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SOLVED PROBLEMS
CHAPTER- 2 : EQUILIBRIUM OF FORCE SYSTEMS
Problem-2.1 Determine the magnitude of F1 and F2 so that particle P is in equilibrium.
Figure for Problem-2.1
Solution:
Soln. Fig. 2.1
Referring to Soln. Fig. 2.1
F
x
 0 yields F1 cos 45o  F2 cos 30 o  500  0
 0.707 F1  0.866 F2  500  0
F
y
...
...
...
...
...
(a)
...
...
...
...
(b)
 0 yields F1 sin 45o  F2 sin 30 o  0
 0.707 F1  0.5 F2  0
...
Subtract Eq. (b) from Eq. (a) to have
0.866F2  0.5F2  500  0
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Chapter- 2 Equilibrium of Force Systems
Page 1
 F2  366.03 N
Putting this value in Eq. (b) we get
0.707F1  0.5366.03  0
 F1  258.86 N
Problem-2.2 A weight W = 5000 N is suspended from a pin B which unites two inclined timbers AB
and CB as shown in the figure. Determine the compressive force on each timber.
Figure for Problem-2.2
Solution:
Soln. Fig. 2.2
Referring to Soln. Fig. 2.2
F
x
 0 yields FAB cos 60 o  FBC cos 60 o  0
 0.5 FAB  0.5 FBC  0
 FAB  FBC
F
y




 0 yields FAB cos 30 o  FBC cos 30 o  5000  0
0.866 FAB  0.866 FBC  5000  0
0.866 FAB  0.866 FAB  5000  0
1.732 FAB  5000  0
FAB  2886.84 N
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Chapter- 2 Equilibrium of Force Systems
Page 2
 FBC  2886.84 N
Problem-2.3 The permissible internal force on timber AB shown in the figure is 20 kN. If   30 o ,
what maximum safe load may be suspended at B?
Figure for Problem-2.3
Solution:
Soln. Fig. 2.3
Referring to Soln. Fig. 2.3
F
x
 0 yields 20 cos 45 o  FBC cos 30 o  0
 14.14  0.866 FBC  0
 FBC  16.328 kN
F
y
 0 yields 20 cos 45 o  FBC cos 60 o  W  0
 200.707  16.3280.5  W  0
 W  22.300 kN
Problem-2.4 The derrick shown diagrammatically in the figure supports a load of W = 2 kips. Find
the tension in the boom cable and the compression in the boom when the angle  is 30o.
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Chapter- 2 Equilibrium of Force Systems
Page 3
Figure for Problem-2.4
Solution:
Soln Fig. 2.4
Referring to Soln. Fig. 2.4, apply cosine rule to have
 AB 2  40 2  70 2  2 4070cos 30 o
 AB  40.62 ft.
Again applying sine rule,
sin 30 o sin  sin 


40.62
40
70
o
   120.50 and   29.50 o
Now,
 Fx  0 gives FBC sin 30 o  FAB cos90 o  29.50 o  30 o   0
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Chapter- 2 Equilibrium of Force Systems
Page 4
 0.5FBC  0.862FAB  0
...
...
...
...
...
(a)
...
(b)
and
 Fy  0 gives FBC cos 30 o  FAB cos29.50 o  30 o   2  0
 0.866FBC  0.507 FAB  2  0
...
...
...
...
Multiply Eq. (a) by 0.866 and (b) by 0.5 and subtract to get
 0.493FAB  1  0
 FAB  2.03 kip
Putting this value in Eq. (a) we get
0.5FBC  0.8622.03  0
 FBC  3.50 kip
Problem-2.5 Determine the magnitude and coordinate direction angles of F in the following figure,
that are required for equilibrium of point O.
Figure for Problem-2.5
Solution:
Referring to figure for problem-2.5 above,
F1 y  400 N,
F1x  0 ,
F2 x  0 ,
F2 y  0 ,
F1z  0
F2 z  800 N
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Chapter- 2 Equilibrium of Force Systems
Page 5
F3 x  700
2
 200 N,
7
F3 y  700
3
 300 N,
7
F3 z  700
6
 600 N
7
Now,
 Fx  0 gives 0  0  200  Fx  0  Fx  200 N
F
F
y
 0 gives 400  0  300  Fy  0  Fy  100 N
z
 0 gives 0  800  600  Fz  0  Fz  200 N
Hence, the magnitude of F is
F
 F    F    F 
2
x

2
y
2
z
2002   1002  2002
 90000  300 N
The coordinate direction angles are
F
200
  cos 1 x 
 48.19 o
F 300
F
 100
y
  cos 1

