Problem 6-101
If a force of magnitude P is applied perpendicular to the handle of the mechanism, determine determine F the magnitude of force for equilibrium. The members are pin-connected at A, B, C, and D. Given: P := 30N a := 625mm b := 100mm c := 125mm d := 100mm e := 125mm f := 125mm g := 750mm Solution:
M A = 0;
FBC⋅ b − P ⋅ a
=
0
P⋅ a
FBC :=
b
FBC = 187.5 N
+
→
Fx = 0;
−Ax + P
0
=
Ax := P
+
↑
Fy = 0;
Ax = 30 N
−Ay + FBC Ay := FBC
M D = 0;
=
0 Ay = 187.5 N
−e⋅ Ax − Ay⋅ ( b + c) + ( g + b + c) ⋅ F
F :=
e⋅ Ax + Ay⋅ ( b + c) g+b+c
=
0
F = 47.1 N
Problem 6-102
The pillar crane is subjected to the load having a mass M . Determine the force developed in the tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tied in the position position shown. Units Used:
kN
3
:= 10 N
Given: M := 500kg a
:= 1.8m
b
:= 2.4m 1
:= 10deg
2
:= 20deg
g
= 9.81
m 2
s Solution:
initial guesses: F CB := 10kN
F AB := 10kN
Given
−M
⋅ g⋅ cos ( 1) − FAB⋅ cos ( 2) + FCB⋅ 2
b 2
⋅ g⋅ sin ( 1) − FAB⋅ sin ( 2) + FCB⋅ 2
⎛ FAB ⎞ ⎝ FCB ⎠
:= Find( F AB , FCB)
⎛ Cx ⎞ ⎝ Cy ⎠
:=
+b
a
−M
=
a 2
a
0
2
− M⋅ g = 0
+b
2
FCB 2
a
+b
⋅ 2
⎛ b ⎞ ⎝ a ⎠
⎛ FAB ⎞ ⎛ 9.7 ⎞ ⎜ ⎟ C ⎜ x ⎟ = ⎜ 11.53 ⎟ kN ⎜ C ⎟ ⎝ 8.65 ⎠ ⎝ y ⎠
Problem 6-103
The tower truss has a weight W and and a center of gravity at G. The rope system is used to hoist it into the vertical position. If rop rop e CB is attached to the top of the shear leg AC and a second rope CD is attached to the truss, determine the required tension in BC to hold the truss in t he position shown. The base of the truss and t he shear shear leg bears against the stake at A , which can be considered as a pin. Also, compute the compressive force acting along the shear leg. Given: W
:= 2300N
:= 40deg a
:= 1.5 m
b
:= 0.9 m
c := 3m d
:= 1.2 m
e := 2.4m Solution: Entire system: TBC⋅ cos ( ) ⋅ ( d TBC := TBC CA
ΣMA = 0;
+ e) − TBC⋅ sin ( ) ⋅ a − W⋅ ( a + b)
=
0
W⋅ ( a
+ b) cos ( ) ( d + e) − sin ( ) ⋅ a
= 3.08 kN
is a tw two-forc e member.
:= atan initial guesses:
At
C :
⎛ e ⎞ ⎝ b + c ⎠ FCA := 500N
TCD := 300N
Given
ΣFx = 0;
FCA⋅
a 2
a
ΣFy = 0;
FCA⋅
d 2
a
⎛ FCA ⎞ ⎝ TCD ⎠
+ ( d + e)
2
+e
+ ( d + e)
:= Find(F CA , TCD)
2
+ TCD⋅ cos ( ) − TBC⋅ cos ( )
− TCD⋅ sin ( ) − TBC⋅ sin ( )
TCD
= 1.43 kN
=
=
FCA
0
0
= 2.96 kN
Problem 6-104
The constant momen momen t M is is applied to t he crank shaft. Determine the compressive force force P that is exerted on the piston for equilibrium as a function of θ . Plot the results of P (ordinate) versus ≤ 90deg . θ (abscissa) for 0deg ≤ Given: a
:= 0.2 m
b
:= 0.45m
M := 50N⋅ m Solution: a⋅ cos ( )
b⋅ sin ( )
=
=
−M + FBC⋅ cos ( − ) ⋅ a F BC
P
=
=
=
asin
⎛ a ⋅ cos ( ) ⎞ ⎝ b ⎠
0
M a⋅ cos (
− )
F BC⋅ cos ( )
=
M⋅ cos ( ) a⋅ cos (
− )
This function goes to infinity at θ = 90 de g, so we will only plot it to θ = 80 deg.
