# MODEL ANSWERS OF ASSIGNMENT NO.(8) EQUILBRIUM OF SYSTEM OF

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```					MODEL ANSWERS OF ASSIGNMENT NO.(8)

EQUILBRIUM OF SYSTEM OF RIGID BODIES

FRAMES & MACHINES

BY

DR. FAROUK ELBARKI

1
Problem 1
In each case, determine the force P required to maintain equilibrium. The block
weighs 100 N (≈ 10 kg)

P

(a)

2
(a) SOLUTION
Ey

E
For pulleys C and D
A
∑ Fy = 0

4P–Q=0

P = 100 / 4                                        B

P                        P
P = 25 N
P                P
P

P           P

C
P                    P

D

Q = 100 N

3
(b) SOLUTION

For pulley B
P

∑ Fy = 0

3P–Q=0

3 P – 100 = 0

P = 100 / 3

P = 33.33 N

(b)

A

P              P

P      P
P

B

Q= 100 N

4
(c) SOLUTION
In fig. 1                                                       A         P
D
∑ Fy = 0

3P–T=0

T=3P            (1)                                                               C

In fig. 2

∑ Fy = 0

3T–Q=0
B

3 T – 100 = 0                               P
P

T = 100 / 3

T = 33.3 N               Fig 1
(c)
From (1)
T
∑ Fy = 0
T

T=3P
T                  T
P = 33.3 / 3

P = 11.1 N

B
Fig 2

Q = 100 N

5
Problem 2:
Determine the force P needed to support the 100N (10Kg) weight. Each pulley has a
weight of 10N . Also, what are the cord reactions at A and B?

A                  C

50 mm

P

C
50 mm
B                         D

50 mm

P

6
Solution

For pulley D:                               TC

C

∑Fy = 0

TB – 2P – 10 = 0

TB = 2P + 10 → 1
10 N
TB            TB
TB

For pulley B:

D

∑Fy = 0

TB + 2P –100 – 10 = 0 → 2

Sub. from 1 in 2
10 N
P           P
4P = 100
P   TB        P
P = 25 N

TB = 60 N

TA = 25 N                                            B

10 N

100 N

7
Problem 3

The principles of a differential chain block are indicated schematically in the figure.
Determine the magnitude of force P needed to support the 800 N force. Also, find the
distance x where the cable must be attached to bar AB so the bar remains horizontal.
All pulleys have a radius of 60 mm.

D
C                                        x

C                                     B

800 N
180 mm       240 mm

T   T       T   T
P

C                       D

T
800 N

x

A                                             B

P           P       P           P    P

8
For the bar CD

∑ Fy = 0

4 T – 800 = 0

T = 800 / 4

T = 200 N

For the bar AB

∑ Fy = 0

T–5P=0

200 – 5 P = 0

P = 200 / 5

P = 40 N

∑ MA = 0

5 P (x) – P (2 r + 4 r + 6 r + 8 r) = 0

Where r = 0.06 m

x = 20 r / 5

x=4r

x = 4 (0.06)

x = 0.24 m (240 mm)

9
Problem 4:
Determine the force P needed to support the 20Kg mass using the Spanish Burton
rig. Also, what are the reactions at supporting hooks A,B and C?

A           B       C

H           G       F

P
E

D

10
Solution

W=m*g
TA               TB                   TC
= 20 * 9.8
= 196 N                    H                G                        F

For pulley D:
∑Fy = 0
P        P       P        P
3T – W = 0
T                T
T = 65.3 N
P       P        P

For pulley E:
E
∑Fy = 0
3P – T = 0
T       T            T
P = 21.8 N
T

For pulley H:
∑Fy = 0                                                                D

TA – 2P = 0
TA = 43.6 N
For pulley G:
W
∑Fy = 0
TB – 2P = 0
TB = 43.6 N
For pulley F:
∑Fy = 0
TC – 2T = 0
TA = 130.6 N

11
Problem 5
The compound beam is pin-supported at C and supported by a roller at A and B.
there is a hinge (pin) at D. Determine the reactions at the supports. Neglect the
thickness of the beam.