 109.47 o
F
300
F
200
  cos 1 z 
 48.19 o
F 300
Problem-2.6 Determine the tension in cords AB and AD for equilibrium of the 10-kg block shown in
the figure.
Figure for Problem- 2.6
Solution:
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Chapter- 2 Equilibrium of Force Systems
Page 6
Soln Fig. 2.6
Referring to Soln. Fig. 2.6,
 Fx  0 gives TAB cos 30o  TAD  0
 0.866TAB  TAD  0
...
...
...
...
...
...
(a)
and,
 Fy  0 gives TAB sin 30o  98.1  0
 TAB  196.2 N
Putting this value in Eq. (a) we get
0.866196.2  TAD  0
 TAD  169.91 N
Problem-2.7 The ends of the 7-m-long cable AB are attached to the fixed walls. If a bucket and its
contents have a mass of 10 kg and are suspended from the cable by means of a small pulley as
shown, determine the location x of the pulley for equilibrium.
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Chapter- 2 Equilibrium of Force Systems
Page 7
Figure for Problem-2.7
Solution:
Soln Fig. 2.7
The cable is a continuous one, which passes over a frictionless pulley, so the tension-force
developed, T has a constant magnitude. Now referring to Soln. Fig. 2.7,
 Fx  0 yields T cos1  T cos 2  0
 1   2   (say).
Now,
cos  
x
AC
 AC 
x
cos 
...
...
...
...
...
...
...
(a)
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Chapter- 2 Equilibrium of Force Systems
Page 8
Again,
4 x
BC
4 x
 BC 
cos 
cos  
...
...
...
...
...
...
...
(b)
Adding Eqs. (a) and (b) we get
x
4 x
AC  BC 

cos  cos 
x4 x
 7
cos 
4
 cos  
7
4
   cos 1  55.15 o
7
Now,
h
4 x
 h  4  x tan   1.436 4  x
tan  
h 1
x
1.436 4  x   1
1.436 
x
1.436x  1.4364  1.436 x  1
2.872 x  6.744
6.744
x
 2.35 ft
2.872
Again, tan  




