these equations, we obtain
T OB OB = 3.60 kN,
T OC 1 .94 kN, OC = 1.
T OD OD = 2.02 kN.
y T OA
10° x
T OB T OC
T OD
Problem 3.70 The small sphere A weighs 20 lb, and its coordina coordinates tes are (4, 0, 6) ft. It is supported supported by two smooth flat plates labeled 1 and 2 and the cable AB . The unit vector e 1 = 94 i + 97 j + 94 k is perpendicular to 9 2 6 i + 11 j + 11 k plate 1, and the unit vector e2 = 11 is perpendicular to to plate 2. What is the tension tension in the cable?
y B
−
(0, 4, 0) ft
2 1
e2
e1
x A
Solution:
A and B are located at A (4, 0, 6), B (0, 4, 0) feet.
The vector locations of the points A and B are: rA = 4 i + 0 j + 6 k,
rB = 0 i + 4 j + 0 k.
The unit vector parallel to the tension acting from A toward B is r −r eAB = |rB −rA | . B
A
The weight is W = 0 i
− |W| j j0k = 0 i − 20 j + 0k.
The unit vectors are eAB =
−0.4851i + 0. 0 .4851 j − 0.7276k
0 .4444i + 0. 0 .7778 j + 0. 0.4444k e1 = 0. e2 =
0 .1818 j + 0. 0 .5455k −0.8182i + 0.
where the values of the last two were given by the problem statement. The forces are expressed in terms of the unit vectors, TAB = TAB eAB ,
|
|
N1 = N1 e1 ,
| |
N2 = N2 e2 .
| |
The equilibrium conditions are
F = 0 = TAB + N1 + N2 + W = 0
0.4444 N1 Fx = ( 0.4851 TAB + 0.
Substitute the force vectors and collect like terms,
| | − 0.8182|N2|)i = 0 = (0. (0 .4851|TAB | + 0. 0 .7778|N1 | − 0.1818|N2 | − 20) j = 0 = ( −0.7276|TAB | + 0. 0.4444|N1 | − 0.5455|N2 |)k = 0 −
Fy Fz
|
|
An electronic calculator was used to solve these equations. The solution is:
|TAB | = 12. 12 .34 lb , |N1 | = 17. 17 .51 lb , |N2 | = 2. 2 .19 lb . y
(0, 4, 0) ft B
x
2 1
e2
e1 A
z
z
plane Problem 3.71 The 1350-kg car is at rest on a plane surface. The unit vector vector en = 0.231i + 0.923 j + 0.308k is perpendicu perpendicular lar to the surface. surface. The y axis points points upward. upward. Determine the magnitudes of the normal and friction forces the car’s wheels exert on the surface. y en
x z
Solution: W =
The weight force is
(1350)(9.81) j = −13240 j N. −mg j = −(1350)(9.
W normal to the surface is The component of W
F N N = W e = W x ex + W y ey + W z ez = W y ey
·
= ( 13240)(0. 13240)(0.923) =
−
−12220 N.
W tangent to the surface (the friction force) can be The component of W calculated from
F T T =
W 2
− F N 2 =
(13240)2
− (12220)2 = 5096 N.
Thus, F N N = 12220 N and F T T = 5096 N.
y en
x z
system shown shown anchors anchors a stanchion Problem 3.72 The system of a cable-suspended roof. If the tension in cable AB is 900 kN, what are the tensions in cables EF and EG ?
y G
(0, 1.4, –1.2) m
Solution:
From the figure, figure, the coordinates coordinates of the points points (in meme-
E
F
ters) are
A(3. (3.4, 1, 0), 0),
B (1. (1.8, 1, 0), 0),
E (0. (0.9, 1.2, 0), 0),
C (2, (2, 0, 1), 1),
F (0 F (0,, 1.4, 1.2), 2),
and
D(2, (2, 0, 1), 1),
− G(0, (0, 1.4, −1.2). 2).
B
(0, 1.4, 1.2) m
A
(2.2, 0, –1) m D
The unit vectors are of the form eIK =
((x ((xI
xK )i + (y ( yI
− − (xI
xK
)2
(3.4, 1, 0) m
(2, 1, 0) m
(1, 1.2, 0) m
( zI − zK )k) − yK ) j + (z , 2 + (y (yI − yK ) + (z ( zI − zK )2
z
x
(2.2, 0, 1) m
C
where I K takes takes on the values BA , BC , BD , BE , EF , and EG . y
We need need to findunitvectors findunitvectors eBA , eBC , eBD , eBE , eEF , and eEG .