12 KN
8 KN
5
15 KN. m                                  4
A                 D
3

B                                 C

30˚
4 KN
6m            4m     2m              8m                  8m                    8m

12 (4/5) KN
NB
NA           8 KN                                                                  Cy

15 KN. m           12 (3/5) KN
4 sin 30˚
Cx
A               D                    B                                         C

4 cos 30˚
6m             4m     2m              8m                  8m                     8m
Fig. (1)

NA         8 KN
Dy

4 sin 30˚                                                   Dx

A               D

4 cos 30˚
Fig. (2)

12
In fig. (2)

∑ MD = 0

4 cos 30˚ (12) – NA (6) + 8 (2) = 0

NA = 9.59 KN

In fig (1)

∑ Fx= 0

- 4 sin 30˚ - 12 (3/5) + Cx = 0

- 9.2 + Cx = 0

Cx = 9.2 KN

∑ MC = 0

12 * 8 (4/5) – NB (16) + 8 (26) – 9.59 (30) + 4 cos 30˚ (36) + 15
=0

NB = 8.55 KN

∑ Fy= 0

- 4 cos 30˚ + 9.59 – 8 + 8.55 – 12 (4/5) + Cy = 0

Cy = 2.92 KN

13
Problem 6:
Determine the greatest force P that can be applied to the frame if the largest force
resultant acting at A can have a magnitude of 2 KN.

0.1 m

0.5 m

A
0.75 m            0.75 m

Solution

0.1 m               P
T

Ay
0.5 m

A

Ax
0.75 m            0.75 m

For entire frame:
P

∑Fy = 0    , Ax – T = 0 ,                    Ax = T→ ( 1 )

∑Fy = 0 , Ay – P = 0 ,                       Ay = P → ( 2 )

∑MA = 0 , -P (1.5) + T (0.6) = 0 ,       T = 2.5P →( 3 )

FA = (Ax 2 + Ax 2 )0.5   = 2.692P

FA ≤ 2 KN , 2.692P ≤ 2 KN       P ≤ 0.743 KN

14
Problem 7
Determine the horizontal and vertical components of force at C which member ABC
exerts on member CEF.

C

4m

E                    r=1m
B

D

4m

A
F

30 KN

6m                       3m

15
C

Fig. (1)

4m

E                                 r=1m
B

D
Ay
4m
NF
Ax
F
A
30 KN

6m                              3m

By                     Ey

30 KN
Cy                                                                                            D
Bx                                    Ex

Cx                                                  B                       E
C

Fig (3)
30 KN
30 KN
Ex
E

Ey
Fig (2)

F

NF

16
For the entire frame (in fig (1)):

∑ MA = 0

NF (6) – 30 (10) = 0

NF = 300 /6

NF = 50 KN

In fig (3)

∑ MB = 0

Ey (3) – 30 (7) + 30 (1) = 0

Ey = 60 KN

In fig (2):

∑ ME = 0

Cx (4) – 30 (1) = 0

Cx = 7.5 KN

∑ Fy= 0

Cy – Ey + NF = 0

Cy – 60 +50 = 0

Cy = 10 KN

17
Problem 8
Determine the force that the smooth roller C exerts on beam AB. Also, what are
the horizontal and vertical components of reaction at pin A? Neglect the weight of
the frame and roller.

D
60 N .m                C
0.2 m
A
B

1m                         1.2 m

Ay
D           ND
60 N .m         C
0.2 m
Ax

A                                                                  B

Fig. (1)
1m                      1.2 m

NC
By
Ay
60 N .m
Ax
Bx
A                                C
B

Fig. (2)
1m                          1.2 m

18
In fig (1):

∑ MA = 0

ND (0.2) – 60 = 0

ND = 60 / 0.2

ND = 300 N

∑ Fx = 0

Ax – ND = 0

Ax = 300 N

∑ Fy= 0

Ay = 0

In fig (2):

∑ MB = 0

- Ay (2.2) + NC (1.2) – 60 = 0

NC = 60 / 1.2

NC = 50 N

19
Problem 9:

Determine the horizontal and vertical components of force at each pin. The
suspended cylinder has a weight of 800N (80Kg).

1.5 m

E

2m

0.5 m

2500 N
A            B          C

2m

D

3m                1m

20
tan θ = 2/3 , θ = 33.7°                                                                      0.5 m
800 N
B                  C
NA
For fig (1):                              A        Bx                 θ                         2500 N

∑MC = 0                                                 By            T
800(0.5) -800 (1) - By(1.5) = 0
By = -266.7 N
∑Fy = 0
By – T sin θ – 800 = 0
Fig. 1
-266.7– T sin θ – 800 = 0                                                               800 N
Ey
T = -1922.5 N
Cx= T cos θ     = -1599.4N
Cy= T sin θ     = 1066.7N                     EX
E
Dx= T cos θ     = -1599.4N
Cy= T sin θ     = 1066.7N                                                          2m