:= 0 , 0.1 .. 80 asin ( ) :=
⎛ a ⋅ cos ( ⋅ deg) ⎞ ⎝ b ⎠
P( )
:=
M⋅ cos ( ( ) ) a⋅ cos (
⋅ deg − ( ) )
1500 s n o t ( w P e N
1000
) 500 0
0
20
40
Degrees
60
80
Problem 6-105
Five coins are stacked in the smooth plastic container shown. If each coin has weight W , determine the normal reactions reactions of the bottom coin on the container at points A and B. Given: W :=
0.094N
a := 3 b :=
4
Solution:
All coins : ΣFy =
0;
NB := NB =
5W 0.470 N
Bottom coin : ΣFy =
0;
b
NB − W − N0⋅
=
a
ΣFx =
0;
N := 0
(NB − W)⋅
N0 =
0.47 N
+ b
a
2
a
0.282 N
+ b
2
2
2
+ b
b
a
NA := N0⋅
NA =
2
2
0
Problem 6-106
Determine the horizontal and vertical c omponents omponents of force at pin B and the normal force the pin at C exerts on the sm ooth slot. Also, determine the moment and horizontal and vertical reactions of force at A. There is a pulley at E. Given: F
:= 50kN
a
:= 4m
b
:= 3m
Solution: Guesses Bx
:= 1kN
By
:= 1kN
NC
:= 1kN
Ax
:= 1kN
Ay
:= 1kN
MA := 1kN⋅ m
Given a
⋅ NC − F
=
0
⋅ NC − F
=
0
(F + Bx)⋅ a − (F + By)⋅ b
=
0
=
0
Bx
+
2
a By
b
−
2
a
F
+b
+b
a
−
2
a
+b
b 2
a
2
+b
2
2
2
⋅ NC − Ax
⋅ NC − Ay
a
−F⋅ 2a +
2
a
+b
2
=
0
⋅ NC⋅ a + MA
=
0
⎛ Bx ⎜ By ⎟ ⎜ ⎟ ⎜ NC ⎟ ⎜ ⎟ := Find(Bx , By , NC , Ax , Ay , MA) ⎜ Ax ⎟ ⎜A ⎟ ⎜ y⎟ ⎝ MA ⎠
NC
= 20 kN
⎛ Bx ⎞
=
⎛ 34 ⎞ kN ⎝ 62 ⎠
=
⎛ 34 ⎞ kN ⎝ 12 ⎠
⎝ By ⎠ ⎛ Ax ⎞ ⎝ Ay ⎠ MA
= 336 kN⋅ m
Problem 6-107
A force F is applied to the handles of the vise grip. Determine the compressive force force developed on the smooth bolt shank A at the jaws. Given: F := 20N
b := 25mm
a := 37.5 mm
c := 75mm
d := 18.5mm
e := 25mm
:= 20deg Solution:
From FBD (a)
ΣME = 0; F⋅ ( b + c) − FCD⋅ ⎡ ⎢
⎤⋅ b 2 2⎥ c d e + + ( ) ⎣ ⎦ d+e
2
c + ( d + e)
FCD := F ⋅ ( b + c) ⋅
ΣFx = 0;
b⋅ ( d + e)
=
0
2
F CD = 159.5 N
c
Ex := FCD⋅
2
c + ( d + e)
2
Ex = 137.9 N
From FBD (b)
ΣMB = 0;
NA⋅ sin ( ) ⋅ d + NA⋅ cos ( ) ⋅ a − Ex⋅ ( d + e)
NA := Ex⋅
d+e sin ( ) ⋅ d + cos ( ) ⋅ a
NA = 144.3 N
=
0
Problem 6-108
If a force of magnitude P is applied to the grip of the clamp, determine the compressive force force F that the wood block exerts on the clamp. clamp. Given: P := 40N a := 50mm b := 50mm c := 12mm d := 18mm e := 38mm
Solution:
:= atan
Define