(0, 1.4, −1.2) m
G
Substitution of the coordinates of the points yields the following six unit vectors:
E
F
eBA = 1 i + 0 j + 0 k,
(1, 1.2, 0) m B
eBC = 0. 0 .140i eBD eBE eEF and eEG
(3.4, 1, 0) m A
(2, 1, 0) m
− 0.707 j + 0. 0 .707k, = 0. 0 .140i − 0.707 j − 0.707k, = −0.981i + 0. 0 .196 j + 0 k, = −0.635i + 0. 0 .127 j + 0. 0 .762k, = −0.635i + 0. 0 .127 j − 0.762k.
(0, 1.4, 1.2) m
(2, 0, −1) m
D
x
C
(2, 0, 1) m
z
The forces are of the form TIK = TIK eIK where IK takes on the same values values as above. above. The known known forcemagnitude forcemagnitude TBA = 900 kN. Thus,
|
y
|
E
TBA = TBA eBA = 900(1i + 0 j + 0 k) kN = 900i kN. The vector equation of equilibrium at point B (see the first free body diagram) is
(0, 1.4, −1.2) m
G
F
T BE (2, 1, 0) m
(1, 1.2, 0) m
B
T BA A
T BD
(0, 1.4, 1.2) m T BC
T BA 0. BA + T BC BC + T BD BD + T BE BE = 0.
D
C
Usethe unitvecto unitvectors rs as TBA above above to write write thisequation thisequation in component component form, and then solve the resulting linear equations for the three scalar unknowns TBC , TBD , and TBE .
(3.4, 1, 0) m
(2, 0, −1) m x
(2, 0, 1) m z
y
The result is
(0, 1.4, −1.2) m
G
T EG
127.3 kN, TBC = 127.
127.3 kN, TBD = 127.
and
917.8 kN. TBE = 917. F
Once we know TBE , we can use the second free body diagram and the equilibrium equation at point E to to solve for the tensions TEF and equilibrium equation equation at point E (see the second TEG . The vector equilibrium free body diagram) is TBE + TEF + TEG = 0. Using the unit vectors as above and solving for T EF and T and T EG , we get T get T EF = 708.7 kN. TEG = 708.
T EF
E
(1, 1.2, 0) m
−T BE (2,
1, 0) m (3.4, 1, 0) m
B
−
D C z
The cable cabless of the system system in in ProbProbProblem Problem 3.73 lem 3.72 will each safely support a tension of 1500 kN. Based on this criterion, what is the largest safe value of the tension in cable AB ?
A
(0, 1.4, 1.2) m
Solution:
(2, 0, −1) m x
(2, 0, 1) m
The largest largest load found in the solution solution of Problem Problem 3.72 is
T BE 917.8 kN. The scale factor, scaling this force up to 1500 kN is BE = 917 f f = (1500 (1500/ /917. 917.8 ) = 1. 1.634. The largest largest safe value value for the load in cable cable AB is T AB = T f f = (900)(1. (900)(1 . 634) = 1471 kN. AB max BA BA
Problem 3.74 The 200-kg slider at A is held in place on the smooth vertical bar by the cable AB . (a) Determine the tension tension in the cable. cable. (b) Determine the force force exerted exerted on the slider by the the bar.
y
2m B
A
5m
2m
x
2m z
Solution:
The coordinates coordinates of the points points A, B are A(2, (2, 2, 0), y
B (0, (0, 5, 2). The vector positions positions 2m rA = 2 i + 2 j + 0 k,
rB = 0 i + 5 j + 2 k
The equilibrium conditions are:
B
0. F = T + N + W = 0.
Eliminate the slider bar normal force as follows: follows: The bar is parallel to the y axis,hencetheunitvect axis,hencetheunitvectorparal orparalleltothe leltothe baris eB = 0i +1 j+0k. The dot product of the unit vector and the normal force vanishes: e B N = 0. Take Take the dot product of e of eB with the equilibrium conditions: eB N = 0 .
A
·
2m
5m
x
·
2m
0. eB F = eB T + eB W = 0.
·
·
·
z
The weight is eB W = 1 j ( j W ) =
·
· −| |
(200)(9.81) = −1962 N. −|W| = −(200)(9.