Problem-2.8 If the cylinder at A in the figure has a weight of 20 lb, determine the weight of B and
the force in each cord needed to hold the system in the equilibrium position shown.
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Chapter- 2 Equilibrium of Force Systems
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Figure for Problem- 2.8
Solution:
Soln. Fig. 2.8a
Soln. Fig. 2.8b
Referring to Soln. Fig. 2.8a,
 Fx  0 yields TEG cos 60 o  TEC cos 45o  0
 0.5TEG  0.70 TEC  0
F
y
...
...
...
...
...
...
(a)
...
...
...
(b)
 0 yields TEG cos 30 o  TEC cos 45 o  20  0
 0.866 TEG  0.707 TEC  20  0
...
...
Subtract Eq. b from a we get  0.366 TEG  20  0
 TEG  54.64 lb
Putting this value in Eq. a we get
0.554.64  0.707 TEC  0
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Chapter- 2 Equilibrium of Force Systems
Page 10
 TEC  38.64 lb
3
 36.87 o , and, referring to Soln. Fig. 2.8b,
4
 Fx  0 yields TEC cos 45o  TCD cos  0
Now,   tan 1
 38.640.707   TCD cos 36.87 o  0
 TCD  34.150 lb
F
y
 0 yields TEC sin 45 o  TCD sin 36.87 o  WB  0
 38.640.707  34.150.6  WB  0
 27.318  20.49  WB  0
 WB  47.81 lb
Problem-2.9 Two weights are suspended from a flexible cable as shown in the figure. Determine the
internal forces in the various parts of the cable and the weight W.
Figure for Problem- 2.9
Solution:
Soln. Fig. 2.9b
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Chapter- 2 Equilibrium of Force Systems
Page 11
Soln. Fig. 2.9a
Referring to Soln. Fig. 2.9a,
 Fx  0 yields TAB sin 30 o  TBC cos15o  0
 0.5TAB  0.966 TBC  0
F
y
...
...
...
...
...
...
(a)
...
...
...
(b)
 0 yields TAB cos 30 o  TBC cos 75 o  10  0
 0.866 TAB  0.259 TBC  10  0
...
...
Multiply Eq. a by 0.866 and Eq. b by 0.5 and subtract to get
 0.707 TBC  5  0
 TBC  7.07 kN
Putting this value in Eq a, 0.5TAB  0.9667.07  0
 TAB  13.66 kN
Again referring to Soln. Fig. b
 Fx  0 gives TCD cos 45o  TBC cos15o  0
 0.707 TCD  7.07 0.966  0
 TCD  9.66 kN
F
y
 0 gives TCD sin 45 o  TBC sin 15 o  W  0
 9.660.707  7.070.259  W  0
 6.83  1.83  W  0
 W  8.66 kN
Problem-2.10 Determine the required length of cord AC in the figure, so that the 8-kg lamp is
suspended in the position shown. The unstretched length of the spring AB is l AB  0.4 m, and the
spring has a stiffness of k AB  300 N/m.
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Page 12
Figure for Problem- 2.10
Solution:
Soln Fig. 2.10
Referring to Soln. Fig. 2.10,
 Fx  0 gives TAB  TAC cos 30 o  0
 TAB  0.866TAC  0
F
y
...
...
...
...
...
(a)
 0 gives T AC cos 60 o  W  0
 0.5TAC  78.48  0
 TAC  156.96 N
Putting this value in Eq. a we get
TAB  0.866156.96  0
 TAB  135.93 N
The stretch in spring AB is
TAB  k x
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Chapter- 2 Equilibrium of Force Systems
Page 13
 135.93  300 x
 x  0.453 m
 The stretched length of the spring AB is
l AB  l AB  x
 l AB  0.4  0.453  0.853 m
The horizontal distance from C to B requires 2  l AC cos 30 o  0.853
 0.866 l AC  0.853  2  0
 l AC  1.324 m
Problem-2.11 A 25-kN sphere rests on a smooth plane inclined at an angle   45o with the
horizontal and against a smooth vertical wall. What are the reactions at the contact surfaces A and
B?
Figure for Problem- 2.11
Solution:
Soln Fig. 2.11
Referring to Soln. Fig. 2.11,
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Chapter- 2 Equilibrium of Force Systems
Page 14
F
x
 0 yields FA  FB cos 45 o  0
 FA  0.707FB  0
...
...
...
...
...
(a)
F
 0 yields FB sin 45o  25  0
25
 35.36 kN
 FB 
sin 45 o
y
Putting this value in Eq. a
FA  0.70735.36  0
 FA  25 kN
Problem-2.12 A 12.5-kN wheel with a radius of 90 cm is acted upon by a force F, as shown in the
figure, which tends to pull the wheel over the obstruction at A. At the instant the wheel is about to
move, the pressure between the wheel and the ground is zero. What is the magnitude of the force F
at this instant if   30o?
Figure for Problem- 2.12
Solution:
Soln Fig. 2.12
Referring to Soln. Fig. 2.12,
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Chapter- 2 Equilibrium of Force Systems
Page 15
  cos 1
F
x
45
 60 o
90
 0 yields F cos 30 o  FA sin 60 o  0
 0.866F  0.866FA  0
 FA  F
F
y
...
...
...
...
...
...
(a)
 0 yields F sin 30 o  FA cos 60 o  12.5  0
 0.5F  0.5 FA  12.5  0
 0.5F  0.5 F  12.5  0
 F  12.5 kN
Problem-2.13 A 2.5-kN cylinder A rests on a smooth inclined plane, as shown in the figure. For a
tension in the rope of 1.25 kN find the inclination of the plane and the plane reaction.
Figure for Problem- 2.13
Solution:
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Chapter- 2 Equilibrium of Force Systems
Page 16
Soln Fig. 2.13
Referring to Soln. Fig. 2.13,
 Fx  0 gives 1.25  2.5sin   0
1.25
 0.5
2.5
   sin 1 0.5  30 o
 sin  
F
y
 0 giveds FR  2.5 cos  0
 FR  2.5 cos 30 o  2.16 kN
Problem-2.14 Two smooth spheres, each of radius 25 cm and weight 500 N rest in a horizontal
channel having vertical walls, the distance between which is 90 cm. Find the pressure exerted on the
walls and floor at the points of contact A, B and D.
Figure for Problem- 2.14
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Chapter- 2 Equilibrium of Force Systems
Page 17
Solution:
Soln. Fig. 2.14b
Soln. Fig. 2.14a
Soln. Fig. 2.14d
Soln. Fig. 2.14c
  cos 1
40
 36.87 o [Soln. Fig. 2.14b]
50
Referring to Soln. Fig. 2.14c,
 Fx  0 gives RC cos 36.87 o  R A  0
 0.8RC  R A  0
...
...
...
...
...
...
(a)
F
 0 gives RC sin 36.87 o  500  0
500
 833.33 N
 RC 
sin 36.87 o
y
Putting this value in Eq. a we get,
0.8833.33  RA  0
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Page 18
 RA  666.66 N
Now referring to Fig. 2.14d,
 Fx  0 gives RB  RC cos 36.87 o  0
 RB  833.330.8  0
 RB  666.66 N
F
y
 0 gives RD  500  RC sin 36.87 o  0
 RD  1000 N
Problem-2.15 Two spheres are at rest against smooth surfaces, as shown in the figure. Sphere A
weighs 3200 Ib and sphere B weighs 400 Ib. Let F = 1000 Ib and   75o, and find the reactions at
C, D and E.
Figure for Problem- 2.15
Solution:
Soln. Fig. 2.15b
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Chapter- 2 Equilibrium of Force Systems
Page 19
Soln. Fig. 2.15a
Referring to Soln. Fig. 2.15a,
 Fx  0 gives 1000  RC cos15o  RO cos 60 o  0
 1000  0.966RC  0.5RO  0
F
y
...
...
...
...
...
(a)
...
...
...
(b)
 0 gives RC sin 15 o  RO sin 60 o  3200  0
 0.259 RC  0.866 RO  3200  0 ...
...
Multiply Eq. a by 0.259 and Eq. b by 0.966 and subtract to get
3350  0.965RO  0
 RO  3471.50 lb
Putting this value in Eq a, we get
1000  0.966RC  0.53471.50  0
 0.966RC  735.75  0
 RC  761.65 lb
Now referring to Soln. Fig. 2.15b
 Fx  0 gives RO cos 60 o  RE  0
 3471.500.5  RE  0
 RE  1735.75 lb
F
y
 0 gives 400  RO sin 60 o  RD  0
 400  3471.500.866  RD  0
 RD  3406.32 lb
Problem-2.16 Determine the moment of force F about points A and B of the beam shown in the
figure.
Figure for Problem-2.16
Solution:
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Chapter- 2 Equilibrium of Force Systems
Page 20
M A  F  d  2002  400 N.m
By the right-hand rule, the moment is directed into the page, since the force tends to rotate the beam
in a clockwise direction about an axis passing through A.
M B  F  d  2000.5  100 N.m
This moment is clockwise and hence directed out of the page.
Problem-2.17 The crank shown in the figure has acting on it a 1 kN and a 1.5 kN force as shown.
What is the moment of these forces about the axis A and in what direction will the crank turn?
Figure for Problem-2.17
Solution:
Compute the components of 1 kN force as
Fx  1 cos 45 o   0.707 kN, and


