8–1.
The mine car and its contents have a total m ass of 6 Mg and a center of gravity gravity at at G. If the coefficient coefficient of static friction friction between the wheels and the tracks is ms = 0.4 when the wheels are locked, find the normal force acting on the front wheels at B and the rear wheels at A when the brakes at both A and B are locked locked.. Does the the car car move? move?
10 kN
0.9 m
G
B
SOLUTION
0.6 m
Equations of Equilibrium: The normal reactions acting on the wheels at ( A and B) are independent independent as to whether whether the wheels wheels are locked locked or not. Hence Hence,, the normal normal reactions acting on the wheels are the same for both cases.
a + © MB = 0;
1 2
1 2
1 2
NA 1.5 + 10 1.05 - 58.86 0.6 = 0 NA = 16. 16.544 544 kN = 16 16.5 .5 kN
+ c © Fy = 0;
A
Ans.
NB + 16.544 - 58.86 = 0 NB = 42. 42.316 316 kN = 42 42.3 .3 kN
Ans.
1 2
1 2 1 2
When both wheels at A and B are lock locked, ed, the then n FA max = msNA = 0.4 16.544 6.6176 176 kN and FB max = msNB = 0.4 42.316 = 16.9264 kN. Since FA max = 6.6 kN,, the wheels do not slip. Thus Thus,, the mine car does + FB max = 23.544 kN 7 10 kN not move. Ans.
1 2
1
2
1.5 m
0.15 m
8–2. P
Determine the maximum force P the connection can support so that no slipping occurs between the plates. There are four bolts used for the connection and each is tightened so that it is subjected to a tension of 4 kN.The coefficient of static friction between the plates is ms = 0.4 .
2 P
2
SOLUTION Free-Body Diagram: The normal reaction acting on the contacting surface is equal to the sum total tension of the bolts. Thus, N = 4(4) kN = 16 kN. When the plate is on the verge of slipping, the magnitude of the friction force acting on each contact contact surface can be computed using the friction formula F = msN = 0.4(16) kN. As indicated on the free-body diagram of the upper plate, F acts to the right since the plate has a tendency to move to the left. Equations of Equilibrium:
+ © Fx = 0;
:
0.4(16) -
P = 0 2
p = 12.8 kN
Ans.
P
If the coefficient of static friction at A is m s= 0.4 and the collar at B is smooth so it only exerts a horizontal force on the pipe, determine the minimum distance x so that the bracket can support the cylinder of any mass without slipping. Neglect the mass of the bracket. 8–3.
100 mm x
B C
200 mm A
If th the e co coef effi fici cien entt of stati staticc fr fric icti tion on at co cont ntac acti ting ng sur urfa face ce betw betwe een bloc blocks ks A and and B is m , and th tha at betw twe een block blo ck B and and bot bottom tom is 2 m , det deter ermin mine e the inc incli linat nation ion u at which the identical blocks, each of weight W , begin to slide slide.. 8–4.
s
s
A B
u
2 μs(2W cos cos θ ) – W sin sin θ = 0 sinθ – – 7 μ s cosθ = 0 7 μ s
F = 2µ s sN ʹ ʹ
ʹ ʹ
The coefficients of static and kinetic friction between the drum and brake bar are ms = 0.4 and mk = 0.3, respectively.. If M = 50 N # m and P = 85 N determine the respectively horizontal and vertical components of reaction at the pin O. Neglect the weight and thickness of the brake. The drum has a mass of 25 kg. 8–5.
300 mm
700 mm
B
125 mm
O
500 mm
M P
.
.
A
The coefficient of static friction between the drum and brake bar is ms = 0.4. If the moment M = 35 N # m , determine the smallest force P that needs to be applied to the brake bar in order to prevent the drum from rotating. Also determine the corresponding horizontal and vertical components of reaction at pin O. Neglec Neglectt the weight weight and and thickness of the brake bar. The drum has a mass of 25 kg. 8–6.
300 mm
700 mm
B
125 mm
O
500 mm
M P
A
. .
.
8–7.
The block brake consists of a pin-connected lever and friction block at B.The coefficient of static friction between the wheel and the lever is ms = 0.3, and a torque of 5 N # m is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P = 30 N, (b) P = 70 N.
5N
150mm 50 mm
m
O P
A B
SOLUTION
200 mm
To hold lever: lev er: a + © MO = 0;
FB (0.15) - 5 = 0;
FB = 33.333 N
Require NB =
33.333 N = 111 111.1 .1 N 0.3
Lever, a + © MA = 0;
PReqd. (0.6) - 111.1(0.2) - 33.333(0.05) = 0
39.8 .8 N PReqd. = 39 a) P = 30 N 6 39.8 N
No
Ans.
b) P = 70 N 7 39.8 N
Yes
Ans.
400 mm
8–8.
The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel and the lever is ms = 0.3 , and a torque of 5 N # m
5N
m
is applied to the wheel. Determin Determine e if the brake can hold
the wheel stationary when the force applied to the lever is (a) P = 30 N, (b (b)) P = 70 N .
150 15 0 mm 50 mm
O P
A B
SOLUTION
200 mm
To hold lever: a + © MO = 0;
- FB(0.15) + 5 = 0;
33.333 333 N FB = 33.
Require NB =
33.3 33 .333 33 N = 111.1 N 0.3
Lever, a + © MA = 0;
PReqd. (0.6) - 111.1(0.2) + 33.333(0.05) = 0
PReqd. = 34. 34.26 26 N
a) P = 30 N 6 34.26 N
No
Ans.
b) P = 70 N 7 34.26 N
Yes
Ans.
400 mm
8–9.
The uniform hoop of weight W is suspended from the peg at A and a horizontal force P is slowly applied at B at B.. If the hoop begins to slip at A at A when the angle is determine the coefficient of static friction between the hoop and the peg. Given:
30 deg
Solution:
F x = 0;
N A cos
P N A
P cos
F y = 0;
N A sin
0
0
N sin N A
W N A
W sin sin
sin
cos
cos N A
= 0; W r sin r sin P r r cos r cos W sin W sin
0
P 1 cos
sin sin cos sin sin cos 1 cos
sin 1
cos
0.27
Ans.
8–10
.
The uniform hoop of weight W is suspended from the peg at A and a horizontal force P is slowly applied at B. If the coefficient of static friction between the hoop and peg is s, determine if it is possible for the hoop to reach an angle before the hoop begins begins to slip. slip. Given: s
0.2 30 deg
Solution:
F x =
0;
N A cos
P N A
P cos
F y =
0;
N A sin
W sin =
0;
W r sin
W sin
sin
W N A cos
If s
0
0
1 cos
sin 1
cos N A
P r r cos P
0
sin N A
sin cos sin
cos
sin cos 1 cos
0.27
0.20 < 0.27 then it is not possible to reach 30.00 30.00 deg.
Ans.
8–11.
The fork lift has a weight
W 1 and
center of gravity at G. If the rear wheels are powered,
whereas the front wheels are f ree to roll, determine the maximum number of crates, each of weight W 2 that the f ork lift can push forward. The coefficient coefficient of static s tatic friction between the wheels and the ground is s and between between each crate and and the ground is
' ' s .
Given: W 1
12kN
W 2
150(9.81 (9.81)) N
s
' s
0.4 0.35
a
0.75m
b
0.375 m
c
1.05m
Solution:
Fork lift: MB =
0;
W c N ( b c) 1 A
NA W1 Fx =
0;
s NA P P
c b c =
=
0
NA
8.8 kN
0
N s A
P
3.54 kN
Crate: Fy =
0;
N W c 2
=
0
N W c 2 Fx =
0;
P'
' N s c
P'
Thus
n
N c
=
's Nc
P P'
1.471 kN
0 P'
n
0.515 kN
6.87
n f lo loor ( n)
n
6
Ans.
8–12.
If a torque of M = 300 N # m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD to prevent the flywheel flywheel from rotating. rotating. The coefficient of static friction between the friction pad at B and the flywheel is ms = 0.4 .
D
0.6 m
SOLUTION
30
1m B
C
Free-BodyDiagram: First we will consider the equilibrium of the flywheel using the Free-BodyDiagram: free-body diagram shown in Fig. a. Here, the frictional force FB must act to the left to produce the counterclockwise moment opposing the impending clockwise rotational motion caused by the 300 N # m couple moment. Since the wheel is required to be on the ver verge ge of sli slippi pping ng,, the then n FB = msNB = 0.4 NB . Sub Subseq sequen uently tly,, the fre free-b e-body ody diagram of member ABC shown in Fig. b will be used to determine FCD. Equations of Equilibrium: We have
a + © MO = 0;
0.4 NB(0.3) - 300 = 0
NB = 2500 N
Using this result, a + © MA = 0;
FCD sin 30°(1.6) + 0.4(2500)(0.06) - 2500(1) = 0 FCD = 3050 N = 3.05 kN
Ans.
60 mm M 300 Nm
0.3 m O
A
8–13.
The cam is subjected to a couple moment of 5 N # m. Determine the minimum force P that should be applied to the follower in order to hold the cam in the position shown. The coefficient of static friction between the cam and the follower is ms = 0.4. The guide at A is smooth.
P
10mm
B
A
60mm O
SOLUTION
5N m
Cam: a + © MO = 0;
5 - 0.4 NB (0.06) - 0. 0.01 01 (NB) = 0 NB = 14 147. 7.06 06 N
Follower:
+ c © Fy = 0;
147.06 - P = 0 P = 147 N
Ans.
.
8–15.
The car has a mass of 1.6 Mg and center of mass at G. If the coefficient of static friction between the shoulder of the road and the tires is ms = 0.4, determine the greatest slope u the shoulder can have without causing the car to slip or tip over if the car travels along the shoulder shoulder at constant velocity. velocity.
0.75m G B
1.5m A
SOLUTION
θ
Tipping: a + © MA = 0;
1 2
1 2
- W cos u 0.75 + W sin u 0.75 = 0 tan u = 1 0.75m
u = 45° 0.75m
Slipping:
Q + ©Fx = 0;
0.4 N - W sin u = 0
a + ©Fy = 0;
N - W cos u = 0
tan u = 0.4 u = 21.8°
Ans.
(Car slips before it tips.)
8–16.
If the coefficient of static friction between the collars A and B and the rod is m 0.6, determine determine the maximum maximum angle u for the system to remain in equilibrium, regardless of the weight of cylinder D. Links Links AC AC and BC have negligible weight and are connected together at C by a pin. s
=
A
B
15
15
u
u
C
D
15°, determine the minimum coefficient of If u static friction between the collars A and B and the rod required for the system to remain in equilibrium, regardless AC and BC have of the weight of cylinder D. Links AC negligible weight and are connected together at C by a pin. 8–17.
=
A
B
15
15
u
u
C
D
8–18.
The 5-kg cylinder is suspended from two equal-length cords. The end of each cord is attached to a ring of negligible mass that passes passes along a horizontal horizontal shaft. If the rings can be separated by the greatest distance d = 400 mm and still supportt the cylinder, suppor cylinder, determ determine ine the coefficient coefficient of static friction between each ring and the shaft.
d
600 mm
SOLUTION Equilibrium of the Cylinder: Referring to the FBD shown in Fig. Fig. a,
+ c © Fy = 0;
B ¢ 2 ≤ R
2 T
32 3 2 6
- m(9.81) = 0
T = 5.2025 m
Equilibrium of the Ring: Since the ring is required to be o on n the verge to slide, the frictional force can be computed using friction formula F f = mN as indicated in the FBD of the ring shown in Fig. b. Using the result result of I ,
+ c © Fy = 0; + © Fx = 0;
:
N - 5.2025 m
¢ 2 ≤ 32 3 2 6
m(4.905 m) - 5.2025 m
N = 4.905 m
= 0
¢≤ 2 6
= 0 m = 0.354
Ans.
600 mm
8–19.
The 5-kg cylinder is suspended from two equal-length cords.. The end cords end of each cord cord is attached attached to a ring of negligible mass, which passes along a horizontal horizontal shaft. If the coefficient of static friction between each ring and the shaft is ms = 0.5, determine the greatest distance d by which the rings can be separated and still support the cylinder.
d
600 mm
SOLUTION Friction: When the ring is on the verge to sliding along the rod, slipping will have to occur. Hence, F = mN = 0.5N. From the force diagram (T is the tension tension developed developed by th the e co cord rd))
tan u =
N = 2 0.5N
u = 63.43°
Geometry:
1
2
d = 2 600 cos 63.43° = 53 537 7 mm
Ans.
600 mm
8–20.
.
8–21.
The uniform pole has a weight W and length L. Its end B is tied to a supporting supporting cord, cord, and end A is placed against the wall, wal l, for whic which h the coeffi coefficie cient nt of stati staticc friction friction is ms. Determine the largest angle u at which the pole can be placed without slipping.
C
L
SOLUTION a + © MB = 0;
- NA (L cos u) - msNA (L sin u) + W
+ © F = 0; x
NA - T sin
:
u
2
a 2 sin b = 0 L
u
A
(1) u
= 0
(2)
L
B
+ c © Fy = 0;
msNA - W + T cos
Substitute Substi tute Eq. (2) into Eq. (3): ms T sin
a
W = T cos
u
2
u
2 u
2
= 0
(3)
- W + T cos
+ m s sin
u
2
u
2
= 0
b
(4)
Substitute Substi tute Eqs. Eqs. (2) and (3) into Eq. (1): T sin
u
cos u - T cos
2
u
2
sin u +
W sin u = 0 2
(5)
Substitute Substi tute Eq. (4) into Eq. (5): sin
u
2
- sin
cos
u
+
2
u
2
cos2
u
2
ms sin
tan
u
cos u - cos
u
2
a
b
u
2
+ m s sin
2
u u 1 1 cos sin u + m s sin sin u = 0 2 2 2 2
1 u u cos sin u = 0 + m s sin 2 2 2
+ m s sin
u
2
sin u +
cos
u
2
u
2
1
=
cos
cos
= sin2
u
2
u
2
= 1
u
2
= m s
u = 2 tan- 1m s
Ans.
Also, because we have a three – force member,
a
L L L sin u = cos u + tan f 2 2 2
b
1 = cos u + m s sin u ms =
u 1 - cos u = tan sin u 2
u = 2 tan-1 m s
Ans.
8–22.