⎛ b ⎞ ⎝ d ⎠
= 70.20 deg
From FBD (a),
ΣMB = 0;
FCD⋅ cos ( ) ⋅ c − P ⋅ ( a + b + c) FCD :=
+
↑
Fy = 0;
P ⋅ ( a + b + c) cos ( ) ⋅ ( c)
FCD⋅ sin ( ) − By By := FCD⋅ sin ( )
=
=
0
FCD = 1102.2 N
0 By = 1037.0 N
From FBD (b),
ΣMA = 0;
By⋅ d − F⋅ e F :=
By⋅ d e
=
0 F = 491.2 N
Problem 6-109
The hoist supports the engine of mass M . Determine the force in member DB and in the hydraulic cylinder H of member FB. Units Used:
kN
:=
3
10 N
Given: M := 125kg
d
a
:=
1m
e := 1m
b
:=
2m
f := 2m
c := 2m
g
1m
:=
9.81
=
m 2
s Solution:
Member Σ ME
GFE
= 0;
−F
: c+d
FB⋅
( c + d)
Σ Fx
FFB
:=
FFB
=
= 0;
M⋅ g⋅
a
+
b⋅ ( c
2
+
b
+
b
(b
−
⋅
( c + d)
d)
e)
+
M⋅ g ⋅ ( a
+
b) = 0
2 2
+
(b
−
e)
2
1.94 kN
Ex
−
b
FFB⋅
( c + d) Ex
:=
−
2
b
FFB⋅
( c + d)
e
=
+
(b
−
e
2
+
−
(b
e)
−
0
2
e)
2
Member EDC: ΣΜc =
0;
Ex⋅ ( c + d)
−
e
F DB⋅
2
e FDB := Ex⋅
c+d e⋅ d
⋅
2
e
⋅d =
+
+
d
d
0
2
2
FDB
=
2.6 kN
Problem 6-110
The flat-bed trailer has weight W 1 and center of gravity at GT .It is pin-connected to the cab at D. The cab has a weight W 2 and center of gravity at G C Determine the range of values x for the position of the load L of weight W 3 so that no axle is subjected to a force greater than F Max. The load has a center of gravity at G L. Given: W1
:= 35kN
a
:= 1.2m
W2
:= 30kN
b
:= 1.8m
W3
:= 10kN
c := 0.9m
Fmax
:= 27.5kN d := 3m e := 3.6m
Solution:
Ca se 1:
As sume
Guesses
Ay x
Given
Ay
Ay
:= Fmax
:= 1m
:= Fmax By
Dy
:= F max
Cy
:= Fmax
:= Fmax
+ By − W2 − Dy
=
0
−W2⋅ a − Dy⋅ ( a + b) + By⋅ ( a + b + c) Dy
− W1 − W3 + Cy
W3⋅ x
=
=
0
0
+ W1⋅ e − Dy⋅ ( c + d + e)
=
0
⎛ By ⎞ ⎜ Cy ⎟ ⎜ ⎟ := Find(By , Cy , Dy , x) ⎜ Dy ⎟ ⎝ x ⎠ ⎛ Ay ⎞ ⎛ 27.5 ⎞ ⎜ ⎟ ⎜ By ⎟ = ⎜ 31.7 ⎟ kN ⎜ ⎟ 15.8 ⎝ Cy ⎠ ⎝ ⎠ Ca se 2:
As sume
Guesses
Ay
x1
:= x
x1
= 9.28 m
Since B y > Fmax then this solution is no good. By
:= Fmax
:= Fmax By
:= F max
Cy
:= Fmax
:= 1m
x Given
Ay
Dy
:= Fmax
+ By − W2 − Dy
0
=
−W2⋅ a − Dy⋅ ( a + b) + By⋅ ( a + b + c) Dy
− W1 − W3 + Cy
W3⋅ x
=
0
0
+ W1⋅ e − Dy⋅ ( c + d + e)
⎛ Ay ⎞ ⎜ Cy ⎟ ⎜ ⎟ := Find(Ay , Cy , Dy , x) ⎜ Dy ⎟
=
=
0
⎛ Ay ⎞ ⎛ 26.2 ⎞ ⎜ ⎟ ⎜ By ⎟ = ⎜ 27.5 ⎟ kN ⎜ ⎟ 21.3 ⎝ Cy ⎠ ⎝ ⎠
⎝ x ⎠
x2
:= x
x2
= 5.21 m
Since A y < Fmax and Cy < Fmax then this solution is good.