T
The unit vector parallel to the cable is by definition, eAB =
rB − rA |rB − rA | .
N
Substitute the vectors and carry out the operation: W
eAB = (a)
0 .7278 j + 0. 0 .4851k. −0.4851i + 0.
The tensio tension n in the cable is T = T eAB . Substitute Substitute into the modified equilibrium condition
| |
(0 .7276 T eB F = (0.
| | − 1962) = 0.0.
2696.5 N from which the tension vector is Solve: T = 2696.
| |
T = T eAB =
| |
(b)
−1308i + 1962 j + 1308k.
The equilibrium conditions are F =0 = T+N+W =
0. −1308i + 1308k + N = 0.
Solve for the normal force: N = 1308i tude is N = 1850 N.
| |
− 1308k. The magni-
Note: For this specific configuration, the problem can be solved without eliminating the slider bar normal force, since it does not appear in the y -component of the equilibrium equation (the slider bar is parallel to the y-axis). However, in the general case, the slider bar will not be parallel to an axis, and the unknown normal force will be projected onto all components of the equilibrium equations (see Problem 3.75 below). In this general situation, it will be necessary to eliminate the slider bar normal force by some procedure equivalent to that used above. End Note.
Problem 3.75 The 100-lb slider at A is held in place on the the smoo smooth th circ circul ular ar bar bar by the the cabl cablee AB . The circular circular bar is contained in the x -y plane. (a) Determine the tension tension in the cable. cable. (b) Determine the normal force exerted exerted on the the slider by the bar.
y
3 ft
B
Solution: Strategy: Strategy: Develop Develop the unitvectors (i) parallel parallel to the cable and (ii) parallel to the slider bar. Apply the equilibrium conditions. conditions. Eliminate the slider bar normal force by taking the dot product of the slider bar unit vector with the equilibrium equilibrium conditions. Solve for the force parallel to the cable. Substitute this this force into the equilibrium condition to find the slider bar normal force.
A
4 ft 20°
Assume that the circular bar is a quarter circle, so that the slider is located on a radius vector (4 ft). With this assumption assumption the coordinates of the points A, B are
x
4 ft z
A(4cos α, 4sin α, 0) = A(3 A (3..76, 76, 1.37, 37, 0), 0), B (0, (0, 4, 3). 3). The vector positions are
y
3 .76i + 1. 1 .37 j + 0k, rA = 3. The equilibrium conditions are:
rB = 0i + 4 j + 3k
3 ft
F = T + N + W = 0.
The normal force is to be eliminated from the equilibrium equations. Thebar is norma normall to theradius theradius vectorat vectorat point point A. Hence Hence theunitvector theunitvector 137.1 lb. parallel to the bar is T = 137.
| |
The dot product with the normal force is zero, eB N = 0. Take Take the dot product of the unit vector and the equilibrium condition:
0. eB F = eB T + eB W = 0.
A
B
20° 4 ft 4 ft z
The weight is eB W = eB ( j W ) =
−| |
(0.9397)(100) 9397)(100) = −94 lb. −0.9397|W| = −(0.
T
N
The unit vector parallel to the cable is by definition, eAB =
rB − rA |rB − rA | .
Substitute the vectors and carry out the operation eAB = (a)
0 .4801 j + 0. 0 .5472k. −0.6856i + 0.
The tensio tension n in the cable is T = T eAB . Substitute Substitute into the modified equilibrium condition
| |
(0 .6854 T eB F = (0.
| | − 94) = 0.0.
Solve: T = 137. 137.1 lb, from which the tension vector is
| |
T = T eAB =
| |
(b)
65 .8 j + 75 k −94i + 65.
Substi Substitut tutee T into the original equilibrium conditions, F
=0 = T+N+W = +75k + N
65 .8 j −94i + 65.
0. − 100 j = 0.
Solve for the normal force exerted by the bar on the slider
34 .2 j N = 94 i + 34.
− 75k (lb)
W
x
The cable AB keeps the 8-kg collar Problem 3.76 A in place on the smooth bar C D. The y axis points upward. What is the tension in the cable?
y
0.15 m
0.4 m
B
C
Solution:
The coordinates of points C and and D are C (0.4, (0.4, 0.3, 0), and D (0.2, 0, 0.25). The unit vector from C toward toward D is given by eCD = eCDx i + eCDy j + eCDz k =
0 .570k. −0.456i − 0.684 j + 0.