If the clamping force is F = 200 N and each board has a mass of 2 kg, determine the maximum number of boards the clamp can support.The coefficient of static friction between the boards is ms = 0.3, and the coefficient of static friction between the boards and the clamp is ms ¿ = 0.45.
F
SOLUTION Free-Bod y Diagra Free-Body Diagram: m: The boards could be on the verge of slipping between the two boards boards at the ends or between between the clamp. clamp. Let n be the number of boards between the clamp. clamp. Thus Thus,, the number of boards boards between the two boards at the ends is n - 2 . If the board boardss slip betwee between n the two end boards boards,, then F = msN = 0.3(200) = 60 N. Equations of Equilibrium: Referring to the free-body diagram shown in Fig. a, we have
+ c © Fy = 0;
2(60) - (n - 2)(2)(9.81) = 0
n = 8.12
If the end boa boards rds sli slip p at the cla clamp mp,, the then n F ¿ = ms ¿ N = 0.45(200) = 90 N. By referring to the free-body diagram shown in Fig. b, we have have a + c © Fy = 0;
2(90) - n(2)(9.81) = 0
n = 9.17
Thus, the maximum number of boards that can be supported by the clamp will be Thus, the smallest value of n obtain obtained ed above, which gives n = 8
a + © Mclamp = 0;
Ans.
60 - (2)(9.81)(n - 1)2 = 0 60 - 9.81(n - 1) = 0 n = 7.12 n = 7
Ans.
F
8–23.
A 35-kg disk rests on an inclined surface surface for which ms = 0.3. Determine the maximum vertical force P that may be applied to link AB without causing the disk to slip at C .
P 200 mm
200 mm C
SOLUTION
°
30
Equations of Equilibrium: From FBD (a),
a + © MB = 0;
1 2
1 2
P 600 - A y 900 = 0
A y = 0.6667P
From FBD (b),
+ c © Fy = 0
NC sin 60° - FC sin 30° - 0.6667P - 343.35 = 0
a + © MO = 0;
FC 200 - 0.6667P 200 = 0
1 2
1 2
(1) (2)
Friction: If the disk is on the verge of moving, moving, slipping would have to occur occur at point C . He H ence, FC = ms NC = 0.3NC . Substituting this value into Eqs. (1) and (2) and solving, solving, we have P = 371 371.4 .4 N NC = 82 825. 5.3 3N
Ans.
A
300 mm
600 mm
B
.
8–25. 25.
The block brake is used to stop the wheel f rom rotating when the wheel is subjected to a couple moment M 0 If the coefficient of static friction between the wheel and the block is s, determine the smallest force P that should be applied.
Solution:
M C = 0;
P a
N
M O = 0;
N b s N c
P a b
s c
s N r M O
s P b
a r
s c
P
0
M O b
0
M O
s c
s r a
Ans.
8–26. 26.
The block brake is used to stop the wheel f rom rotating when the wheel is subjected to a couple moment M 0 If the coefficient of static friction between the wheel and the block is s , show that the brake is self locking, i. e., P 0 , provided
b c
s
Solution:
M C = 0;
P a
N
M O = 0;
N b s N c P a b
s c
s N r M O s P a r b
s c
P
P < < 0 if b
0
M O b
0
M O
s c Ans.
s r a
s c
0 i.e. if
b c
s
Ans.
8–27. 27.
The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple moment M 0 If the coefficient of static friction between the wheel and the block is s , determine the smallest force P that should be applied if the couple moment MO is applied counterclockwise .
Solution:
M C = 0;
P a
N
M O = 0;
s
N b s N
b
0
s c
N r M O
s c
P
P a
s P a r b
c
M O b
0
M O
s c
s r a
Ans.
8-28.
The 10-kg
.
8–29.
The friction pawl is pinned at A and rests against the wheel at B. It allows freedom freedom of movement movement when when the wheel is rotating counterclockwise about C . Clockw Clockwise ise rotati rotation on is is prevented due to friction of the pawl which tends to bind the wheel. If ms B = 0.6, determ determine ine the design angle u which will prevent clockwise motion for any value of Hint: t: Neglect the weight of the pawl so applied moment M. Hin that it becomes a two-force member.
A
12
θ
B
°
20
M
C
SOLUTION Friction: When the wheel is on the verge of rotating, slipping would have to occur. Hence, FB = mNB = 0.6NB . From the force diagram (FAB is the force developed in the two force member AB)
1
2
tan 20° + u =
0.6NB = 0.6 NB
u = 11.0°
Ans.
8–30.
If u = 30° determine the minimum coefficient of static friction at A and B so that equilibrium of the supporting frame is maintained regardless of the m ass of the cylinder C. Neglect the mass of the rods.
C
u
L
u
L
SOLUTION Free-Body Diagram: Due to the symmetrical loading and system, ends A and B of the rod will slip slip simultaneously. simultaneously. Since end B tends to move to the right, the friction force FB must act to the left as indicated on the free-body diagram shown in Fig. a. Equations of Equilibrium: We have
+ © Fx = 0;
FBC sin 30° - FB = 0
FB = 0 .5FBC
+ c © Fy = 0;
NB - FBC cos 30° = 0
NB = 0.8660 FBC
:
Therefore, to prevent slipping the coefficient of static friction ends A and B must be at least ms =
FB NB
=
0 .5FBC = 0.577 0.8660FBC
Ans.
A
B
8–31.
If the coefficient of static friction at A and B is ms = 0.6, determine the maximum angle u so that the frame remains in equilibrium, equilibrium, regard regardless less of the mass mass of the cylinder. cylinder. Neglect the mass of the rods.
C
u
L
u
L
SOLUTION Free-Body Diagram: Due to the symmetrical loading and system, ends A and B of the rod will slip simultaneously. simultaneously. Since end B is on the verge of sliding to the right, the friction force FB must act to the left such that FB = msNB = 0.6NB as indicated on the free-body diagram shown in Fig. a. Equations of Equilibrium: We have
+ c © Fy = 0; + © Fx = 0;
:
NB - FBC cos u = 0
NB = FBC cos u
FBC sin u - 0.6(FBC cos u) = 0
tan u = 0 .6 u = 31.0°
Ans.
A
B
8–32.
The semicylinder of mass m and radius r lies on the rough inclined plane for which f = 10° and the coefficient of static friction is ms = 0.3. Determine if the semicylinder slides down the plane, and if not,find not, find the angle of tip u of its AB base .
B
A
SOLUTION φ
Equations of Equilibrium:
a + © MO = 0;
+ © F = 0; x
:
+ c © Fy = 0
12
F r - 9.81m sin u
a 34 b = 0 r p
(1)
F cos 10° - N sin 10° = 0
(2)
F sin 10° + N cos 10° - 9.81m = 0
(3)
Solving Eqs. (1), (2) and (3) yields N = 9.661m
F = 1.703m
u = 24.2°
Ans.
Friction: The maximum friction force that can be developed between the semicylinder and the inclined plane is F max = m N = 0. 0.3 3 9. 9.66 661 1m = 2.898m. Since Fmax 7 F = 1.703m, the semicylinder will not slide down the plane . Ans.
θ
r
8–33.
The semicylinder of mass m and radius r lies on the rough inclined inclin ed plane plane.. If the inclina inclination tion f = 15°, determ determine ine the smallest coefficient of static friction which will prevent the semicylinder from slipping.
B
A
SOLUTION φ
Equations of Equilibrium:
+ Q Q © © Fx¿ = 0;
F - 9.81m sin 15° = 0
a+ © Fy¿ = 0;
N - 9.81m cos 15° = 0
F = 2.539m N = 9.476m
Friction: If the semicylinder semicylinder is on the verge of moving, moving, slipping would would have to occur.. Hence occur Hence,, F = ms N
1
2
2.539m = ms 9.476m ms = 0.268
Ans.
θ
r
8–34.
The coefficient of static friction between the 150-kg crate and the ground is ms = 0.3, while the coefficient of static friction between the 80-kg man’s shoes and the ground is mœs = 0.4. Determine if the man can move the crate. crate.
30
SOLUTION Free -- Body Diagram Free Diagram:: Since P tends to move the crate to the right, right, the frictional force FC will act to the left as indi cated on the free - body diagram shown in Fig. a. Since the crate is required to be on the verge of sliding the magnitude of FC can be computed using the friction formula, i. i .e. FC = msNC = 0.3 NC . As As indicated on the free - body diagram di agram of the man shown in Fig. b , the frictio frictional nal force force Fm acts to the right since force P has the tendency to cause the man to slip to the left. Equations of Equilibrium: Referring to Fig. a ,
+ c © Fy = 0; + © F = 0; x
:
NC + P sin 30° - 150(9.81) = 0 P cos 30° - 0.3NC = 0
Solving, P = 43 434. 4.49 49 N NC = 125 1254.2 4.26 6N
Using the result of P and referring to Fig. b, we have
+ c © Fy = 0; + © F = 0; x
:
Nm - 434 434.49 .49 sin 30° - 80(9.81) = 0
Nm = 100 1002. 2.04 04 N
Fm - 434.49 cos 30° = 0
Fm = 376.28 N
Since Fm 6 Fmax = ms ¿ Nm = 0.4(1002.04) = 40 400. 0.82 82 N, the man does not slip slip.. Thus Thus,, he can move the crate. Ans.
8–35.
If the coefficient of static friction between the crate and the ground groun d is ms = 0.3, determ determine ine the minimum coeffi coefficient cient of static friction between the man’s shoes and the ground so that the man can move the crate.
30
SOLUTION Free - Body Diagram: Since force P tends to move the crate Free crate to the right, right, the frictional force FC will act to the left as indicated on the free - body diagram shown in Fig. a. Since the crate is required to be on the verge of sliding, FC = msNC = 0.3 NC. As indicated on the free - body diagram of the man shown in Fig. b, the frictional frictional force Fm acts to the right since force P has the tendency to cause the man to slip to the left. Equations of Equilibrium: Referring to Fig. a,
+ c © Fy = 0; + © F = 0; x
:
NC + P sin 30° - 150(9.81) = 0 P cos 30° - 0.3NC = 0
Solving yields 434. 4.49 49 N P = 43 NC = 124 1245.2 5.26 6N
Using the result of P and referring to Fig. b,
+ c © Fy = 0; + © F = 0; x
:
Nm - 434. 434.49 49 sin 30° - 80(9.81) = 0
Nm = 100 1002.0 2.04 4N
Fm - 434.49 cos 30° = 0
Fm = 376.28 N
Thus, the required minimum coefficient of static friction between the man’s shoes Thus, and the ground is given by ms ¿ =
Fm Nm
=
376.28 = 0.376 1002.04
Ans.
8–36.
The thin rod has a weight W and rests against the floor and wall for which the coefficients of static friction are mA and mB, respe respectively ctively.. Determ Determine ine the smalles smallestt value of u for which the rod will not move.
B L
SOLUTION A
Equations of Equilibrium:
+ © F = 0; x
FA - NB = 0
(1)
+ c © Fy = 0
NA + FB - W = 0
(2)
a + © MA = 0;
NB (L sin u) + FB (cos u)L - W cos u
:
a2b = 0 L
(3)
Friction: If the rod is on the verge of moving, moving, slipping will have to occur at points A and B. Hence, FA = mANA and FB = mBNB. Substituting these values into Eqs. (1), (2), and (3) and solving we have NA =
W 1 + mAmB u = tan - 1
NB =
a 1 -2
mA W
1 + mAmB
mAmB
mA
b
Ans.
u
8–37. 37.
Determine the magnitude of force P needed to start towing the crate of mass M . Also determine A . the location of the resultant normal force acting on the crate, measured from point A. Given: M 40 kg
a
0.3 400 mm
b
d e c
s
200 mm 3 4
800 mm
Solution: Initial guesses:
N C
P 50 N
200 N
Given
F x = 0;
P N 0 s C 2 2 d e
F y = 0;
N C M g
d
e P 2
d
N C P
2
0
e
Find N C P
N C 280. 280.2 2 N
M O = 0;
P 140 N
s N C N 1 x
b 2 0 2 2 d e
a
2
x
1
2
s N C a
2
Thus, the distance from A from A is is
Ans.
d
2
e eP 2
N C
d
x
A
e P
2
b
x
123.51 123.51 mm
e b 2
A
523.51 523.51 mm
Ans.
8–38.
Determine the friction force on the crate of mass M , and the resultant normal force and its position x position x,, measured from point A point A,, if the force force is P. Given: M 40 kg
s
0.5
0.2
a
400 mm
k
b
800 mm
d 3
c
200 mm
e
25 N
4
P 300 N Solution: Initial guesses:
F C
N C
100 N
Given
F x = 0;
P
F y = 0;
F 0 C 2 2 d e d
N C M g
P
F C N C
Find F C N C
180.00 00 N Since F C 180.
F Cmax
M O = 0;
76.1 76.13 3 N
0 2 2 d e
a P 2 2 d e
P
0.39 m
F C
then the crate slips
<
s N C
k N C
e
F C 30.5 N N 152.3 C
Ans.
c 0 2 2 d e d
e a d c
N C
F Cmax
>
N C x P
x
Since x
e
b
2 Then the block does not tip.
2 2 d e
0.40 m
x1
a
x
x1
0.79 m
Ans.
8–39.
Determine the smallest force the man must exert on the rope in order order to move move the 80-kg 80-kg crate. crate. Also Also,, what is the the angle u at this moment? The coefficient of static friction between the crate and the floor is ms = 0.3. 45
30
u
SOLUTION Crate:
+ © F = 0; x
:
+ c © Fy = 0;
0.3NC - T ¿ sin u = 0
(1)
NC + T ¿ cos u - 80(9.81) = 0
(2)
Pulley:
+ © F = 0; x
:
+ c © Fy = 0;
- T cos 30° + T cos 45° + T ¿ sin u = 0 T sin 30° + T sin 45° - T ¿ cos u = 0
Thus, T = 6.29253 T ¿ sin u T = 0.828427 T ¿ cos u u = tan - 1
a 0.828427 b = 7.50° 6.29253
T = 0.82134 T ¿
Ans. (3)
From Eqs. Eqs. (1) and (2), NC = 239 N T ¿ = 550 N
So that T = 452 N
Ans.
8–40. 40.