Ca se 3:
As sume
Guesses
Ay x
Given
Ay
Cy
:= Fmax
:= Fmax
:= 1m
By Dy
:= Fmax
Cy
:= Fmax
:= Fmax
+ By − W2 − Dy
0
=
−W2⋅ a − Dy⋅ ( a + b) + By⋅ ( a + b + c) Dy
− W1 − W3 + Cy
W3⋅ x
0
0
+ W1⋅ e − Dy⋅ ( c + d + e)
⎛ Ay ⎞ ⎜ By ⎟ ⎜ ⎟ := Find(Ay , By , Dy , x) ⎜ Dy ⎟ ⎝ x ⎠
=
=
=
0
⎛ Ay ⎞ ⎛ 24.8 ⎞ ⎜ ⎟ ⎜ By ⎟ = ⎜ 22.7 ⎟ kN ⎜ ⎟ 27.5 ⎝ Cy ⎠ ⎝ ⎠
Since A y < Fmax and By < Fmax then this solution is good. We conclude that x3 = 0.52 m < x < x2 = 5.21 m
x3
:= x
x3
= 0.52 m
Problem 6-111
Determine the force created in the hydr aulic cylinders EF and and AD in order to hold the shovel in equilibrium. The shovel load has a mass mass W and a center ce nter of gravity at G. All joints are pin connected. Units Used: 3
Mg := 10 kg 3
kN := 10 N Given: a := 0.25 m
1 := 30deg
b := 0.25m
2 := 10deg
c := 1.5m
3 := 60deg
d := 2m
W := 1.25Mg
e := 0.5m Solution:
Assembly FHG :
(
( ))
ΣMH = 0; −[ W⋅ g⋅ ( e) ] + FEF⋅ c ⋅ sin 1 FEF := W⋅ g⋅
e c⋅ sin
( 1)
=
0
FEF = 8.17 kN (T)
Assembly CEFHG :
ΣMC = 0; FAD⋅ cos
(
)
1 + 2 ⋅ b − W⋅ g⋅ ( a + b + c) cos
FAD := W⋅ g⋅
cos
( 2) + e
( 2)⋅ a + cos ( 2)⋅ b + cos ( 2)⋅ c + e cos ( 1 + 2) ⋅ b
FAD = 158 kN (C)
=
0
Problem 6-112
The aircraft-hangar door opens and closes slowly by means of a motor which draws in the cable and length AB. If the door is made in two sections ( bifold) and each section has a uniform weight W and L, determine the force in the cable as a function of the door's position θ . The sections are pin-connected at C and and D and the bottom is attached to a roller that travels along the vertical track.
Solution: L
ΣMD = 0;
2W
ΣΜC = 0;
T⋅ L⋅ cos
cos
2
⎛ ⎞ − 2L⋅ sin ⎛ ⎞ N ⎝ 2 ⎠ ⎝ 2 ⎠ A
=
0
⎛ ⎞ − N ⋅ L⋅ sin ⎛ ⎞ − W⋅ L ⋅ cos ⎛ ⎞ A 2 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠
N A
=
0
W =
2
cot
⎛ ⎞ ⎝ 2 ⎠
T
=
W
Problem 6-113
A man having weight W attempts attempts to lift himself using one of the t wo methods shown. Determine the the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglect the weight w eight of the platform. Given:
W
:= 750N
Solution:
(a) Bar:
+
↑
Fy = 0;
2⋅
⎛ F ⎞ − 2⎛ W ⎞ ⎝ 2 ⎠ ⎝ 2 ⎠
F
:=
0
=
W
F
= 750 N
Man:
+
↑
Fy = 0;
⎛ F ⎞ ⎝ 2 ⎠
NC
−W−2
N C
:=
W
=
+F
0 N C
= 1500 N
( b) Bar:
+
↑
Fy = 0;
2⋅
⎛ W ⎞ − 2⋅ F 2 ⎝ 4 ⎠
F
:=
=
0
W
F
2
= 375 N
Man:
+
↑
Fy = 0;
NC
− W + 2⋅
NC
:=
W
⎛ F ⎞ ⎝ 2 ⎠
−F
=
0
NC
= 375 N
Problem 6-114
A man having weight
W 1 attempts
to lift himself using one of the two methods shown. Determine the
total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has weight W 2.