The location of point A is given by x A = xC + d CA eCDx , with similar equations for y A and z A . From the the figure, figure, d CA = 0.2 m. From this, we find the coordinates of A are A (0.309, 0.162, 0.114). From the figure, the coordinates of B are B (0, 0.5, 0.15). The unit vector from A toward B is then given by eAB = eABx i + eABy j + eABz k =
O
x
0.25 m D
0.2 m z
0 .735 j + 0. 0 .079k. −0.674i + 0. y 0.15 m
T AB AB =
B
−0.674T 674T AB 0 .735T 735T AB 0 .079T 079T AB AB i + 0. AB j + 0. AB k.
0.4 m C
From the free body diagram, the equilibrium equilibrium equations are:
F Ny Ny + T AB AB eABy
and F Nz Nz + T AB AB eABz = 0. z
·
0.2 m
D
0.4 m W C
B
T AB
0.5 m
F Nx 38.9 N, Nx = 38.
−4.53 N. z
Problem 3.76, determine determine the magnimagniProblem 3.77 In Problem tude of the normal force exerted on the collar A by the smooth bar. Solution:
The solution to Problem 3.76 3.76 above provides provides the magnitudes of the components of the normal force exerted on the collar at A.
|F N N | =
2 2 2 (F Nx ( F Ny (F Nz Nx ) + (F Ny ) + (F Nz ) .
Substituting in the values found in Problem 3.77, we get
53 .2 N. |F N N | = 53.
x
y
We now have four equations in our four unknowns. Substituting in the numbers and solving, we get
F Ny 36 .1 N, and F Nz Ny = 36. Nz =
0.25 m
0.15 m
0. FNx eCDx + F Ny Ny eCDy + F Nz Nz eCDz = 0.
T AB 57 .7 N, AB = 57.
0.2 m 0.3 m
A
0.5 m
mg = 0, − mg =
We have have three equation in four unknowns. We get another equation from the condition that the bar C D is smooth. This means that the normal force has no component parallel to CD . Mathematically, Mathematically, this can be stated as FN eCD = 0. Expanding this, this, we get
0.3 m
0.5 m
The tension force in the cable can now be written as
F Nx Nx + T AB AB eABx = 0 ,
0.2 m
A
0.2 m
F N
D
A
0.2 m 0.3 m
0.25 m
x
Problem 3.78 The 10-kg collar A and 20-kg collar B are held in place on the smooth bars by the 3-m cable from A to B and the force F acting on A. The force force F is parallel to the bar. Determine F .
y
(0, 5, 0) m
(0, 3, 0) m
F
Solution:
The geometry is the first part of the the Problem. To ease our work, let us name the points C , D , E , and G as shown in the figure. The unit vectors from C to to D and from E to to G are essential to the location of points A and B . The diagram shown contains contains two free bodies plus the pertinent geometry. geometry. The unit vectors from C to to D and from E to to G are given by eCD = erCDx i + eCDy j + eCDz k,
3m
A B
x
(4, 0, 0) m (0, 0, 4) m
z
and eEG = erEGx i + eEGy j + eEGz k. y
Using the coordinates of points C , D , E , and G from the picture, the unit vectors are eCD =
(0, 5, 0) m
0 .781 j + 0 k, −0.625i + 0.
F
(0, 3, 0) m
0.6 j + 0. 0.8k. and eEG = 0 i + 0.
3m
A
The location of point A is given by
xA = x C + CA eCDx ,
B
yA = y C + CA eCDy ,
(0, 0, 4) m
z
and z A = z C + CA eCDz , where CA = 3 m. From these equations, we find that the location of point A is given by A (2.13, 2.34, 0) m. Once we know the location of point A , we can proceed to find the location of point B . We have two ways to determine the location of B . First, B is 3 m from point A along the line AB (which we do not know). Also, B lies on the line EG . The equations for the location of point B based on line AB are:
xB = x A + AB eABx ,
y
D (0, 5, 0) m F
G (0, 3, 0) m
yB = y A + AB eABy ,
N A
and z B = z A + AB eABz .
T AB
B m Bg
yB = y E + EB eEGy ,
and z B = z E + EB eEGz .