The spool of wire having a mass M rests rests on the ground at A and against the wall at B at B.. P Determine the force required to begin pulling the wire horizontally off the spool. The coefficient of static friction between the spool and its points of contact is s. Units Used:
kN
3
10 N
Given: M 150 kg s
0.25
a
0.45 m
b
0.25 m
Solution: Initial guesses:
P 100 N
F A
N A
10 N
20 N
N B
30 N
F B
10 N
Given
F y = 0;
N A
F B M
F x = 0;
F A
N B P
M B = 0;
b M P F A
P F A F B N A N B
ga
s N A
g 0 0
N A a F A a F B
Find P F A F B N A N B
0
s N B
F A N A F B N B
0.28 1.12 kN 0.36 1.42
P 1.14 1.14 kN
Ans.
8– 41. 41.
The spool of wire having a mass M rests rests on the ground at A and against the wall at B at B.. Determine the forces acting on the spool at A and B and B for the given force P . The coefficient coefficient of static friction between the spool spool and the ground at point A is s. The wall at B at B is smooth. Units Used: kN
3
10 N
Given: P 800 N
a
0.45 m
M 150 kg
b
0.25 m
s
0.35
Solution ion:
Assume no no sl slipping
Initial guesses : F A
N A
10N
10N
N B
10N
F Amax
10N
Given
F x = 0;
F A
N B P
F y = 0;
N A
M g
M 0 = 0;
P b F A a F Amax
F A F Amax N A N B
0
0 0
s N A
F A F Amax
Find F A F Amax N A N B If F A
444 N 515
444 N < F Amax
515 N
Ans.
then our no-slip assumption is good.
N A 1.47 kN F A 0.44
N B
1.24 1.24 kN
Ans.
8–42.
The friction hook is made from a fixed frame which is shown colore colored d and a cylinder cylinder of negligible weight. weight. A piece of paper is placed between the smooth wall and the cylinder. cylin der. If u = 20°, determ determine ine the smalles smallestt coeffi coefficient cient of m static friction at all points of contact so that any weight W of pap paper er p can be held. u p
SOLUTION Paper:
W
+ c © Fy = 0;
F = 0.5W
F = mN;
F = mN N =
0.5W m
Cylinder: a + © MO = 0;
+ © F = 0; x
:
F = 0.5W N cos 20° + F sin 20° -
0.5W m
= 0
+ c © Fy = 0;
N sin 20° - F cos 20° - 0.5 W = 0
F = mN;
m2 sin 20° + 2m cos 20° - sin 20° = 0 m = 0.176
Ans.
8–43.
The uniform rod has a mass of 10 kg and rests on the inside of the smooth ring at B and on the ground at A. If the rod is on the verge of of slipping, determine the coefficient of static friction between the rod and the ground.
C B 0.5 m
SOLUTION 0.2 m
a + © mA = 0;
NB(0.4) - 98.1(0.25 cos 30°) = 0 NB = 53.10 N
+ c © Fy = 0;
30
NA - 98.1 + 53.10 cos 30° = 0
A
NA = 52.12 N
+ © Fx = 0;
:
m(52.12) - 53.10 sin 30° = 0 m = 0.509
Ans.
8–44.
The rings A and C each weigh W and rest on the rod, rod, which m has a coefficient of static friction of s. If the suspended ring at B has a weight of 2W , determ determine ine the largest largest distance distance d between A and C so that no motion occurs occurs.. Neglec Neglectt the weight of the wire. wire. The wire is smooth smooth and has a total length leng th of l .
d
SOLUTION
B
Free-Body Diagram: The tension developed in the wire can be obtained by considering the equilibrium of the free-body diagram shown in Fig. a.
+ c © Fy = 0;
2T sin u - 2w = 0
T =
w
sin u
Due to the symmetrical loading and system, rings A and C will slip simultaneously. Thus,, it’s sufficient to consider the equilibrium of either ring. Here, the equilibrium Thus of ring C will be considered. considered. Since ring C is required to be on the verge of sliding to the left, the friction friction force FC must act to the right such that FC = msNC as indicated on the free-body diagram of the ring shown in Fig. b. Equations of Equilibrium: Using the result of T and referring to Fig. b, we have have
+ c © Fy = 0; + © Fx = 0;
:
C
A
c sin d sin = 0 d cos = 0 (2 ) - c sin W u
NC - w -
W u
ms w
tan u =
u
NC = 2w
u
1 2ms
a 2 b - a 2 b 2 A = = l
From the geometry of Fig. c , we find find tha thatt tan u
2
d
d 2
2
l 2 - d2 . d
Thus,
2 l - d 2
d d =
2
1 2ms
=
2msl
2 1 + 4
ms 2
Ans.
.
8–46.
The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass of 50 kg and negligible negligible thickness thickness.. Determ Determine ine the minimum force P needed to move the post.The coefficients of sta static tic frict friction ion at B and C are mB = 0.4 and mC = 0.2, respectively.
800 N/ m A
B P
2m
5
400 mm
4
300 mm C
SOLUTION Member AB: a + © MA = 0;
- 800
a 43 b +
NB (2) = 0
NB = 533.3 N
Post: Assume slipping occurs at C ; FC = 0.2 NC a + © MC = 0;
+ © F = 0; x
:
+ c © Fy = 0;
4 - P(0.3) + FB(0.7) = 0 5 4 P - FB - 0.2NC = 0 5 3 P + NC - 533.3 - 50(9.81) = 0 5 P = 355 N
Ans.
NC = 81 811. 1.0 0N FB = 121. 121.6 6N
(FB)max = 0.4(533.3) = 213.3 N 7 12 121. 1.6 6N
(O.K.!)
3
8–47.
The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass of 50 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P = 15 150 0 N, the post slips slips at both both B and C simultaneously.
800 N/ m A
B P
2m
5
400 mm
4
300 mm C
SOLUTION Member AB: a + © MA = 0;
- 800
a 43 b +
NB (2) = 0
NB = 53 533. 3.3 3N
Post:
+ c © Fy = 0;
NC - 533.3 + 150
a 35 b - 50(9.81) = 0
NC = 933 933.8 .83 3N
a + © MC = 0;
4 - (150)(0.3) + FB (0.7) = 0 5 51.429 429 N FB = 51.
+ © F = 0; x
:
4 (150) - FC - 51.429 = 0 5 FC = 68. 68.57 571 1N
mC =
mB =
FC NC FB NB
=
68.571 = 0.0734 933.83
Ans.
=
51.429 = 0.0964 533.3
Ans.
3
8–48.
The beam AB has a negligible mass and thickness and is subjected to a force of 200 N. It is supported at one end by a pin and at the other end by a spool having a mass of 40 kg. If a cable is wrappe wrapped d around around the inner inner core core of the spool, spool, determine the minimum cable force P needed to move the spool. The coefficient coefficientss of static friction at B and D are mB = 0.4 and mD = 0.2, respectively.
200 N 2m
B
0.1 m
Equations of Equilibrium: From FBD (a),
12
12
NB 3 - 200 2 = 0
133. 3.33 33 N NB = 13
From FBD (b),
+ c © Fy = 0 + © F = 0; x
:
a + © MD = 0;
ND - 133.33 - 392.4 = 0
ND = 525.73 N
P - FB - FD = 0
(1)
1 2 1 2
(2)
FB 0.4 - P 0.2 = 0
1
2
Friction: Assuming the spool slips at point B, then FB = ms BNB = 0.4 133.33 = 53.33 N. Substituting this value into Eqs. (1) and (2) and solving, we have FD = 53 53.3 .33 3N
106. 6.67 67 N = 107N P = 10
1m
A
SOLUTION
a + © MA = 0;
1m
Ans.
Since FD max = ms DND = 0. 0.2 2 52 525. 5.73 73 = 10 105. 5.15 15 N 7 FD, the spool does not slip at point D. Therefore the above assumption is correct. correct.
D
0.3 m
P
.
.
8–51.
The block of weight W is being pulled up the inclined plane of slope a using a force P. If P acts at the angle f as shown, show that for slipping to occur, P = W sin a + u cos f - u , where u is the angle of friction; u = tan-1 m.
1
2> 1
P
f
2
a
SOLUTION Q +© Fx = 0;
P cos f - W sin a - mN = 0
+ a© Fy = 0;
N - W cos a + P sin f = 0 P cos f - W sin a - m(W cos a - P sin f) = 0 P = W
sin a cos
a + m cos a f + m sin f
b
Let m = tan u P = W
a + u)
sin ( a cos (
f - u)
b
(QED)
8–52.
Determine the angle f at which P sh shou ould ld ac actt on th the e bl bloc ock k so th that at th the e ma magn gnit itud ude e of P is as sm smal alll as po poss ssib ible le to be begi gin n pushing the block up the incline. What is the corre co rrespo spondi nding ng val value ue of P ? The bl blo ock wei eigh ghss W and th the e a slope is known.
P f
a
SOLUTION Slipping occurs when P = W
a + u)
sin ( a cos (
f - u)
b where
u is the angle of f riction
u = tan - 1u.
dP = W df
a sin ( cos+ () sin- ( ) - ) b = 0 a
u
2
f
f
u
u
sin (a + u) sin (f - u) = 0 sin (a + u) = 0 a = -u
P = W sin (a + f)
or
sin (f - u) = 0 f = u
Ans. Ans.
.
.
8–54.
The uniform beam has a weight W and length 4a. It rests on the fixed rails at A and B. If the coefficient of static friction at the rails is ms, determine the horizontal force P , app applied lied perpendicular to the face of the beam, which will cause the beam to move.
3a a B A P
SOLUTION From FBD (a),
+ c © F = 0;
NA + NB - W = 0
a + © MB = 0;
- NA 3a + W 2a = 0
1 2
NA =
1 2
2 W 3
NB =
1 W 3
Support A can sustain twice as much static frictional force as support B. From FBD (b),
+ c © F = 0;
P + FB - FA = 0
a + © MB = 0;
- P 4a + FA 3a = 0
1 2
FA =
1 2
4 P 3
FB =
1 P 3
The fricti frictional onal load at A is 4 times as great as at P =
3 FA 4
max
=
B.The
beam will slip at A first.
3 1 ms NA = ms W 4 2
Ans.
8–55.
Determine the greatest angle u so that the ladder does not slip when it supports the 75-kg man in the position shown. The surface is rather slippery, slippery, where the coefficient of static friction at A and B is ms = 0.3.
C
0.25 m
G
u
2.5 m 2.5 m
SOLUTION Free-Body Diagram: Diagram: The slipping could occur at either end A or B of the ladder.W ladder. We will assume that slipping occurs at end B. Thus, FB = msNB = 0.3NB . Equations of Equilibrium: Referring to the free-body diagram shown in Fig. b , we have
+ © Fx = 0;
:
FBC
+ c © Fy = 0;
> sin > 2 = 0.3 cos > 2 = 0 cos > 2 =
FBC sin u 2 - 0.3NB = 0
NB FBC
NB
u
FBC
(1)
u
NB(2)
u
Dividing Dividin g Eq. (1) by Eq. Eq. (2) yields yields
>
tan u 2 = 0.3 u = 33.40° = 33.4°
Ans.
Using this result and referring to the free-body diagram of member AC shown in Fig. a, we have have a + © MA = 0;
FBC sin 33.40°(2.5) - 75(9.81)(0.25) = 0
¢ ≤ ¢ ≤
+ © Fx = 0;
FA - 133.66 sin
+ c © Fy = 0;
NA + 133.66 cos
:
33.40° 2
33.40° 2
= 0
- 75(9.81) = 0
FBC = 133.66 N FA = 38.40 N
NA = 607.73 N
Since FA 6 (FA) max = msNA = 0.3(607.73) = 182.32 N, end A will not slip. Since slip. Thu Thus, s, the above assumption is correct.
A
B
8–56.
The uniform 6-kg slender rod rests on the top center of the 3-kg block. If the coefficients of static friction at the points of contact are mA = 0.4, mB = 0.6, and mC = 0.3, determine the largest couple moment M which can be applied to the rod without causing motion of the rod.
C
800 mm
M
SOLUTION B
Equations of Equilibrium: From FBD (a), :
+ © F = 0; x
FB - NC = 0
(1)
+ c © Fy = 0;
NB + FC - 58.86 = 0
(2)
a+ © MB = 0;
1 2
1 2
1 2
+ © F = 0; x
:
a+ © MO = 0;
(3)
NA - NB - 29.43 = 0
(4)
FA - FB = 0
(5)
1 2
12
12
FB 0.3 - NB x - 29.43 x = 0
(6)
Friction: Assume slipping occurs at point C and the block tips, then FC = msCNC = 0.3NC and x = 0.1 m. Substituting these values into Eqs. (1), (2), (3), (4), (5), and (6) and solving, we have M = 8. 8.56 561 1 N # m = 8. 8.56 56 N # m
50.8 .83 3N NB = 50
1 2 1 2
80.2 .26 6N NA = 80
1 2 1 2
600 mm
100 mm 100 mm
FC 0.6 + NC 0.8 - M - 58.86 0.3 = 0
From FBD (b),
+ c © Fy = 0;
300 mm A
Ans.
FA = FB = NC = 26.75 N
Since FA max = ms A NA = 0.4 80.26 = 32. 32.11 11 N 7 FA , the block does not slip. Also, FB max = ms B NB = 0.6 50.83 = 30 30.5 .50 0 N 7 FB , then slipping does not occur at point B. Therefore, the above assumption is correct.
8–57.
The disk has a weight W and lies on a plane which has a coefficient coeff icient of static frictio friction n m. Determi Determine ne the maximum height h to which the plane can be lifted without causing the disk to slip.
z
h a y
2a
SOLUTION
x
Unit Vector: The unit vector perpendicular to the inclined plane can be determined using cross product. A = (0 - 0)i + (0 - a) j + (h - 0)k = - a j + hk B = (2a - 0)i + (0 - a) j + (0 - 0)k = 2ai - a j
Then
N = A * B =
n =
3
i 0 2a
3
j -a -a
k h = ahi + 2ah j + 2a2k 0
ahi + 2ah j + 2a2k N = N a 5h2 + 4a2
2
Thus cos g =
2a
2 5h
2
sin g =
hence
+ 4a2
2 5
2 5h
h
2
+ 4a2
Equations of Equilibrium and Friction: When the disk is on the verge of sliding down the plane, F = mN.