Given: W 1
:= 750N
W 2
:= 150N
Solution:
(a) Bar:
+
↑
Fy = 0;
Man:
+
↑
Fy = 0;
2⋅
F
−
2
F
:=
F
= 900 N
N C
−
W 1
W 1
N := F C
N C
( b)
(W1 + W2)
+
+
−
W
2
0
2
F =
2 W
=
0
1
= 1650 N
Bar:
+
↑
Fy = 0;
−2 ⋅ ⎛ ⎞ + 2 ⎝ 2 ⎠ F
W 1
+
W2 =
4
0
W F
:=
1
+
W
2
2 F
= 450 N
Man:
+
↑
Fy = 0;
N C
−
W 1
N := W C 1
N C
+
2
F =
2
−
= 300 N
F
0
Problem 6-115
The piston C moves moves vertically between the two smooth walls. If the spring has s tiffness k and and is unstretched when θ = 0 , determine the couple M that must be applied to AB to hold the mechanism in equilibrium. Given: N
k := 15
cm
:= 30deg a := 8cm b := 12cm
Solution: Geometry: bsin ( ) asin ( ) =
⎛ ⎝
asin sin( ) ⋅ := asin
:= 180deg − rAC
b ⎠
−
b
=
sin ( )
a ⎞
sin ( )
= 19.47 deg = 130.53 deg
rAC := b⋅
sin ( )
rAC = 18.242 cm
sin ( )
Free Body Diagram: The solution for t his his problem will be simplified if one realizes that member CB is a two t wo force member. member. Since the sprin g stretches x := ( a + b) − rAC
x = 1.758 cm Fsp := k⋅ x
the spring force is
Fsp = 26.37 N
Equations of Equilibrium: Using the method of joints
+
↑
Fy = 0;
FCB⋅ cos ( ) − Fsp
=
0
FCB :=
Fsp cos ( )
FCB = 27.97 N
From FBD of bar AB +ΣMA = 0;
FCB⋅ sin ( ) ⋅ a − M
=
0
M := F CB⋅ sin ( ) ⋅ a
M = 1.70 N⋅ m
Problem 6-116
The compound shears are used to cut metal parts. Determine the vertical cutting force exerted on the rod R if a force F is applied at the g rip G.The lobe CDE is in smooth contact with the head of the shear blade at E . Given: F := 100N
e
a
:=
0.42m
f := 0.15m
b
:=
0.06m
g
:=
0.15m
h
:=
0.75m
:=
60deg
c := 0.6 m d := 0.225m
:=
0.15m
Solution:
Member AG : ΣMA =
0;
F⋅ ( a
+
b)
F BC := F⋅
−
FBC⋅ b⋅ sin ( ) a
+
b
=
0
FBC = 923.8 N
b⋅ sin ( )
Lobe : ΣMD =
0;
F BC⋅ f − NE⋅ e NE := FBC⋅
=
0
f
NE
e
=
923.8 N
=
1.109 kN
Head : ΣMF =
0;
−N ⋅ ( h +
E
NR
:=
g)
+
h⋅ NR
h
+
g
NE⋅
h
=
0
NR
Problem 6-117
The handle of the sector press is fixed to gear G, which in turn is in mesh with the sector gear C . Note that AB is pinned at its end s to gear C and the underside of the table EF , which is allowed to move vertically due to the smooth guides at E and F . If the gears exert tangential forces between them, determine the compressive force developed on the cylinder S when a vertical force F is is applied to the handle of the press. Given: F :=
40N
a :=
0.5 m
b :=
0.2 m
c :=
1.2m
d :=
0.35m
e :=
0.65m
Solution:
Member ΣMG =
GD :
0;
−F ⋅ a + F
CG⋅ b
=
0
a
F CG := F ⋅
F CG =
b
100 N
Sector gear :
ΣMH =
0;
c
F CG⋅ ( d + e) − F AB⋅
2
c
+ d
2
F
F d e AB := CG⋅ ( + ) ⋅
c
⋅d
+ d
c⋅ d
=
0
2 2 F
AB =
297.62 N