N Ax Ax + T AB AB eABx + F eCDx = 0,
− mA g = 0,
3m
A
m Ag
C (4, 0, 0) m
x
E (0, 0, 4) m
z
B and the disWe have six new equations in the three coordinates of B tance E B . Some of the information information in the equations is redundant. redundant. However, we can solve for E B (and the coordinates of B ). We get that the length E B is 2.56 m and that point B is located at (0, 1.53, 1.96) m. We next write write equilibrium equations equations for bodies A and B . From the free body diagram for A, we get
N Ay Ay + T AB AB eABy + F eCDy
T AB
N B
The equations based on line EG are:
xB = x E + EB eEGx ,
x
(4, 0, 0) m
We now have have two fewer equation than unknowns. Fortunately, Fortunately, there are two conditions we have have not yet invoked. invoked. The bars at A and B are smooth. smooth. This means that the normal force on each bar can have no component along that bar. This can be expressed by using the dot product of the normal force and the unit vector along the bar. The two conditions are NA eCD = N Ax Ax eCDx + N Ay Ay eCDy + N Az Az eCDz = 0
·
for slider A and
and N Az Az + T AB AB eABz + F eCDz = 0.
0. NB eEG = N Bx Bx eEGx + N By By eEGy + N Bz Bz eEGz = 0. From the free body diagram for B , we get
·
Solving the eight equations in the eight unknowns, we obtain
N Bx Bx
and
− T AB AB eABx = 0, N by by − T AB AB eABy − mB g = 0, N Bz 0. Bz − T AB AB eABz = 0.
F = 36. 36 .6 N . Other values obtained in the solution are EB = 2.56 m,
N Ax Ax = 145 N, N Bx Bx =
−122 N,
N Ay Ay = 116 N, N By By = 150 N,
N Az Az = and
−112 N, N Bz Bz = 112 N.
Problem 3.79 The 100-lb crate is held in place on the smooth surface by the rope AB . Determine the tension in therope andthe magnit magnitudeof udeof thenormal thenormal force force exert exerted ed on the crate by the surface.
A
45°
B
Solution: Isolate the crate, crate, and solve the equilibrium equilibrium conditions. conditions. The weight is W = 0i 100 j. j. The angle between the normal force and the positive x axis is (90 30) = 60◦ . The normal force is
−
100 lb
−
(0.5i + 0. 0 .866 j). N = N (i cos60 + j sin60) = N (0.
| |
| |
The angle angle between between the string string tension tension and the positiv positivee x axis axis is (180◦ 45◦ ) = 135 ◦, hence the tension is
30°
−
0.7071 j. T = T (i cos135◦ + j sin135◦ ) = T ( 0.7071i + 0.
| |
| |−
The equilibrium conditions are
y T
F = W + N + T = 0.
Substituting, Substituting, and collecting like terms
Fx = (0. (0 .5 N Fy
| | − 0.7071|T|)i = 0 = (0. (0 .866|N| + 0. 0.7071|T| − 100) j = 0
51 .8 lb, N = 73 .2 lb Solve: T = 51.
| |
| |
Check: Use a coordi coordinat natee systemwith systemwith the x axis axis parall parallel el to theinclined theinclined surface. The equilibrium equation equation for the x -coordinate is
F x W sin sin 30◦
|T| =
from which
− |T| cos cos 15◦ = 0
| |
sin sin 30◦ cos cos 15◦
100 = 51. 51.76 = 51. 51.8 lb.
The equilibrium equation for the y -coordinate is
cos 30◦ + |T| sin sin 15◦ − 0, | | − W cos
F y = N
from which N = 73 .2 lb. check.
| |
A
45° B
100 lb
30°
N
W
β
x
called Russell’s Russell’s Problem 3.80 The system shown is called traction traction.. If the sum of the downward downward forces forces exerted exerted at A and B by the patient’s leg is 32.2 lb, what is the weight W ?
y
Solution: Isolate Isolate the the leg. Express Express the the tensions tensions at at A and B in scalar components. components. Solve the the equilibrium equilibrium conditions. conditions. The pulleys pulleys change the direction but not the magnitude of the force W . The The force at B is
| |
FB = W (i cos cos 60◦ + j sin sin 60◦ ). FB
60°
20°
| | = |W|(0. (0.5i + 0. 0.866 j).