© Fn = 0;
N - W cos g = 0
N = W cos g
© Ft = 0;
W sin g -
N =
mN =
0
W sin g m
(1) (2)
Divide Eq. (2) by (1) yields sin g
= 1
m cos g
1 5
1 5
h
h
m
a 2
2
+ 4a
2
2a 2
2
5h + 4a
h =
2
b
2 5 a
m
= 1
Ans.
8–58.
Determine the largest angle u that will cause the wedge wedge to be self-locking regardless of the magnitude of horizontal force P applie applied d to the blocks. blocks. The coefficien coefficientt of static friction between the wedge and the blocks is ms = 0.3 . Neglect the weight of the wedge.
P
SOLUTION Free-Body Diagram: Fo Forr the wedge to be self-locking self-locking,, the frictional frictional force force F indicated on the free-body diagram of the wedge shown in Fig. a must act downward and its magnitude must be F … msN = 0 .3N . Equations of Equilibrium: Referring to Fig. a, we have
>
+ c © Fy = 0;
>
2N sin u 2 - 2F cos u 2 = 0
>
F = N tan u 2
Using the requirement F … 0 .3N, we obtain
>
N tan u 2 … 0 .3N u = 33.4°
Ans.
u
P
8–59.
If the beam AD is loaded loaded as show shown, n, det determ ermine ine the the horizontal force P which must be applied to the wedge in order to remove it from under the beam.The coefficients of static friction at the wedge’s top and bottom surfaces are mCA = 0.25 and mCB = 0.35, respec respectively tively.. If P = 0, is the wedge self-locking? Neglect the weight and size of the wedge and the thickness of the beam.
4 kN/m
A
D
C B
3m
SOLUTION Equations of Equilibrium and Friction: If the wedge is on the verge of moving to the right, right, then slipping slipping will have have to occur occur at both contac contactt surfaces surfaces.. Thus Thus,, FA = ms A NA = 0.25NA and FB = ms B NB = 0.35NB . From FBD (a),
a + © MD = 0;
12
12 - 6.00122 - 16.0152 = 0
NA cos 10° 7 + 0.25NA sin 10° 7
12.78 78 kN NA = 12. From FBD (b),
+ c © Fy = 0;
1 2
12.78 78 sin 80° 80° - 0.25 12.78 sin 10° = 0 NB - 12. NB = 13. 13.14 14 kN
+ © F = 0; x
:
1 2 - 0.35113.142 = 0
P + 12. 12.78 78 cos 80° 80° - 0.25 12.78 cos 10°
P = 5. 5.53 53 kN
Ans.
Since a force P 7 0 is required to pull out the wedge, the wedge will be self-locking when P = 0. Ans.
4m
°
10
P
8–60.
The wedge has a negligible weight and a coefficient of static friction ms = 0.35 with all contact contacting ing surfaces surfaces.. Determine the largest angle u so that it is “self-loc “self-locking. king.” ” This requires no slipping for any magnitude of the force P applied to the joint.
u –– 2
P
SOLUTION Friction: When the wedge is on the verge of slippin slipping, g, then F = mN = 0.35N . Fro From m the force diagram (P is the ‘locking’ ‘locking’ force.), force.),
tan
u
2
=
0.35N = 0.35 N
u = 38.6°
Ans.
u –– 2
P
8–61.
If the spring is compressed 60 mm and the coefficient of static friction between the tapered stub S and the slider A is mSA = 0.5, determine the horizontal force P needed to move the slider forward. The stub is free to move without friction within the fixed collar C . The coefficie coefficient nt of static friction between A and surface B is mAB = 0.4. Neglect the weights of the slider and stub.
C k
300 N/ m
SOLUTION
S
Stub:
+ c © Fy = 0;
NA cos 30° - 0.5NA sin 30° - 300(0.06) = 0
P
30
A
NA = 29. 29.22 22 N B
Slider:
+ c © Fy = 0;
NB - 29.22 cos 30° + 0.5(29.22) sin 30° = 0 NB = 18 N
+ © F = 0; x
:
29.22 22 sin 30° 30° - 0.5(29.22) cos 30° = 0 P - 0.4(18) - 29. P = 34.5 N
Ans.
8–62.
If P = 250 N, N, determine the required minimum compression in the spring so that the wedge will not move to the right. Neglect the weight of A and B. The coeffic coefficient ient of static static friction for all contacting surfaces is ms = 0.35. Neglect friction at the rollers.
k
15
kN/ m B
SOLUTION P
3
Free-Body Diagram: Free-Body Diagram: The spring force acting on the cylinder is Fsp = kx = 15(10 )x. Since it is required that the wedge is on the verge to slide to the right, the frictional force must act to the left on the top and bottom surfaces of the wedge and their magnitude can be determined using friction formula.
(F f)1 = mN1 = 0.35N1
(F f)2 = 0.35N2
Equations of Equilibrium: Referring to the FBD of the cylinder, Fig. a, N1 - 15(103)x = 0
+ c © Fy = 0;
3
N1 = 15(103)x
4
Thus, (F f)1 = 0.35 15(103)x = 5.25(103)x Referring to the FBD of the wedge shown in Fig. b,
+ c © Fy = 0;
N2 cos 10° - 0.35N2 sin 10° - 15(103)x = 0 N2 = 16.233(103)x
+ © Fx = 0;
:
3
4
250 - 5.25(103)x - 0.35 16.233(103)x cos 10° - 16.233(103)x sin 10° = 0
3
4
x = 0.01830 m = 18.3 mm
Ans.
A
10
8–63.
Determine the minimum applied force P required to move wedge A to the right.The spring is compressed compressed a distance distance of 175 mm. Neglec Neglectt the weight of A and B. The coefficie coefficient nt of static friction for all contacting surfaces is ms = 0.35. Neglect friction at the rollers.
k = = 15 kN/m B
SOLUTION P
Equations of Equilibrium and Friction: Using the spring formula, Fsp = kx = 15 0.175 = 2.625 kN. If the wedge is on the verge of moving to the right, then slipping will have to occur at both contact surfaces. Thus, FA = msNA = 0.35NA and FB = msNB = 0.35NB. From FBD (a),
1 2
+ c © Fy = 0;
NB - 2.625 = 0
2.625 625 kN NB = 2.
From FBD (b),
+ c © Fy = 0;
NA cos 10° - 0.35NA sin 10° - 2.625 = 0 NA = 2. 2.84 841 1 kN
+ © F = 0; x
:
1 2
1 2
P - 0.35 2.625 - 0.35 2.841 cos 10°
- 2.841 sin 10° = 0 P = 2. 2.39 39 kN
Ans.
A °
10
8–64.
.
8–65.
The coefficient of static friction between wedges B and C is ms = 0.6 and between the surfaces of contact B and A and C and D, ms ¿ = 0.4. If the spring is compressed 200 mm when in the position shown, determine the smallest force force P needed to move wedge C to the left. Neglec Neglectt the weight of the wedges.
k
15 A
B
15 15 D
SOLUTION Wedge B:
+ © F = 0; x
:
+ c © Fy = 0;
NAB - 0.6NBC cos 15° - NBC sin 15° = 0 NBC cos 15° - 0.6NBC sin 15° - 0.4NAB - 100 = 0
210.4 .4 N NBC = 210 NAB = 176.4 N
Wedge C :
+ c © Fy = 0;
NCD cos 15° - 0.4NCD sin 15° + 0.6(210.4) sin 15° - 210.4 cos 15° = 0 NCD = 19 197. 7.8 8N
+ © F = 0; x
:
197.8 si s in 15° + 0.4(197.8) cos 15° + 210.4 sin 15° + 0.6(210.4) cos 15° - P = 0 P = 304 N
500 N/ m
Ans.
C
P
8–66.
The co The coeff effic icien ientt of st stati aticc fr fric ictio tion n be betw tween een th the e wed wedge gess B and C is ms = 0.6 and between the surfaces of contact B and A and C and D, ms ¿ = 0.4. If P = 50 N, determine the largest allowable allowabl e compre compressionof ssionof the springwithout causin causing g wedgeC to mo move ve to th the e le left ft.. Ne Negl glec ectt th the e we weig ight ht of th the e we wedg dges es..
k
500 N/ m
15 A
B
15 15 D
SOLUTION Wedge C :
+ © F = 0; x
:
c + © Fy = 0;
(NCD + NBC) si sin n 15° 15° + (0.4NCD + 0.6NBC) co coss 15° 15° - 50 = 0 (NCD - NBC) co coss 15° 15° + ( - 0.4NCD + 0.6NBC) si sin n 15° 15° = 0 34.6 .61 1N NBC = 34 NCD = 32.53 N
Wedge B:
+ © F = 0; x
:
34.61 61 sin 15° 15° = 0 NAB - 0.6(34.61) cos 15° - 34. NAB = 29. 29.01 01 N
c + © Fy = 0;
34.61 cos 15° - 0.6(34.61) sin 15° - 0.4(29.01) - 500x = 0 x = 0.0 0.03290 3290 m = 32. 32.9 9 mm
Ans.
C
P
8– 67. 67.
The vise is used to grip the pipe. If a horizontal horizontal force F force F 1 is applied perpendicular to the end of the handle of length l , determine the compressive force F force F developed developed in the pipe. The square threads have a mean diameter d and and a lead a. How much force must must be applied perpendicular to the handle to loosen the vise? Given: F1 d
100N
37.5mm
s 0.3 L 250mm a
5mm
Solution: r
d
2
atan
a 2 r
2.43 deg
atan
s
16.70 deg
F
3.84 kN
Ans.
P
73.3 N
Ans.
F1 L F
=
F r tan(
F1
)
L r tan (
)
)
To loosen screw, P L
P
=
F r tan(
F r
tan( L
)
8–68
.
Determine the couple forces F forces F that that must be applied to the handle of the machinist’s vise in order to create a compressive compressive forc f orcee F A in the block. Neglect frict ion at the bearing A bearing A.. The guide at B at B is smooth so that the axial force on the screw i s F A. The single si ngle square-threaded screw has a mean radius b and a lead c, and the coefficient of static friction is s. Given: a
125mm
FA 400N b
6mm
c 8mm s 0.27
Solution:
atan
s
15.11 deg
atan
c 2 b
11.981 deg
4.91 N
F 2 a
F
=
FA b tan(
FA
b
2a
tan(
)
)
F
Ans.
8–69.
The column is used to support the upper upper floor. If a force F = 80 N is applied perpendicular to the handle to tighten the screw, screw, determine the compressive force in the column. The square-threaded screw on the jack has a coefficient of static fricti friction on of ms = 0.4, mean diamete diameterr of 25 mm, and a lead of 3 mm.
0.5 m F
SOLUTION
12 1 2 = tan 10.42 = 21.80° 3 = tan c d = 2.188° 2 112.52 8010.52 = 10.01252 tan121.80° + 2.188°2 M = W r tan fs + up fs
up
-1
-1
p
W
W = 7.19 kN
Ans.
8–70.
If the force F is removed from the handle of the jack in Prob. 8-69, determine if the screw is self-locking. self-locking.
0.5 m
SOLUTION
1 2= c 1 2d =
-1
fs = tan
up = tan-1
0.4
F
21.80°
3 2p 12.5
2.188°
Since fs 7 up , the screw is self locking.
Ans.
8–71.
If the clamping force at G is 900 N, N, determine the horizontal F force that must be applied perpendicular to the handle of the lever at E. The mean diameter and lead of both single square-threaded screws at C and D ar are e 25 25 mm an and d 5 mm mm,, respectively.. The coefficient of static friction is ms = 0.3 . respectively
200 mm
200 mm C
G A
B D
SOLUTION
E
Referring to the free-body diagram of member GAC shown in Fig. a, we ha have ve FCD = 900 N © MA = 0; FCD(0.2) - 900(0.2) = 0 L Since the screw screw is being tightened, Eq. 8–3 should be used. Here, u = tan-1 = 2pr 5 tan - 1 = 3.643°; 2p(12.5)
c
d
a b
fs = tan - 1 ms = tan - 1(0.3) = 16.699°; an and d M = F(0.125). Sin Since ce M must overcome
the friction of two screws, M = 2[Wr tan(fs + u)] F(0.125) = 2 [900(0.0125)tan(16.699° + 3.643°)] F = 66.7 N
Note: Since fs 7 u, the screw is self-locking.
Ans.
125 mm
8–72.
If a horizontal force of F = 50 N is applied perpendicular to the handle of the lever at E, determ determine ine the clamping clamping force developed at G. The mean diameter diameter and lead of of the single square-threaded screw at C and D are 25 mm and 5 mm, respectively.The coefficient of static friction is ms = 0.3.
200 mm
200 mm C
G A
B D
SOLUTION E
Since the screw screw is being tightened, tightened, Eq. 8–3 should be used. used. Here Here,, u = tan-1 tan-1
c
5 2p(12.5)
d = 3.643°;
a2 b = L pr
fs = tan-1ms = tan-1(0.3) = 16.699°; and M = 50(0.125) . Sin Since ce M must overcome
the friction of two screws, M = 2[Wr tan(fs + u)]
50(0.125) = 2[FCD(0.0125)tan(16.699° + 3.643°)] FCD = 674.32 N
Using the result of F CD and referring to the free-body diagram of member shown in Fig. a, we have
Ans. GAC
© MA = 0; 674.32(0.2) - FG(0.2) = 0 FG = 674 N
Note: Since fs 7 u, the screws are self-locking.
Ans.
125 mm
8–73.
A turnb turnbuckle uckle,, similar to that that shown shown in Fig. Fig. 8–17, is used used to tension member AB of the truss. truss. The coefficient coefficient of the static friction between the square threaded screws and the turnbuckle is ms = 0.5. The screws have a mean radius of 6 mm and a lead of 3 mm. If a tor torque que of M = 10 N # m is applied to to the turnbuckle turnbuckle,, to draw the the screws screws closer together, determine the force in each each member of the truss. truss. No external forces act on the truss.
D
B
4m M
SOLUTION Frictional Forces on Screw: Here, u = tan-1
a 2 b = tan c 2 3162 d = 4.550°, l
-1
pr
p
M = 10 N # m and fs = tan-1 ms = tan-110.52 = 26.565°. Since f riction at two screws
must be overcome, th t hen, W = 2FAB . Applying Eq. 8– 8 –3, we w e have
1
2
M = Wr tan u + fS
1 2 1 2 1380.6 0.62 2 N 1T2 = 1. 1.38 38 kN 1T2 = 138
10 = 2FAB 0.006 tan 4.550° + 26.565° FAB
Ans.
Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if moment M is removed. Method of Joints:
Joint B:
+ © F = 0; x
:
1380.62
a 35 b -
FBD = 0
12 12 4 - 1380.62 a b = 0 5 1.10 10 kN 1 2 = 1104.50 N 1 2 = 1.
828. 8.37 37 N C = 828 N C FBD = 82
+ c © Fy = 0;
FBC FBC
C
C
Ans.
Ans.
Joint A:
+ © F = 0; x
:
a 35 b = 0 828 8N1 2 = 828.37 N 1 2 = 82 4 1380.62 a b = 0 5 1104.5 4.50 0 N 1 2 = 1. 1.10 10 kN 1 2 = 110 FAC - 1380.62 FAC
+ c © Fy = 0;
C
C
Ans.
FAD
FAD
C
C
Ans.
Joint C :
+ © F = 0; x
:
a 35 b - 828.37 = 0 1380.6 0.62 2 N 1 2 = 1. 1.38 38 kN 1 2 = 138 4 + 1380.62 a b - 1104.50 = 0 5
FCD FCD
+ c © Fy = 0;
Cy
T
T
!) Cy = 0 ( No external applied load. check !)
Ans.
C
A
3m
8–74.
A turnbu turnbuckle ckle,, similar to that that shown shown in in Fig. Fig. 8–17, is used used to tension member AB of the truss. truss. The coefficient of the static friction between the square-threaded screws and the turnbuckle is ms = 0.5. The screws have a mean radius of 6 mm and a lead of 3 mm. Determin Determine e the torque torque M which must be applied to the turnbuckle to draw the screws closer together, togethe r, so that that the compress compressive ive force force of 500 500 N is developed in member BC .
D
B
4m M
SOLUTION Method of Joints: C
A
Joint B :
+ c © Fy = 0;
500 - FAB
a b=0 4 5
3m
12
FAB = 625 N C
a 2 b = tan c 2 3162 d = 4.550°
l -1 pr p and fs = tan-1ms = tan-1 0.5 = 26.565°. Since friction at two screws must be overcome, then, W = 2FAB = 2 62 5 = 1250 N. Applying Eq. 8–3, we have
Frictional Forces on Screws: Here, u = tan-1
1 2
M
1 2 = tan1 + 2 = 125010.0062 tan14.550° + 26.565°2 Wr
u
= 4.53 N # m
f
Ans.
Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if moment M is removed.
8–75.
The shaft has a square-threaded screw with a lead of 8 mm and a mean radius of 15 mm. If it is in contact with a plate gear having a mean radius of 30 mm, determine the resisting torque M on the plate gear which can be overcome if a torque of 7 N # m is applied to the shaft. The coefficient of static friction at the screw is mB = 0.2. Neglect friction of the bearings located at A and B.
15 mm
B
M
SOLUTION
30 mm
Frictional Forces on Screw: Here, u = tan-1 W = F, M = 7 N # m and fs = tan-1ms = tan-1 we have
a 2 b = tan c 2 18152 d = 4.852°, 10.22 = 11.310°. Applying Eq. 8–3, l pr
-1
p
1 2 10.0152 tan 14.852° + 11.310°2
A
M = Wr tan u + f
7N·m
7 = F
F = 1610.29 N
Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if force F is removed. Equations of Equilibrium:
a+© MO = 0;
1 2
1610.29 0.03 - M = 0 M = 48 48.3 .3 N # m
Ans.
.
8–77.
The fixture clamp consist of a square-threaded screw having a coefficient of static friction of ms = 0.3, mean diameter of 3 mm, and a lead of of 1 mm. The five five points points indicated indicated are pin connections. Determine the clamping force at the smooth blocks D and E when a torque of M = 0.08 N # m is applied to the handle of the screw screw..
30 mm
E
45°
B
D
40 mm 30 mm C
40 mm
A
40 mm
SOLUTION
¢ ≤
B
= 0.08 N · m M =
R
1 l Frictional Forces on Screw: Here, u = tan - 1 = tan - 1 = 6.057°, 2pr 2p(1.5) W = P, M = 0. 0.08 08 N # m and fs = tan - 1 ms = tan - 1 (0.3) = 16.699°. Ap Applying Eq. 8–3,we have have M = Wr tan(u + f)
0.08 = P(0.0015) tan (6.057° + 16.699°) P = 127.15 N
Note: Since fS 7 u, the screw is self-locking. It will not unscrew even if moment M is removed. Equation of Equilibrium:
a + © MC = 0;
127.15 cos 45° (40) - FE cos 45°(40) - FE sin 45°(30) = 0 FE = 72 72.6 .66 6 N = 72 72.7 .7 N
Ans.
The equilibrium of the clamped blocks requires that FD = FE = 72 72.7 .7 N
Ans.
8–78.
The braking mechanism consists of two pinned arms and a square-threaded screw with left and righthand threads.Thus when turned, turned, the screw draws draws the two arms together. together. If the lead of the the screw is is 4 mm, the mean diameter diameter 12 12 mm, and m the coefficient of static friction is s = 0.35, determine the tension in the screw when a torque of 5 N # m is applied to tighten the screw. screw. If the coefficient of static friction between the brake pads A and B and the circular shaft is msœ = 0.5, determine the maximum torque M the brake can resist.
5N·m
300 mm 200 mm A
C
a 2 b = tan c 2 4162 d = 6.057°,
l -1 pr p M = 5 N # m and fs = tan-1ms = tan-1 0.35 = 19.290°. Since f riction at two screws must be overcome, t he hen, W = 2P. Applying Eq. 8 –3 –3, we have
1 2
1
2 10.0062 tan16.057° + 19.290°2
M = Wr tan u + f
5 = 2P
P = 879 879.6 .61 1 N = 880 N
Ans.
Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if moment M is removed. Equations of Equilibr Equations Equilibrium ium and Frict Friction: ion: Since the shaft is on the verge to rotate about point O, then, FA = ms ¿ NA = 0.5NA and FB = ms ¿ NB = 0.5NB . From FBD (a),
a + © MD = 0;
1 2
1 2
879.61 0.6 - NB 0.3 = 0
1759. 9.22 22 N NB = 175
From FBD (b), a + © MO = 0;
3 1
2410.22 -
2 0.5 1759.22
M = 0
M = 352N # m
B
300 mm
SOLUTION Frictional Forces on Screw: Here, u = tan-1
M
Ans.
D
8–79.
If a horizontal force of P = 100 N is applied perpendicular to the handle of the lever at A, determ determine ine the compressive compressive force F exerted on the materia material. l. Each single square square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction at all contacting surfaces of the wedges is ms = 0.2, and the coefficient of static friction at the screw is msœ = 0.15.
A
15 B
SOLUTION Since the screws screws are being being tightened, tightened, Eq. 8–3 should be used. used. Here Here,, 7 .5 L u = tan -1 = tan -1 = 5.455°; 2pr 2p(12.5)
a b
c
d
#
fs = tan - 1ms = tan - 1(0.15) = 8.531°; M = 100(0.25 ) = 25 N m; and W = T, where T is the tension in the screw shank. Since M must overcome the friction of
two screws,
4
M = 2[Wr, tan(fs + u)
3
4
25 = 2 T(0.0125) tan (8.531° + 5.455°) T = 4015.09 N = 4.02 kN
Ans.
Referring to the free-body diagram of wedge B shown in Fig. a using the result of T , we have
+ © Fx = 0; + c © Fy = 0; :
4015.09 - 0 .2N ¿ - 0 .2N cos 15° - N sin 15° = 0
(1)
N ¿ + 0 .2N sin 15° - N cos 15° = 0
(2)
Solving, N = 6324.60 N
N ¿ = 5781.71 N
Using the result of N and referring to the free-body diagram of wedge C shown in Fig. b, we have have
+ c © Fy = 0;
3
4
2(6324.60) cos 15° - 2 0.2(6324.60) sin 15° - F = 0 F = 11563.42 N = 11.6 kN
Ans.
C 15
250 mm
8–80.
Determine the horizontal force P that must be applied perpendicular to the handle of the lever at A in order to develop a compressive force of 12 kN on the material. Each single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction at all contacting surfaces of the wedges is ms = 0.2, and the coefficient of static friction at the screw is msœ = 0.15.
A
15 B
SOLUTION Referring to the free-body diagram of wedge C shown in Fig. a, we have have
+ c © Fy = 0;
3
4
2N cos 15° - 2 0 .2N sin 15° - 12000 = 0 N = 6563.39 N
Using the result of N and referring to the free-body diagram of wedge Fig. b, we have
+ c © Fy = 0;
B
shown in
.2(6563.39) 63.39) sin 15° = 0 N ¿ - 6563.39 cos 15° + 0 .2(65 N ¿ = 6000 N
+ © Fx = 0;
:
T - 6563.39 sin 15° - 0 .2(656 .2(6563 3 .39) cos 15° - 0.2(6000) = 0 T = 4166.68 N
Since the screw screw is being tightened, tightened, Eq. 8–3 should be used. used. Here Here,, u = tan - 1
7 .5 c 2 d = tan c 2 (12.5) d = 5.455°; L pr
-1
p
4166.68N.. Since fs = tan-1ms = tan-1(0.15) = 8.531°; M = P(0.25); and W = T = 4166.68N M must overcome the friction of two screws,
3
4
M = 2 Wr tan (fs + u) P(0.25) = 2 4166.68(0.0125) tan (8.531° + 5.455°) P = 104 N
3
4
Ans.
C 15
250 mm
8–81.
Determine the clamping force on the board A if the screw of the “C” “C” clamp is is tightened tightened with with a twist of M = 8 N # m. The single square-threaded screw has a mean radius of 10 mm, a lead of 3 mm, and the coefficient of static friction friction m is s = 0.35.
M
SOLUTION fs = tan - 1(0.35) = 19.29° up = tan - 1
A
3 c 2 (10) d = 2.734° p
M = W(r) tan (fs + up)
8 = P(0.01) tan (19.29° (19.29° + 2.734°) 1978 78 N = 1. 1.98 98 kN P = 19
Ans.
8–82.
If the required clamping force at the board A is to be 50 N, determine the torque M that must be applied to the handle of the “C” clamp to tighten it down. The single square-threade square-threaded d screw scr ew has has a mean mean radius radius of 10 mm, a lead of 3 m m, and the coefficient of static friction is ms = 0.35.
M
SOLUTION fs = tan - 1(0.35) = 19.29° up = tan - 1
A
3 a 2 b = tan c 2 (10) d = 2.734° l pr
-1
p
M = W(r) ta tan n (f s + u p)
0.20 202 2N#m = 50(0.01) tan (19.29° + 2.734°) = 0.
Ans.
8–83.
A cylinder having a mass of 250 kg is to be supported by the cord which which wraps over the pipe. pipe. Determi Determine ne the smallest vertical force F needed to support the load if the cord passes (a) once ov over the pipe, b = 180°, and (b) two times b over the pipe, = 540°. Take ms = 0.2 .
SOLUTION Frictional Force on Flat Belt: Here, T1 = F and T2 = 250(9.81) = 2452.5 N. Applying Applyi ng Eq. Eq. 8–6, we have have
a) If b = 180° = p rad T2 = T1 e mb
2452.5 = Fe 0.2p F = 1308 1308.38 .38 N = 1. 1.31 31 kN
Ans.
b) If b = 540° = 3 p rad T2 = T1 e mb
2452.5 = Fe 0.2(3p) F = 372.38 N = 372 N
Ans.
F
8–84.
A cylinder having a mass of 250 kg is to be supported by the cord which which wraps over the the pipe. Determi Determine ne the largest largest vertical force F that can be applied to the cord without moving the cylinder. The cord passes (a) once over the pipe, b = 180°, and (b) two times over the pipe, b = 540° . Ta Take ms = 0.2.
SOLUTION
F
Frictional Force on Flat Belt: Here, T1 = 250(9.81) = 245 2452. 2.5 5 N and T2 = F . Applying Applyin g Eq. Eq. 8–6, we have
a)
If b = 180° = p rad T2 = T1 e mb F = 2452.5e 0.2 p F = 459 4597.1 7.10 0 N = 4.60 kN
b)
Ans.
If b = 540° = 3 p rad T2 = T1e mb F = 2452.5e 0.2(3 p) F = 16 15 152. 2.32 32 N = 16.2 kN
Ans.
8–85
.
The cord supporting the cylinder of mass M passes passes around three pegs, A, A , B, B, C, where the coefficient of friction is s. Determine the range of values for the magnitude magnitude of the horizontal force P for which the cylinder will not move up or down. Given: M 6 kg
s
45 deg 0.2
g 9.81
m 2
s Solution: Total angle Forces
Answer
P min
P min
5 2
4
s
Mge
270.00 270.00 deg P max
15.9 N < P < P < < P max
s
Mge
217.4 7.4 N
Ans.
8–86.
A force of P = 25 N is just sufficient to prevent the 20-kg cylinder from descending. Determine the required force P to begin lifting the cylinder. The rope passes over a rough peg with two and half turns.
SOLUTION The coefficient of static friction ms between the rope and the peg when the cylinder is on the verge of descending requires T2 = 20(9.81) N, T1 = P = 25 N and b = 2.5(2p) = 5p rad. Thus, P
T2 = T1emsb
20(9.81) = 25ems(5p) In 7.848 = 5pms ms = 0.1312
In the case of the cylinder ascending T 2 = P and T 1 = 20(9.81) N. N. Using the result of ms, we can write T2 = T1emsb P = 20(9.81)e0.1312(5p)
= 1539.78 N = 1.54 kN
Ans.
8–87.
The 20-kg cylinder A and 50-kg cylinder B are connected together using a rope that passes around a rough peg two and a half turns. If the cylinders are on the verge of moving, determine the coefficient of static friction between the rope and the peg.
SOLUTION
75 mm
In this case, T 1 = 50(9.81)N 5 0(9.81)N,, T 2 = 20 20(9 (9.8 .81) 1)N N an and d b ‚ = 2. 2.5( 5(2 2p) = 5p ra rad.Th d.Thus us,,
A
T1 = T2emsb
50(9.81) = 20(9.81)ems(5p)
B
In 2 .5 = ms(5p) ms = 0.0583
Ans.
8–88
.