For fig (2):                                                 Fig. 2
800 N
∑ME = 0                                                                        0.5 m
Bx
Bx (2) – By (1.5) + 800 (1.5) = 0                                     B

Bx = - 800 N
By
∑Fx = 0
Bx – Ex + 800 = 0
Ex = 0
∑Fy = 0
Ey – By = 0
Ey = -266.7N
For fig (1):
∑Fx = 0
NA – Bx – T Cos θ – 800 - 2500 = 0
NA = 900 N
21
Ey

ANOTHER SOLUTION
1.5 m
Ex
E

For entire frame:
2m
∑MD = 0
0.5 m
-800 (4) +Ex (4) – NA (2)
NA
A             B    C               2500 N
+2500(2) = 0
Dy
2m
NA = 900 N

Dx
D
∑Fx= 0
3m       1m
NA – Dx – Ex - 2500 = 0

Dx = - 1600N                                                          800N

∑Fy = 0

-800 + Dy + Ey = 0

Dy = 1066.7N

22
Problem 10

The compound arrangement of the pan scale is shown. If the mass on the pan
is 4 kg, determine the horizontal and vertical components at pins A, B, and C
and the distance x of the 25 g mass to keep the scale in balance.

75 mm

100 mm
300 mm                        350 mm

F                                 E

C
x
B

G
50 mm
25 g

A                                 D

4 kg

23
75 mm

100 mm
300 mm                                          350 mm
Cy
F                               Cx                   E

C
By
x

Bx

B                 G
Ay                  50 mm
0.025(9.81) N
Fig (1)
Ax            A                                                    D

4 (9.81) N

Ay                                                   T1
50 mm

Ax
Fig (2)
A                                                    D

4 (9.81) N

75 mm
300 mm
Cy

F                                                                E       Fig (3)
Cx        C

T2                                                       T1

By             T2
x

Bx
B              G

Fig (4)
0.025(9.81) N

24
In fig (2):

∑ MA = 0 :       - 4 * 9.81* 0.05 + T1 (0.375) = 0

T1 = 1.962 / 0.375

T1 = 5.232 N

∑ Fy = 0:      Ay – (4 * 9.81) + T1 = 0

Ay = 39.24 – 5.232 ,       Ay = 34 N

∑ Fx = 0 : Ax = 0

In fig (3):

∑ MC = 0 :     - T1 (0.075) + T2 (0.3) = 0

- 5.232 (0.075) + T2 (0.3) = 0

T2 = 1.308 N

∑ Fy = 0:

- T2 + Cy – T1 = 0

- 1.308 + Cy – 5.232 = 0

Cy = 6.54 N

∑ Fx = 0:

Cx = 0

25
In fig (4)

∑ Fx = 0

Bx = 0

∑ MB = 0

T2 (0.1) – (25 * 10-3 * 9.81) (0.825 – x) = 0

x = 0.292 m (292mm)

∑ Fy = 0

By + T2 – (25 * 10-3 * 9.81) = 0

By = 0.24525 – 1.308

By = - 1.06 N

By = 1.06 N ↓

26
Problem 11

Determine the reaction components at the hinges (A), (B) , (C) of the shown
frame ABC.
B

20 kN
2m
20 kN
2m

2m

30 kN                 4m
C
1m
A
5m             1.5 m 1.5 m

By
B
Bx         B
20 kN
20 kN
2m
2m
20 kN
2m                    Cy

2m                                                                          Cy
Cx                     2m
30 kN            4m                        C                                                       C    Cx
Ax                      1m
A
5m          1.5 m 1.5 m
1.5 m 1.5 m
Ay

ENTIRE SYSTEM (Fig 1)                                            F.B.D OF (BC) Fig(2)

27
For fig (1):

∑MA = 0

- 30(1) – 20 (3) – 20(6.5) - Cx (1) + Cy (8) = 0

Cx   - 8 Cy = - 220 …………………..              (1)

For fig (2):

∑MB = 0

- 20(1.5) + Cx(4) + Cy (3) = 0

4 Cx + 3 Cy = 30 …………………….. (2)

Solving (1) & (2) :

Cx= - 12 kN                     ,      Cy = 26 kN

For fig (1):

∑Fx = 0

Ax +Cx + 30 + 20 = 0        ,       Ax = - 38 kN

∑Fy = 0

Ay + Cy - 20 = 0           ,        Ay = - 6 kN

For fig (2):

∑Fx = 0

Cx – Bx = 0             ,           Bx = - 12 kN

∑Fy = 0

Cy + By - 20 = 0       ,            By = - 6 kN

28

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