25°
B A
The angles at A relative to the positive x axis are: 180◦ and 180◦ 25◦ = 155◦ . The force at A is the sum of the two forces:
−
W
FA = W (i cos180◦ + j sin180◦ ) + W (i cos155◦ + j sin155◦ ) FA
| | = |W|(−1.906i + 0. 0.4226 j).
| |
x
The total force exerted by the patient’s leg is FP = F H 32. 32.2 j, j, H i where F H equilibrium conditions are H is an unknown component. The equilibrium
−
0, F = FA + FtB + FP = 0, 60°
20°
from which: and
25°
(0 .5 W FX = (0. FY
Solve for the weight:
| | − 1.906|W| + F H H )i = 0 = (0. (0 .866|W| + 0. 0.4226|W| − 32. 32.2) j = 0. 0.
B A
W 32.2 = 25 lb . |W| = 132. .2886
Problem 3.81 A heavy heavy rope rope used as as a hawser hawser for for a cruise ship sags as shown. If it weighs 200 lb, what are the tensions in the rope at A and B ?
55°
A B
40°
Solution: Resolve the tensions tensions at A and B into scalar components. Solve the equilibrium equilibrium equations. The tension at B is A
cos 40◦ + j sin sin 40◦ ) TB = TB (i cos TB
| | = |TB |(0. (0.7660i + 0. 0.6428 j).
55° B
The angle at A relative to the positive x axis is 180◦
40°
− 55◦ = 125◦.
The tension at A: TA = TA (i cos125◦ + j sin125◦ ) = TA ( 0.5736i + 0. 0.8192 j).
| |
The weight is: W = 0 i
| |−
from which
− 200 j. j. The equilibrium conditions conditions are
0, F = TA + TB + W = 0,
Solve:
(0 .766 TB Fx = (0. Fy
| | − 0.5736|TA|)i = 0 = (0. (0 .6428|TB | − 0.8192|TA | − 200)i = 0. 0.
115.1 lb, |TA | = 153. 153.8 lb. |TB | = 115.
cable AB is horizo horizonta ntal, l, andthe box Problem Problem 3.82 The cable on the right weighs 100 lb. The surfaces are smooth. (a) What is the tension tension in the cable? cable? (b) What is the weight of the box on the left? left?
A
B
20° 40°
Solution:
Isolate the right right hand box, resolve resolve the forces into into components, and solve the equilibrium equilibrium conditions. Repeat for the box on the left. A
(a)
j. The angle For right right hand hand box. The weight weight is is W = 0i 100 j. betweenthe betweenthe normal normal forceand thepositive thepositive x axisis (90◦ 40◦ ) = 50◦ . The force:
−
−
20° 40°
cos 50◦ + j sin sin 50◦ ) = N (0. (0.6428i + 0. 0 .7660 j). N = N (i cos
| |
| |
The cable tension is T = tions are
from which
Fx
Fy
and
equilibrium condi −|T|i + 0 j. j. The equilibrium
= T + N + W = 0, 0,
F
T
= (0. (0 .6428 N
| | − |T|)i = 0 = (0. (0 .7660|N| − 100) j = 0
N
40°
W
Solve: T = 83. 83 .9 lb
| |
(b)
W = 0i W j. j. The angle For left hand box: The weight weight W angle between between the normal normal forceand the positiv positivee x axis axis is (90◦ +20 ◦ = 110◦ . The normal force:
−| |
y
T
0.9397 j). N = N ( 0.3420i + 0.
| |−
The cable tension is: T = T i + 0 j. equilibrium condi j. The equilibrium tions are:
| |
F
= W + N + T = 0, 0,
from which: and
F y
F x
= ( 0.342 N + 83. 83 .9)i = 0
− | | = ( −0.940|N| − |W|) j = 0. 0.
Solving for the weight of the box, we get
230.6 lb. |W| = 230.
B
20°
W
N
x
concrete bucket bucket used at a construcProblem 3.83 A concrete tion site is supported by two cranes. The 100-kg bucket contains 500 kg of concrete. Determine the tensions in the cables AB and AC .
(1.5, 14) m y
B
C
(3, 8) m
(5, 14) m
A
x
Solution:
We need unit vectors eAB and eAC . The coordinates of A are A, B , and C are
eAB eAC
= 0.243i + 0. 0 .970 j = 0. 0 .316i + 0. 0.949 j
−
The forces are
TAB TAC W
F x F y
A
= 0.243T 243T AB 0 .970T 970T AB AB i + 0. AB j = 0. 0 .316T 316T AC 0.949T 949T AC AC i + 0. AC j = 5886 j N
− −
=
243T AB 0 .316T 316T AC −0.243T AB + 0. AC = 0 = 0 .970T 970T AB 949T AC AB + 0 .949T AC − 5886 = 0
B
C
(5, 14) m
y
A
(3,8)
W = –mg j mg j = (600)(9.81)N
2 .66 kN Solving, T AB AB = 3.47 kN, T AC AC = 2. (1.5, 14) m
T AC
T AB
(3, 8) m
x
mass of of the suspended object A is Problem 3.84 The mass the pulleys are negligible. negligible. DemA and the masses of the termine the force T necessary for the system to be in equilibrium.