The uniform bar AB bar AB is supported by a rope that passes over a fri ctionless pulley at C and a fixed peg at D at D.. If the coefficient of static friction between the rope and the peg is D, determine the smallest distance x distance x from the end of the bar at which a force F may be placed and not cause the bar to move. Given: F
D
a
20 N
1m
0.3
Solution: Initial guesses: T A
T B
5N
x
10 N
10 m
Given
M A = 0;
F x T B a
F y = 0;
T A
T A
T A T B x x
T B F
T B e
Ans.
0
2
D
Find T A T B x
0.38 m
0
8–89.
The truck, truck, which has has a mass of 3.4 3.4 Mg, is to be lowered lowered down the slope by a rope that is wrapped wrapped around a tree. If the wheels are free to roll and the man at A can resist a pull of 300 N, N, determine the minimum number of turns the rope should be wrapped around the tree to lower the truck at a constant speed. The coefficient of kinetic friction between the tree and rope is mk = 0.3.
A
SOLUTION Q + © Fx = 0;
T2 - 33 354 sin 20° = 0
20
T2 = 11 407.7 T2 = T1 e mb
11 40 407. 7.7 7 = 300 e 0.3 b 12.127 1275 5 rad b = 12. Approx. 2 turns (695°)
Ans.
8–90.
The smooth beam is being hoisted using a rope which is wrapped around the beam and passes through a ring at A as shown. If the end of the rope is subjected to a tension T and the coefficient of static friction between the rope and ring is ms = 0.3, determine the angle of u for equilibrium.
T
A
θ
SOLUTION Equation of Equilibrium:
+ c © Fx = 0;
T - 2T ¿ cos
u
2
= 0
Frictional Force Force on Flat Belt: Here, b = T2 = T1 emb, we have T = T ¿ e0.3
T = 2T ¿ cos u
2
u
2
(1)
, T2 = T and T1 = T ¿ . Applying Eq. 8–6
1 >22 = T¿e0.15 u
u
(2)
1.73104 3104 rad = 99.2° u = 1.7
Ans.
Substituting Eq. (1) into (2) yields 2T ¿ cos
u
2
= T ¿ e0.15 u
e0.15 u = 2 co coss
u
2
Solving by trial and error
The other solution, which starts with T' = Te 0.3(0/2) based on cinching the ring tight, is 2.4326 rad = 139°. Any angle from 99.2° to 139 ° is equilibrium.
8–91
.
A cable is attached to the plate B of mass M B, passes over a fixed peg at C , and is attached to the block at A at A.. Using the coefficients of static friction shown, determine the smallest mass of block A block A so that it will prevent sliding motion of B of B down the plane. Given: M B
20 kg
30 deg
g 9.81
m
A
0.2
B
0.3
C
0.3
2
s Solution: Iniitial guesses:
T 1
1N
T 2
1N
N A
1N
Given Block A Block A::
F x = 0;
T 1
g sin A N A M A g sin
F y = 0;
N A
g cos M A g cos
F x = 0;
T 2
g sin B N M B g sin B A N A
F y = 0;
N B
g cos N A M B g cos
T 2
0
0
Plate B Plate B::
Peg C :
T 1 T 2 N A N B M A M A
C
T 1 e
Find T 1 T 2 N A N B M A
2.22 .22 kg
Ans.
0
0
N B
1N
M A
1 kg
.
.
8–94
.
Block A Block A has has mass m A and rests on surface B surface B for for which the coefficient of static friction is
sAB. If the
coefficient of static friction between the cord and the fixed peg at C is is sC , determine the greatest mass m D of the suspended cylinder D cylinder D without without causing motion. Given: m A
50 kg
sAB
sC
0.25
0.3
a
0.3 m
b
0.25 m
c
0.4 m
d 3 f 4 g 9.81
m 2
s Solu Soluti tion on::
Assu Assume me bloc block k A slip slipss but but doe doess not not tip tip..
The initial guesses:
Given
N B T m D x Since x
N B
m D g
f
T 50 N
100 N
sC
Te
T N 0 sAB B 2 2 f d
Find N B T m D x
51.6 51.6 mm <
b 2
N B T
f d
atan
m D
x
1 kg
10 mm
T m g N 0 A B 2 2 f d
T a 2 2 f d
d
f
413.05 N 129.08
125mm our assumption assumption is correct correct
b T N B x 0 2 2 2 f d d
m D
25.6 5.6 kg
x
0.05 .052 m
m D
25.6 kg
Ans.
8–95
.
Block A Block A rests rests on the surface for which the coefficient of friction is sAB. If the mass of the suspended cylinder is m D, determine the smallest mass m A of block A block A so so that it does not slip or tip. The coefficient of static friction between the cord and the fixed peg at C is is sC . Units Used: g 9.81
m 2
s Given:
sAB m D
4 kg
sC
0.25
0.3
a
0 .3 m
b
0.25 m
c
0.4 m
d 3 f 4 Solu Soluti tion on::
Assu Assume me that that slip slippi ping ng is the the crti crtitic tical al moti motion on
The initial guesses:
m D g
Given
N B T m A x Since x
N B
100 N
T 50 N
sC
Te
f
T N 0 sAB B 2 2 f d
Find N B T m A x
1 kg
m A
f d
atan
x
b 2
10 mm
T m g N 0 A B 2 2 f d d
T a 2 2 f d f
N B 64.63 N T 20.20
T b N x 0 B 2 2 2 f d d
m A x
51.6 51.6 mm <
125mm our assumption assumption is correct correct
m A
7.8 7.82 kg
0.05 .052 m
7.8 7.82 kg
Ans.
.
8–97.
Determine the smallest force P required to lift the 40-kg crate.The coefficient of static friction between the cable and each peg is ms = 0.1.
200 mm
200 mm
A
C
200 mm
SOLUTION
B
Since the crate is on the verge of ascending, T1 = 40(9.81) N and T2 = P . Fr From the geometry shown in Figs. a and b, the total angle the rope makes when in contact with 135° 90° the peg is b = 2 b 1 + b 2 = 2 Thus,, p + p = 2p rad. Thus 180° 180°
a
b a
b
T2 = T1ems b P = 40(9.81)e0.1(2p)
= 736 N
Ans.
P
8–98.
Show that the frictional relationship between the belt tensio ten sions ns,, the coe coeffi fficie cient nt of fri fricti ction on m , and the ang angular ular mb sin(a 2) contacts a and b for the V-belt is T2 = T1e .
> >
Impending motion
a b
SOLUTION
T2
FBD of a section of the belt is shown. Proceeding in the general manner:
© Fx = 0;
coss - (T + dT) co
du + 2
T cos
© Fy = 0;
sin n - (T + dT) si
du 2
T sin
Replace
sin
du du by , 2 2
cos
du by 1, 2
du + 2 dF = 0 2
du a + 2 dN sin = 0 2 2
dF = m dN
Using this and (dT)(du)
:
0, the above relations become
dT = 2m dN
a
T du = 2 dN sin
a
2
b
Combine dT du = m T sin a2
Integrate from u = 0, T = T1 to u = b , T = T2 we get,
¢ ≤ = T e mb
T2
1
sin a2
Q.E.D
T1
8–99.
If a force of P = 200 N is applied to the handle of the bell crank,, determi crank determine ne the maximum torque M that can be resisted so that the flywheel does not rotate clockwise.The coefficient of static friction between the brake band and the rim of the wheel is ms = 0.3.
P
900 mm
SOLUTION
400 mm
Referring to the free-body diagram of the bell crane shown in Fig. flywheel shown in Fig. b, we have have
a
C 100 mm
and the A
a + © MB = 0;
TA(0.3) + TC(0.1) - 200(1) = 0
(1)
a + © MO = 0;
TA(0.4) - TC(0.4) - M = 0
(2)
O
M
300 mm
By considering the friction between the brake band and the rim of the wheel where 270° b = p = 1.5 p rad and TA 7 TC, we can write 180° TA = TCemsb TA = TCe0.3(1.5p) TA = 4.1112 TC
(3)
Solving Solvin g Eqs. Eqs. (1), (2), and (3) (3) yields yields M = 187 N # m TA = 616.67 N
TC = 150.00 N
Ans.
B
8–100.
A 10-kg cylinder D, which is attached attached to a small pulley B, is placed on the cord cord as shown. Determine the largest angle u so that the cord does not slip over the peg at C .The .The cylinder at E also has a mass of 10 kg, and the coefficient of static friction between the cord and the peg is ms = 0.1.
A
u
C
B E
SOLUTION
D
Since pully B is smooth, the tension in the cord cord between pegs A and C remains constant. Referring to the free-body diagram of the joint B shown in Fig. a, we have have
+ c © Fy = 0;
u
2T sin u - 10(9.81) = 0
T =
49.05 sin u
In the case where cylinder E is on the verge of ascending, T2 = T = T1 = 10(9.81) N. Here,
p
2
49.05 and sin u
Thus,, + u, Fig. b. Thus
T2 = T1e msb
49.05 = 10(9.81) e 0.1 sin u ln
a
a2 + b
p 0.5 = 0.1 + u sin u 2
p
u
b
Solving by trial and error, error, yields u = 0.4221 rad = 24.2°
In the case where cylinder E is on the verge of descending, T 2 = 10(9.81) N and p 49.05 T 1 = . Here, + u. Thus, sin u 2 T2 = T1e m s b
10(9.81) =
49.05 0.1 e sin u
ln (2 sin u) = 0.1
a2 + b p
u
a2 + b p
u
Solving by trial and error, error, yields u = 0.6764 rad = 38.8°
Thus,, the range of u at which the wire does not slip over peg C is Thus 24.2° 6 u 6 38.8° umax = 38.8°
Ans.
8–101.
A V-belt V-belt is used to connect the hub A of the motor to wheel B. If the belt can withstand a maximum tension of 1200 N, determine the largest mass of cylinder C that can be lifted and the corresponding torque M that must be supplied to A. The coefficient of static friction between the hub and the belt is between the wheel and the the belt is ms ¿ = 0.20. ms = 0.3, and between Hint: See Prob. Prob. 8–98.
a
60
200 mm 150 mm
C
In this case, case, the maximum tension in the belt is T 2 = 1200 N. N. Referring to the free A, a B body diagram of hub shown in Fig. and the wheel shown in Fig. b, we have have M + T1 (0.15) - 1200(0.15) = 0 M = 0.15(1200 - T1)
a + © MO¿ = 0;
B
M
SOLUTION
a + © MO = 0;
15 A
300 mm
(1)
1200(0.3) - T1(0.3) - MC (9.81)(0.2) = 0
1200 - T1 = 6.54MC If hub A is on the verge of slipping, slipping, then
> >
T2 = T1emsb1 sin(a 2) where b 1 =
+ 75° a 90°180° b
(2)
p = 0.9167p rad
>
1200 = T1e0.3(0.9167p) sin 30° T1 = 213.19 N
Substituting T 1 = 213.19 213.19 N into into Eq. (2), yields MC = 150.89 kg
If wheel B is on the verge of slipping, then
> >
T2 = T1ems¿ b1 sin(a 2) where b 2 =
a 180°180°+ 15° b
p = 1.0833p rad
>
1200 = T1e0.2(1.0833p) sin 30° T1 = 307.57 N
Substituting T 1 = 307.57 307.57 N into into Eq. (2), yields MC = 136.45 kg = 136 kg (controls!)
Ans.
Substituting T 1 = 307.57 N into Eq. Eq. (1), we obtain obtain M = 0.15(1200 - 307.57)
= 134 N # m
Ans.
8–102.
The 20-kg motor has a center of gravity at G and is pinconnected at C to maintai maintain n a tension tension in the the drive drive belt. Determine the smallest counterclockwise twist or torque M that must be supplied by the motor to turn the disk B if wheel A locks and causes the belt to slip over the disk. No slipping occurs at A. The coeffi coefficient cient of of static fricti friction on between the belt and the disk is ms = 0.35.
M
A
B
50 mm
G
50 mm C
100 mm
SOLUTION Equations of Equilibrium: From FBD (a),
a + © MC = 0;
1 2
1 2
1 2
T2 100 + T1 200 - 196.2 100 = 0
(1)
From FBD (b), a + © MO = 0;
1 2
1 2
M + T1 0.05 - T2 0.05 = 0
(2)
Frictional Force on Flat Belt: Here, b = 180° = p rad. Applying Eq. 8– 8 –6, mb T2 = T1 e , we have T2 = T1 e0.3p = 3.003T1
(3)
Solving Eqs. (1), (2), and (3) yields M = 3.93 N # m
39.22 22 N T1 = 39.
T2 = 117.8 N
Ans.
150 mm
8–103.
Blocks A and B have a mass of 100 kg and 150 kg, respectively.. If the coefficient of static friction between A respectively and B and between B and C is ms = 0.25 and between the ropes and the pegs D and E m ¿ s = 0.5 determine the smallest force F needed to cause motion of block B if P = 30 N.
E
D A
45 B
C
SOLUTION
P
Assume no slipping between A and B. Peg D : T2 = T1 emb;
p
FAD = 30 e0.5( 2 ) = 65.80 N
Block B :
+ © Fx = 0;
:
- 65.80 - 0.25 NBC + FBE cos 45° = 0
+ c © Fy = 0;
NBC - 981 + FBE sin 45° - 150 (9.81) = 0 FBE = 768.1 N NBC = 1909.4 N
Peg E : T2 = T1emb;
3p
F = 768.1e0.5( 4 ) = 2.49 kN
Note: Since B moves to the right,
(FAB)max = 0.25 (981) = 245.25 N p
245.25 = Pmax e0.5( 2 ) Pmax = 112 N 7 30 N
Hence, no slipping occurs occurs between A and B as originally assumed.
Ans.
F
8–104.
Determine the minimum coefficient of static friction m s between the cable and the peg and the placement d of the 3-kN force for the uniform 100-kg beam to maintain equilibrium.
B
3 kN
d
SOLUTION A
Referring to the free-body diagram of the beam shown in Fig. a, we have have
+ © Fx = 0;
:
TAB cos 45° - TBC cos 60° = 0
+ c © Fy = 0;
TAB sin 45° + TBC sin 60° - 3 -
a + © MA = 0;
TBC sin 60°(6) -
6m
100(9.81) = 0 1000
100(9.81) (3) - 3d = 0 1000
Solving, d = 4.07 m
Ans.