Table: ΣFy =
0;
F
c AB⋅
2
c
− F
+ d
F s := F AB⋅
2
s
c
2
c
+ d
2
=
0
Fs =
286 N
Problem 6-118
The mechanism is used to hide kitchen appliances under a cabinet by allowing the s helf to rotate downward. If the mixer has weight W , is centered on the shelf, and has a mass center at G, determine the stretch in the spring necessary to hold the shelf in the equilibr ium position shown. There is a similar mechanism on each side of the shelf, so that each each mechanism supports half of the load W . The springs each have stiffness k . Given: W := 50N k := 1
a := 50mm
N
b := 100mm
mm
:= 30deg
c := 375mm
:= 30deg
d := 150mm
Solution: W
ΣMF = 0;
FED :=
+
→
2
⋅ b − a⋅ FED⋅ cos ( )
W⋅ b
2⋅ a ⋅ cos ( )
Fx = 0;
+
↑
Fy = 0;
Fy :=
W
2
FED = 57.74 N
−Fx + FED⋅ cos ( )
Fx := F ED⋅ cos ( )
−W 2
0
=
=
0
Fx = 50.00 N
+ Fy − FED⋅ sin ( )
+ FED⋅ sin ( )
=
0
Fy = 53.87 N
Member FBA:
ΣMA = 0;
Fy⋅ ( c + d) ⋅ cos ( ) − Fx⋅ ( c + d) ⋅ sin ( ) − Fs⋅ sin ( Fs :=
Fs = ks;
Fs
=
Fy⋅ ( c + d) ⋅ cos ( ) − Fx⋅ ( c + d) ⋅ sin ( ) d⋅ sin (
k⋅ x
x :=
Fs k
+ )
x = 87.50 mm
+ )⋅d
=
0
Fs = 87.50 N
Problem 6-119
If each of the thr ee links of the mechanism has h as a weight W, determine the angle θ for equilibrium.The spring, which always remains horizontal, is unstretched when θ = 0°. Given: W
:= 25kN kN
k
:= 60
a
:= 4m
b
:= 4m
m
Solution:
:= 30deg
Guesses Bx
:= 10kN By := 10kN
Cx
:= 10kN Cy := 10kN
Given a
−W⋅ ⋅ sin ( ) − Cy⋅ a ⋅ sin ( ) + Cx⋅ a ⋅ cos ( ) 2 −W⋅ By
b
+ Cy⋅ b
2
+ Cy − W
−Bx + Cx
=
=
=
=
0
0
0
0
a
a
a
−Bx⋅ a⋅ cos ( ) − By⋅ a ⋅ sin ( ) − W⋅ ⋅ sin ( ) + k⋅ ⋅ sin ( ) ⋅ ⋅ cos ( ) 2 2 2
⎛ Bx ⎞ ⎜ By ⎟ ⎜ ⎟ ⎜ Cx ⎟ := Find(Bx , By , Cx , Cy , ⎜ ⎟ ⎜ Cy ⎟ ⎝ ⎠ Cx
T
=
=
1 2
k⋅
⋅ sin ( ) ⋅
W
+ 2⋅ C y
cos ( )
⎛ a ⋅ sin ( ) ⎞ ⎝ 2 ⎠
=
0
⎛ Bx ⎞
)
⎛ 16.58 ⎞ ⎜ By ⎟ ⎜ 12.5 ⎟ kN ⎜ ⎟=⎜ ⎟ 16.58 C ⎜ x⎟ ⎝ 12.5 ⎠ Cy ⎝ ⎠
⎛ ⎞ ⎜ T ⎟ := Find ( , T , Cx) ⎝ Cx ⎠
= 33.6 deg
Problem 6-120
Determine the required force P that must be applied at the blade of the pruning shears so that the blade exerts a normal normal force F on on the twig at a t E. Given: F
:= 100N
a
:= 12.5 mm
b
:= 100mm
c := 18.75mm d
:= 18.75mm
e := 25mm Solution:
initial guesses: Ax
:= 1N
Ay
Dy
:= 10N
P
:= 1N
Dx
:= 20N
:= 10N
FCB := 20N
Given
−P⋅ ( b + c + d) − Ax⋅ a + F⋅ e Dy
− P − Ay − F
Dx
− Ax
=
0
=
0
−Ay⋅ d − Ax⋅ a + ( b + c) ⋅ P Ax
− FCB⋅
c 2
c Ay
=
+ P − FCB⋅
=
0
0
=
2
+a
a 2
a
=
2
+c
0
0
⎛ Ax ⎜ ⎟ A ⎜ y ⎟ ⎜D ⎟ ⎜ x ⎟ := Find A , A , D , D , P , F ( x y x y CB) ⎜D ⎟ ⎜ y ⎟ ⎜ P ⎟ ⎜F ⎟ ⎝ CB ⎠
⎛ Ax ⎞ ⎛ 66.67 ⎞ ⎜ Ay ⎟ ⎜ ⎟ ⎜ 32.32 ⎟ ⎜ Dx ⎟ = ⎜ 66.67 ⎟ N ⎜ D ⎟ ⎜ 144.44 ⎟ ⎜ y ⎟ ⎝ 80.12 ⎠ FCB ⎝ ⎠
P
= 12.12 N