T
A
Solution: Break the system system into four free body diagrams diagrams as shown. Carefully label the forces to ensure ensure that the tension in any single cord is uniform. The equations of equilibrium for the four ob jects, starting with the leftmost leftmost pulley pulley and moving moving clockwise, are: S
− 3T = 0, 0,
and 2T + + 2S + + 2R 2R
R
− 3S = = 0,
F
F
− − 3R = 0,
R
− mA g = 0. R
We want to eliminate S , R, and F from our result and find find T in in terms m A and g . From the first of m first two equations equations,, we get S = 3T , and R = 3S = 9T . Substituting Substituting these into the last equilibrium equilibrium equation results in 2T 2 T + + 2(3T 2(3T )) + 2(9T 2(9T )) = m A g .
R R
S S S
Solving, we get T = m A g/26 g/ 26 .
S
T T
S
S
R
R
T
T T A m Ag
T
A
T . The Note: We did not have to solve for F to find the appropriate value of T final final equati equation on wouldgiv wouldgivee us thevalue thevalue of F in in terms terms of mA and g. We wouldget wouldget assembly, F = 27m 27 mA g/26 g/ 26. If we then drew a free body diagram of the entire assembly, the equation of equilibrium would be F T mA g = 0. Substituting Substituting in the known values for T and and F , we see that this equation is also satisfied. Checking the equilibrium solution by using the “extra” free body diagram is often a good procedure.
− − − −
Problem 3.85 The assembly A, including the pulley, weighs 60 lb. What force F is necessary for the system to be in equilibrium?
F
A
Solution:
From the free body diagram diagram of the assembly assembly A , we
have 3F 3F
− − 60 = 0, or
F
A
F F F F
F
F F
60 lb.
F = 20 lb
Problem 3.86 The mass of block A A is 42 kg, and the mass of block B B is 50 kg. The surfaces are smooth. If the blocks are in equilibrium, what is the force F ? B F
45° A
20°
Solution:
Isolate the top block. Solve the the equilibrium equilibrium equations. equations.
Theweight Theweight is. Theangle Theangle betwee between n thenormal thenormal force force N1 and the positiv positivee axis is. The normal normal force force is. The force force N 2 is. The equilibr equilibrium ium x axis conditions are
B F
from which
F = N1 + N2 + W = 0 20°
Fx = (0 .7071 N1 Fy
| | − |N2|)i = 0 = (0 .7071|N1 | − 490. 490.5) j = 0. 0. y
693.7 N, N1 = 693.
Solve:
45° A
|N2 | = 490. 490.5 N B
Isolate the bottom block. The weight is
W = 0i
N2
(42)(9.81) j = 0 i − 412. 412.02 j (N). − |W| j j = 0 i − (42)(9.
N1
W
α x
The angle between the normal force N1 and the positive x axis is 45◦ ) = 225◦ .
(270◦
−
y
The normal force:
N1
N1 = N1 (i cos225◦ + j sin225◦ ) = N1 ( 0.7071i
| |
| |−
− 0.7071 j).
F
α
A
β
The angle between the normal force N 3 and the positive x -axis is 20◦ ) = 70◦ .
(90◦
−
The normal force is
cos 70◦ + j sin sin 70◦ ) = N3 (0. (0.3420i N1 = N3 (i cos
| |
| |
− 0.9397 j).
j. The equilibrium conditions The force is . . . F = F i + 0 j. conditions are
| |
0, F = W + N1 + N3 + F = 0,
from which:
Fx = ( Fy
0 .3420|N3 | + |F|)i = 0 −0.7071|N1 | + 0. = ( −0.7071|N1 | + 0. 0 .9397|N3 | − 412) j = 0
693.7 N from above: For N1 = 693.
| |
x N3
|F| = 162 N
W
Problem 3.87 Cable AB is attached to the top of the vertical 3-m post, and its tension is 50 kN. What are the tensions in cables AO , AC , and AD ?
y
5m 5m C D
Solution:
Get the unit vectors vectors parallel parallel to the cables using using the coordinat coordinates es of the end points. Express Express the tensions tensions in terms of these unit vectors, and solve solve the equilibrium conditions. conditions. The coordinates of points A , B , C , D , O are found from the problem sketch: The coordinates of the points are A(6 A(6,, 2, 0), B(12 B (12,, 3, 0), C (0, (0, 8, 5), D(0, (0, 4, 5), O(0 O (0,, 0, 0).