TBC = 2.914 kN
TAB = 2.061 kN
Using the results for T BC and T AB and considering the friction between the cable 45° + 60° and the peg, where b = p = 0.5833p rad, we have 180°
c a
bd
TBC = TAB emsb
2.914 = 2.061 ems(0.5833p) ln 1.414 = ms(0.5833p) ms = 0.189
60
45
Ans.
C
8–105.
A convey conveyer er belt is used to transfer transfer granular material material and the frictional resistance on the top of the belt is F = 50 500 0 N. Determine the smallest stretch of the spring attached to the moveable axle of the idle pulley B so that the belt does not slip at the drive pulley A when the torque M is applied. What minimum torque M is req requir uired ed to keep keep the the belt belt moving? The coefficient of static friction between the belt and the wheel at A is ms = 0.2.
0.1 m A
SOLUTION Frictional Force on Flat Belt: Here, b = 180° = p rad and T2 = 500 + T and Applying ng Eq. Eq. 8–6, we have have T1 = T. Applyi T2 = T1 emb
500 + T = Te0.2p 571.7 .78 8N T = 571 Equations of Equilibrium: From FBD (a),
1 2 1
a + © MO = 0;
21 2
M + 571.78 0.1 - 500 + 578.1 0.1 = 0 M = 50 50.0 .0 N # m
Ans.
From FBD (b),
+ © F = 0; x
:
1
2
Fsp - 2 578.71 = 0
Fsp = 1143.57 N
Thus,, the spring stretch is Thus x =
Fsp k
=
1143.57 286 6 mm = 0.2859 m = 28 4000
M
Ans.
F = = 500 N
0.1 m B k = = 4 kN/m
8–106.
The belt on the portable dryer wraps around the drum D, idler pulley A, and motor pulley pulley B. If the motor can develop develop # a maxi maximum mum tor torque que of M = 0.80 N m, det determi ermine ne the smallest spring tension required to hold the belt from slipping. The coefficient of static friction between the belt and the drum and motor pulley is ms = 0.3. Ignore the size of the idler pulley A.
A
SOLUTION
B
1 2
1 2
a + © MB = 0;
- T1 0.02 + T2 0.02 - 0.8 = 0
T2 = T1 emb;
T2 = T1 e
30°
50 mm
= 0.8 N⋅m M =
45° C
50 mm
20 mm
10.321 2 = 2.5663T1 p
T1 = 25.537 N T2 = 65.53 N
a + © MC = 0; Fs = 85.4 N
1 2 1
21
2
1
2
- Fs 0.05 + 25.537 + 25.537 sin 30° 0.1 cos 45° + 25.537 cos 30° 0.1 sin 45° = 0 Ans.
D
P = 2000 2000 N.
50 mm 75 mm
R 2 = 37.5 mm , R1 = 25 mm , µ 3 = 0.3 P = 2000 N, N, we have
=
2 3
37.5 (0.3)(2000) 37.5
3
3
25 2
2
25
=19 000 000 N mm = 19 N m
Ans.
P =
M
3.6 N·m
2000 N
= 3.6 (1000) = 3600 N·mm and
P
R 2 = 37.5 mm, R 1 = 25 mm, = 2000 N, we have
50 mm
2 3600 =
μ k
3
37.5 )(2000) 37.5
3
(
= 0.0568
k
2
3
25
2
25
75 mm
Ans.
4000 N. If µ s = 0.35, 0.35,
2 18.75 mm
=
50 mm
3
3 3 50 25 2 2 50 25
(0.35)(4000)
P = = 4000 N 25 mm
= 54 44 444 N mm
M
= 54 54.4 N m
Ans.
8–110.
The shaft is supported by a thrust bearing A and a journal bearing B. Determine the torque M required to rotate the shaft at constant angular velocity.The coefficient of kinetic friction at the thrust bearing is mk = 0.2. Neglect friction at B.
M A
a B
a
P
SOLUTION 75 mm
150 mm
0.075 0.15 Applying Eq. 8–7 with R1 = = 0.0375 m, R2 = = 0.075 m ms = 0.2 2 2 and P = 4000 N, we have ,
Section a-a
¢
R23 - R13 2 M = msP 3 R22 - R12
=
≤
2 0.0753 - 0.03753 0.2 4000) 3 0.0752 - 0.03752
1 21
a
= 46.7 N # m
b Ans.
4
kN
8–111.
The thrust bearing supports an axial load of P = 6 kN. kN. If a # M torque of = 150 N m is required to rotate the shaft, determine the coefficient of static friction at the constant surface.
200 mm
100 mm
SOLUTION Applying Appl ying Eq. 8–7 with R1 =
0.1 m 0.2 m = 0.05 m, R2 = = 0.1 m M = 150 N # m 2 2
M
,
and P = 6000 N, we have
¢
R 23 - R 13 2 M = msP 3 R2 2 - R 1 2
150 =
≤
2 0.13 - 0.053 ms 6000) 3 0.12 - 0.052
1
a
ms = 0.321
P
b Ans.
8–112.
Assuming that the variation of pressure at the bottom of the pivot bearing is defined as p = p0 R2 r , determ determine ine the the torque M needed to overcome friction if the shaft is subjected to an axial force P. The coeff coefficient icient of static static m friction fricti on is s. For the solutio solution, n, it is neces necessary sary to determ determine ine p0 in ter terms ms of P and the bearing dimensions R1 and R2 .
P
1 >2
M
R2
SOLUTION © Fz = 0;
R1 2p
L L = L L a b
P =
L
R2
dN =
0
A
2p
R2
p0
0
R1
pr dr du
R1
R2 r
r dr du p0
= 2p p0 R2 (R2 - R1) Thus, p0 =
p
P
C 2pR2 (R2
- R1) D 2p
© Mz = 0;
R2
L L L = L L a b
M =
ms pr 2 dr du
r dF =
0
A
2p
0
R1
R2
ms p0
R1
= ms (2p p0) R2
R2 r
r 2 dr du
1 A R22 - R21 B 2
Using Eq. (1): M =
r
1 m s P (R2 + R1) 2
Ans.
p0 R2 r
8–113.
The plate clutch consists of a flat plate A that slides over the rotating shaft S. The shaft is fixed to the driving plate gear B. If the gear C , which is in mesh with B, is subjected to a torque of M = 0.8N # m, determine the smallest force P , that must be applied via the control arm, to stop the rotation. The coefficient of static friction between the plates A and D is ms = 0.4. Assume the bearing pressure between A and D to be uniform.
D A F
100 mm S
125 mm P
SOLUTION
200 mm
150 mm E
0.8 F = = 26.667 N 0.03
15 0 m m
B
30 mm
M = 26.667(0.150) = 4. 4.00 00 N # m R32 - R31 2 M = m P ¿ 3 R22 - R21
a
4.00 =
C
b
(0.125)3 - (0.1)3 2 (0.4) (P ¿ ) 3 (0.125)2 - (0.1)2
a
M
b
P ¿ = 88.525 N
a + © MF = 0; P = 118 N
88.525(0.2) - P(0.15) = 0 Ans.
0. 8 N
m
8–114.
The conical bearing is subjected to a constant pressure distribution at its surface of contact. contact. If the coefficient of static friction is ms, determine the torque M required to overcome friction if the shaft supports an axial force P.
P M
R
SOLUTION The differential area (shaded) dA = 2pr
P =
L
p cos u dA =
P = ppR2
dN = pdA =
M =
L
rdF =
p = P pR
2
L
L
p cos u
¢ ≤ ¢ ≤ dr cos u
=
2prdr cos u
2prdr cos u
= 2pp
L
u
R
rdr
0
P pR 2
¢
2prdr cos u
≤
ms rdN =
=
=
2P rdr R cos u 2
2ms P R 2 cos u
L
R
r 2 dr
0
2ms P R 3 2ms PR = 2 3 co coss u R cos u 3
Ans.
8–115.
The pivot bearing is subjected to a pressure distribution at its surface surface of contac contactt which varies as as shown. shown. If the coefficient of static friction is m, determine the torque M required to overcome friction if the shaft supports an axial force P.
P
M
R
SOLUTION
a2 b = L cos a 2 b = L a cos a 2 b b L pr
dF = m dN = m p0 cos
M
pr
rm p0
R
R
0
dr
BA
2r
pr
cos
p 2 2R
R3
p
p0 2p
du
0
A a b + 2 B ¢ 16 ≤ c a 2 b - 2 d
= m p0 = mp0
pr
r2
dA
r dr du
R
A
m p0
R
r
R
B r2 A p2R B 3
p 2 2R
2
sin
pr
R
R
0
a 2 b d 12 2 p
2
p2
= 0.7577m p0 R3 R
2p
L a cos a 2 b b L 1 cos a b + A B sin a 2 b R 12 2 = B 2 A B a1 - 2 b = 4
P =
L
dN =
0
A
p0
pr
p0
R
pr
R
p 2 2R
p0 R 2
r
p 2R
rdr
du
0
pr
R
R
0
p
p
= 1.454p0 R2 Thus,
M = 0.521 P mR
Ans.
p = p0 cos
π r
2 R
8–116.
A 200-mm diameter post is driven 3 m into sand for which ms = 0.3. If the normal pressure pressure acting acting completely around the post varies linearly linearly with depth as shown, determine the frictional torque M that must be overcome to rotate the post.
M
200 mm
3m
SOLUTION Equations of Equilibrium and Friction: The resultant normal force on the post is 1 600 + 0 3 p 0.2 = 180p N. Since the post is on the verge of rotating, N = 2 F = ms N = 0.3 180p = 54.0p N.
1
a + © MO = 0;
21 21 21 2 1 2
1 2
M - 54.0p 0.1 = 0 M = 17 17.0 .0 N # m
Ans.
600 Pa
8–117.
A beam having a uniform weight W rests on the rough horizontal surface having a coefficient of static friction ms . If the horizontal force P is applied perpendicular to the beam’s length, length, determi determine ne the location location d of the point O about which the beam begins to rotate.
O d 1 3 L
2 3 L
P
SOLUTION w =
msN
L
© Fz = 0;
N = W
© Fx = 0;
P +
© MOz = 0; msW(L - d)2
2L
ms Nd
L
-
ms N(L - d)2
2L
+
msWd2
2L
-
ms N(L - d)
L
+
a 23
L
ms Nd2
2L
- d
ba
= 0
- P
a 23
L
- d
ms W(L - d)
L
b=0
-
ms Wd
L
b=0
3(L - d)2 + 3d2 - 2(2L - 3d)(L - 2d) = 0 6d2 - 8Ld + L2 = 0 Choose the root 6 L. d = 0.140 L
Ans.
a 20-mm-di 20-mm-diameter ameter pin at B and and to the the cran crank k shaf shaftt by a 50-mm50-mm-dia diamet meter er bearin bearing g A. If the piston is moving
= =
50(0.2) 2 20(0.2) 2
= 5mm
Ans.
= 2mm
Ans.
. .
8–120
.
A pulley of mass M has has radius a and the axle has a diameter D. D. If the coefficient of kinetic friction between the axle and the pulley is k determine the vertical force P on the rope required to lift the block of mass M B at constant velocity. Given:
a
120 mm
M 5 kg D
k
40 mm
0.15
M B
80 kg
Solution: k r f
atan k
D 2 sin k
M p = 0; M B g a
P
r f M g r f P a r f
M B g a
r f M g r f a r f
0
P 826 N
Ans.
8–121
.
D. If the coefficient of kinetic A pulley of mass M has has radius a and the axle has a diameter D. friction between the axle and the pulley is k determine the force P on the rope required to lift the block of mass M B at constant velocity. velocity. Apply the force P horizontally horizontally to the right (not as shown in the figure). Given: a
120 mm
M 5 kg
D
k
40 mm
0.15
M B
80 kg g 9.81
Solution: k r f
2
2
s
atan k
D
m
sin k
Guesses P
1N
R
1N
1 deg
Given R cos P
M B g M g
R sin
M B g a
P R
0
0
P a R r f
Find P R
0
P 814 N
Ans.
8–122
.
The collar fits loosely around a fixed shaft that has radius r. If the coefficient of kinetic friction between the shaft and the collar is k , determine the force P on the horizontal segment of the belt so that the collar rotates c ounterclockw ounterclockwise ise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt is R.
Given: r 50mm k 0.3 R 56.25 mm F 100N
k
k 16.699 deg
k
rf 14.3674 mm
k atan
Solution:
rf r sin Equilibrium :
Fy = 0;
Fx = 0;
R y F
=
0
R y F
P R x
=
0
R x
R
=
2
R x R y
2
P 1N
Given
P F rf F R P R
2
P 2
P F
=
Guess
2
=
R y 100.00 N
=
0
2
P Find ( P)
. .97 N P 68
Ans.
8–123
.
The collar fits loosely around a fixed shaft that has radius r. If the coefficient of kinetic friction between the shaft and the collar is k , determine the force P on the horizontal segment of the belt so that the collar rotates c lockwise with a constant angular velocity. Assume that the belt does not slip on t he collar; rather, rather, the collar slips sl ips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt is R.
Given: r 50mm k 0.3 R 56.25mm F 100N
atan k
Solution:
rf r sin
k
k 16.699 deg
k
rf 14.367 mm
Equilibrium :
Fy = 0;
Fx = 0;
R y F
=
0
R y F
P R x
=
0
R x
R
Guess Given
=
2
R x R y
2
=
R y 100.00 N
P 2
P F
=
2
P 1N 2
2
P F rf F R P R
=
0
P Find ( P)
P 145.0 N
Ans.
8–124.
A pulley having a diameter of 80 mm and mass of 1.25 1.25 kg is supported loosely on a shaft having a diameter of 20 mm. Determine the torque M that must be applied to the pulley to cause it to rotate with constant motion.The coefficient of kinetic friction between the shaft and pulley is mk = 0.4. Also calculate the angle u which the normal force at the point of contact makes with the horizontal. The shaft itself cannot rotate.
M
40 mm
SOLUTION Frictional Force on Journal Bearing: Here, fk = tan-1 mk = tan-10.4 = 21.80°. Then the radius of friction circle is r f = r sin fk = 0.0 0.01 1 sin 21.80° 21.80° = 3.714 10 -3 m. The angle which the normal force makes with horizontal is
1 2
u = 90° - fk = 68.2°
Ans.
Equations of Equilibrium:
+ c © Fy = 0;
R - 12.2625 = 0
a + © MO = 0;
12.2625 3.714 10-3 - M = 0
R = 12. 12.262 2625 5N
1 21 2
M = 0.0455 N # m
Ans.
8–125.
The 5-kg 5-kg skateboard skateboard rolls down the 5° slope at constant constant speed. If the coefficient coefficient of of kinetic friction friction between between the 12.5 mm diameter axles and the wheels is mk = 0.3, determine the radius of the wheels. wheels. Neglect rolling resistance of the wheels on the surface. surface. The center of mass for the skateboard is at G.
75 mm G
5 250 mm
SOLUTION Referring to the free-body diagram of the skateboard shown in Fig. a, we have have
© Fx¿ = 0;
Fs - 5(9.81) sin 5° = 0
Fs = 4.275 N
© Fy¿ = 0;
N - 5(9.81) cos 5° = 0
N = 48.86 N
The effect of the forces acting on the wheels can be represented as if these forces are acting on a single wheel as indicated on the free-body diagram shown in Fig. b. We have
© Fx¿ = 0;
Rx¿ - 4.275 = 0
Rx¿ = 4.275 N
© Fy¿ = 0;
48.86 - Ry¿ = 0
Ry¿ = 48.86 N
Thus, Thu s, the magnitude magnitude of R is R =
2 R
x¿
2
+ Ry ¿ 2 =
2 4.275 4.275
2
+ 48.862 = 49.05 N
fs = tan-1 ms = tan-1(0.3) = 16.699°. Thu Thus, s, the moment arm of R from point O is
(6.25 sin 16.699 16.699°) mm. Using these these results and and writing the moment moment equation equation about O b point , Fi Fig. g. , we have have a + © MO = 0;
4.275(r) - 49.05(6.25 sin 16.699° = 0) r = 20.6 mm
Ans.
300 mm
Determine the force P required to overcome rolling resistance and pull the 50-kg roller up the inclined plane with constant velocity. velocity. The coefficient of rolling resistance is a = 15 mm . 8–126.
P
300 mm
30
30
Determine the force P required to overcome rolling resistance and support the 50-kg roller if it rolls down the inclined inclined plane plane with constant constant velocity. velocity. The coefficient of rolling resistance is a = 15 mm . 8–127.
P
300 mm
30
30
8–128
.
The lawn roller has mass M . If the arm BA is held at angle from from the horizonta horizontall and the coefficien coefficientt of rolling rolling resistanc resistancee for the roller is r , determine the force P needed needed to push roller roller at consta constant nt speed speed.. Neglec Neglectt frictio friction n develo developed ped at the axle, axle, A, and assume that the P resultant force acting on the handle is applied along arm BA . Given: M 80 kg
30 deg
a 250 mm r 25 mm
Solution: 1
asin
1
5.74 deg 5.74
r
a
M 0 = 0; r M g P sin r P cos a cos 1 0
P
r M g
sin r cos a cos 1
P 96.7 N
Ans.
8–129
.
A machine of mass M is is to be moved moved over over a level level surface using a series series of rollers rollers for which which the coefficie coefficient nt of rolling rolling resistance resistance is a g at the the grou ground nd and and am at the bottom surface of the machine. Determine the appropriate diameter of the rollers so that the machine can be pushed forward with a horizontal force P. Hint: Use the result of Prob. 8-131. Units Used: Mg 1000 kg Given: M 1.4 Mg a g 0.5 mm am 0.2 mm P 250 N Solution: P
M g a g am 2 r
a g am 2 P
r M g
r 19.2 19.2 mm
d 2 r
d 38.5 38.5 mm
Ans.
8–130.
The hand cart has wheels with a diameter of 80 mm. If a crate having a mass of 500 kg is placed on the cart so that each wheel carries an equal load, determine the horizontal force P that must be applied to the handle to overcome the rolling resistance. The coefficient of rolling resistance is 2 mm. Neglec Neglectt the mass mass of the cart. cart.
P
SOLUTION P L
Wa r
1 2 a 402 b
= 500 9.81 P = 245 N
Ans.
8–131.
The cylinder is subjected to a load that has a weight W . If the coeff coefficients icients of rolling rolling resistance resistance for the cylind cylinder’s er’s top and bottom surfac surfaces es are aA and aB, respec respectively tively,, show that a horizontal force having a magnitude of P = [W(aA + aB)] 2r is required to move the load and thereby roll the cylinder forward. Neglect the weight of the cylinder.
W P
>
A
r
SOLUTION
B
+ © F = 0; x
(RA)x - P = 0
(RA)x = P
+ c © Fy = 0;
(RA)y - W = 0
(RA)y = W
:
a + © MB = 0;
P(r cos fA + r cos fB) - W(a A + aB) = 0
(1)
Since fA and fB are very small, cos fA - cos fB = 1 . Hence, from Eq. (1) P =
W(aA + aB)
2r
(QED)
8–132.
A large crate crate having a mass of 200 kg is moved moved along the floor using a series of 150-mm-diameter rollers for which the coefficient of rolling resistance is 3 mm at the ground and 7 mm at the bottom surface of the crate. Determine the horizontal force P needed to push the crate forward at a constant speed. Hint: Use the result result of Prob. 8–131.
P
SOLUTION Rolling Resistance: Applying the result obtained in Prob. 8–131. P =
1
W aA + aB
2,
2r with a A = 7 mm, aB = 3 mm mm,, W = 200 9.81 = 19 1962 62 N, and r = 75 mm mm,, we have
1 2
P =
1962 7 + 3 2 75
130.8 .8 N = 131 N = 130
Ans.
Each of the cylinde cylinders rs has a mass of 50 kg. If the coefficients of static friction at the points of contact are mA = 0.5, mB = 0.5, mC = 0.5, and mD = 0.6, determine the smallest couple moment M needed to rotate cylinder E. 8–133.
D
A 300 mm
B
.
M
E
300 mm
C
8–134. The clamp is used to tighten the connection between two concrete concrete drain pipes. pipes. Determine the least coefficient of static friction at A or B so that the clamp does not slip regardless of the force in the shaft CD.
100 mm
A
250 mm
C
D B
8–135.
If P = 900 N is applied to the handle of the bell crank, determine the maximum torque M the cone clutch can transmit. The coefficient of static static friction at the contacting surface is ms = 0.3.
15
250 mm
300 mm C
M
200 mm
B
SOLUTION Referring to the free-body diagram of the bellcrank shown in Fig. a ,we have a + ©MB = 0;
900(0.375) - FC(0.2) = 0
+ © F = 0; x
2
¢
N sin 15° 2
≤
- 1687.5 = 0
N = 652 6520.0 0.00 0N
The area of the differential element shown shaded in Fig. c is dr 2p Thus,, dA = 2pr ds = 2pr = r dr. Thus sin 15° sin 15° A =
L
dA =
A
surface is p =
L
0.15 0. 15 m
0.125 m
2p 0.0834 8345 5 m2. The pressure acting on the cone r dr = 0.0 sin 15°
N 6520.00 = = 78.13(103) N A 0.08345
>m
2
The normal force acting on the differential element d A is 2p dN = p dA = 78.13(103) r dr = 1896.73(103)r dr. sin 15° Thus, Thu s, the frictional frictional force force acting on this different differential ial element is given by 3 3 dF = msdN = 0.3(1896.73)(10 )r dr = 569.02(10 )r dr. Th The e mom momen entt equ equat ation ion about the axle of the cone clutch gives
c
© M = 0; M -
L
d
rdF = 0
M =
L
rdF = 569.02(103)
M = 270 N # m
L
375 mm P
FC = 1687.5 N
Using this result result and referring to the free-body diagram of the cone clutch shown in Fig. b, :
A
0.15 m
r2 dr
0.125 m
Ans.
8–136.
The lawn roller has a mass of 80 kg. If the arm BA is held at an angle of 30° from the horizontal and the coefficient of rolling resistance resist ance for the roller roller is 30 mm, determ determine ine the force P needed to push the roller at constant constant speed. Neglect friction A, and assume that the resultant force P developed at the axle, A acting on the handle is applied along arm BA.
P
B
250 mm A
SOLUTION u = sin - 1
30 a 250 b = 6.892°
a + © MO = 0;
- 30(784.8) - P sin 30°(30) + P cos 30°(250 cos 6.892°) = 0
Solving, P = 11 117 7.8 N
30mm
Ans.
30
.
8–138.
The uniform 60-kg crate C rests uniformly on a 10-kg dolly D. If the front casters of the dolly at A are locked to prevent rolling while the casters at B are free to roll, determ determine ine the maximum force P that may be applied without causing motion of the crate.The coefficient of static friction between the casters and the floor is m f = 0.35 and between the dolly and the crate, md = 0.5.
0.6 m
P
C
1.5 m
0.8 m
SOLUTION Equations of Equilibrium: From FBD (a),
D
+ c © Fy = 0; + © F = 0; x
:
a + © MA = 0;
Nd - 588.6 = 0
Nd = 58 588. 8.6 6N
0.25 m
P - Fd = 0
+ © F = 0; x
:
a + © MB = 0;
12 1 2
588.6 x - P 0.8 = 0
(2)
NB + NA - 588.6 - 98.1 = 0
(3)
P - FA = 0
(4)
1 2 1 2 - 588.610.952 - 98.110.752 = 0
NA 1.5 - P 1.05
(5)
1 2
Friction: Assum Assuming ing the the crate crate slips on dolly dolly,, then Fd = ms dNd = 0.5 588.6 294.3 .3 N. Substituting this value into Eqs. (1) and (2) and solving, we have = 294
294. 4.3 3N P = 29
0.400 400 m x = 0.
Since x 7 0. 0.3 3 m, the crate tips on the dolly dolly.. If this is the case x = 0. 0.3 3 m. Solving Eqs. (1) and (2) with x = 0. 0.3 3 m yields P = 220.725 N Fd = 220.725 N
Assuming the dolly slips at A, then FA = ms fNA = 0.35NA . Substituting this value into Eqs. Eqs. (3), (4), and (5) and solving, solving, we have NA = 559N
A
(1)
From FBD (b),
+ c © Fy = 0
B
NB = 128 N
P = 19 195. 5.6 6 N = 196 N (Control!)
Ans.
0.25 m 1.5 m
3-Mg four-whee four-wheel-drive l-drive truck truck (SUV) (SUV) has a 8–139. The 3-Mg center of mass at G. Determ Determine ine the maximum maximum mass of the log that can be towed by the truck. The coefficient of static friction between the log and the ground is ms = 0.8, and the coefficient of static friction between the wheels of the truck and the ground is mœs = 0.4. Assume that the engine of the truck is powerful enough to generate a torque that will cause all the wheels to slip.
G B
0.5 m
Ans.
A
1.6 m
1.2 m
8–140. A 3-Mg front-wheel-drive truck (SUV) has a center
of mass at G. Determine the maximum mass of the log that that can be towed by the truck. The coefficient coefficient of static friction between betw een the log and the groun ground d is ms = 0.8, and the coefficient of static friction between the front wheels of the truck and the ground is mœs = 0.4. The rear wheels are free to roll. Assume that the engine of the truck is powerful enough enough to generate a torque that will cause the front wheels to slip.
G B
0.5 m
Ans.
A
1.6 m
1.2 m
A roofer, having a mass of 70 kg, kg, walks slowly slowly in an upright position down along the surface of a dome that has a radius of curvature of r 20 m. If the coefficient of static friction between his shoes and the dome is ms 0.7, determine the angle u at which he first begins to slip. slip. 8–141.
=
=
u
20 m
60
.
8–142.
Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is ms = 0.25.
P
30
SOLUTION Free-Bo dy Diagr Free-Body D iagram: am: When the crate is on the verge of sliding down the plane, the frictional force F will act up the plane plane as indicated indicated on the free-body diagram of the crate shown in Fig. a. Equations of Equilibrium: a +©Fy¿ =
0;
N
Q +©Fx¿ =
0;
P cos 30°
-
P sin 30° +
-
50(9.81) cos 30°
0.25N
-
=
0
50(9.81) sin 30°
=
0
Solving P
=
140 14 0N
N
=
494. 49 4.94 94 N
Ans.
8–143.
Determine the minimum force P required to push the crate up the plane plane.. Th The e crate crate has has a mass mass of 50 kg kg and the the coefficient of static friction between the crate and the plane is ms = 0.25.
P
30
SOLUTION When the crate is on the verge of sliding up up the plane, the frictional force F ¿ will act down the plane as indicated on the free-body diagram of the crate shown in Fig.b. a +© Fy¿ = Q +© Fx¿ =
0; N ¿
-
P sin 30°
0; P cos 30°
-
50(9.81) cos 30°
-
0.25N ¿
-
=
0
50(9.81) sin 30°
=
0
Solving, P
=
N¿
474 47 4N
=
661.9 661 .92 2N
Ans.
8–144.
A horizo horizontal ntal force of P = 100 N is just suffic sufficient ient to hold the crate from sliding sliding down the plane plane,, and a horizo horizontal ntal force of P = 350 N is required to just push the crate up the plane. Determine the coefficient of static friction friction between the plane and the crate, and find the mass of the crate.
P
30
SOLUTION Free-Bo dy Diagram: Free-Body Diagr am: When the crate is subjected to a force of P = 10 100 0 N, it is on the verge of slipping down the plane.Thus, the frictional force F will act up the plane as indica indicated ted on the the free-body free-body diagram of the crate shown in Fig. Fig. a. When P = 350N , it will cause the crate to be on the verge of slipping up the plane, and so the frictional force force F ¿ acts down the plane as indicated on the free - body diagram of the crate shown in Fig. b. Thus, F = msN and F ¿ = msN ¿ . Equations of Equilibrium: + a © Fy¿ = + Q© Fx¿ =
0; N
-
0; msN
100 sin 30° +
m(9.81) cos 30°
-
100 cos 30°
-
=
m(9.81) sin 30°
0 =
0
Eliminating N , ms =
4.905m - 86.603 8.496m + 50
(1)
Also by referring to Fig, b, we can can write write + a© Fy¿ =
0; N ¿
+ Q© Fx¿ =
0; 35 350 0 cos 30 30°°
-
m(9.81) cos 30° -
-
350 sin 30°
m(9.81) sin 30°
-
=
0
msN ¿ = 0
Eliminating N ¿ , ms =
303.11 - 4.905m 175 + 8.496m
(2)
Solving Eqs. (1) and (2) yields m
=
36.5 kg
ms = 0.256
Ans. Ans.