4m 8m
−
(6, 2, 0) m O
The vector locations of these points are: z
rA = 6 i + 2 j + 0k,
rB = 12i + 3 j + 0 k,
rD = 0 i + 4 j
rO = 0i + 0 j + 0k.
− 5k,
3m
rC = 0 i + 8 j + 5 k,
12 m x
The unit vector parallel to the tension acting between the points A, B in the direction of B is by definition y
eAB
− rA . = |rB − rA | rB
5m
5m
Perform this for each of the unit vectors
D
4m
C
0 .1644 j + 0k eAB = +0 .9864i + 0. 8m
eAC = eAD eAO
−0.6092i + 0. 0 .6092 j + 0. 0.5077k = −0.7442i + 0. 0 .2481 j − 0.6202k = −0.9487i − 0.3162 j + 0k
TAD
| | = |TAD |eAD ,
TAC = TAC eAC ,
|
|
TAO = TAO eAO .
|
|
The equilibrium conditions are
F = 0 = TAB + TAC + TAD + TAO = 0.
Substitute and collect like terms,
O
(6, 2, 0) m A
The tensions in the cables are expressed in terms of the unit vectors,
TAB = TAB eAB = 50 eAB ,
B
A
(0 .9864(50) Fx = (0.
− 0.6092|TAC | − 0.7422|TAD | −0.9487|TAO |)i = 0 (0 .1644(50) + 0. 0.6092|TAC | + 0. 0.2481|TAD | Fy = (0. −0.3162|TAO |) j = 0 (+0.5077|TAC | − 0.6202|TAD |)k = 0 . Fz = (+0.
This set of simultaneous equations in the unknown forces may be solved using any of several standard algorithms. The results are:
6 .8 kN, |TAD | = 5. 5 .5 kN. |TAO | = 43 .3 kN, |TAC | = 6.
plane Problem 3.88 The 1350-kg car is at rest on a plane surface surface with its brakes brakes locked. The unit vector vector en = perpendic dicula ularr to the surfac surface. e. 0.231i +0 .923 j + 0.308k is perpen The y axis points upward. The direction cosines of the cablefrom A to B are cos θx = 0.816, cos θy = 0.408, cos θz = 0.408, and the tension in the cable is 1.2 kN. Determine the magnitudes of the normal and friction forces the car’s wheels exert on the surface.
y en B
−
−
ep
x z
Solution: Assume that all all forces act at the center of mass of the car. The vector equation of equilibrium equilibrium for the car is
y
ep
Writing these forces in terms of components, we have W =
en
B
0. FS + TAB + W = 0.
A
(1350)(9.81) = −13240 j N, −mg j = −(1350)(9.
FS = FSx i + FSy j + F Sz Sz k,
x
and TAB = TAB eAB ,
z
where eAB = cos θx i + cos θy j + cos θz k =
y
0 .408 j − 0.408k. −0.816i + 0.
Substituting these values into the equations of equilibrium and solving for the unknown components of F of FS , we get three scalar equations of equilibrium. equilibrium. These are:
F Sx Sx
and
− T ABx ABx = 0 , F Sz Sz − T ABz ABz = 0.
F Sy Sy
F Sx 979.2 N, Sx = 979. 489 .6 N. and F Sz Sz = 489.
The next step is to find the component of FS normal to the surface. This component is given by
F N N = FN en = F Sx Sx eny + F Sx Sx eny + F Sz Sz enz .
·
Substitution yields
F N N = 12149 N . From its components, the magnitude of F of FS is F S = 12800 N. Using the Pythagorean theorem, the friction force is
f =
− F S2
2 = 4033 N. F N
FS F
x W z
F Sy 12, 754 N, Sy = 12,
F N
T AB
− T ABy 0, ABy − W = 0,
Substituting in the numbers and solving, we get
en
"car"