Fy  1 sin 45o  0.707 kN.
Letting the clockwise moment be positive and summing moments about A, we get
M
A
 0.70750  1.525  2.15 kN.cm
Since the answer is negative, the net moment is in the counterclockwise sense, the direction in which
the crank will turn if it moves.
Problem-2.18 Two coplanar forces act on the beam shown in the figure. Determine the moment of
each force about point O.
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Chapter- 2 Equilibrium of Force Systems
Page 21
Figure for Problem-2.18
Solution:
Soln. Fig. 2.18
Each force is resolved into its x and y components. Now
For the force 200 N:
Taking counterclockwise positive, M O  200 cos 60 o 0  200 sin 60 o 0.4
 69.28 N.m
For the force 400 N:
Taking counterclockwise positive, M O  400 sin 30 o 0.2  400 cos 30 o 0.4
 40  138.56  98.56 N.m
The negative sign indicates M O acts clockwise.
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Chapter- 2 Equilibrium of Force Systems
Page 22
Problem-2.19 Four wheels A, B, C and D apply loads of 15, 30, 50 and 75 kN respectively, to a
beam GH as shown in the figure. Find the resultant of these loads.
Figure for Problem-2.19
Solution:
The magnitude of the resultant is obtained as
R  15  30  50  75  170 kN acting downward.
To define R completely, we locate its line of action by using the principle of moment. Choosing any
point for center of moments, say point A, we get (clockwise positive).
 M A  150  301.8  504.2  757.2
 54  210  540  804 kN.m
Since, R r  804
 r  804 170  4.73 m.
Therefore the resultant R  170 kN acts vertically downward through a point 4.73 m from A.
Problem-2.20 Three parallel forces A, B and C are acting as shown in the figure. Determine their
resultant.
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Figure for Problem-2.20
Solution:
The magnitude of the resultant is R   F  3.0  0.5  105  2.0 kN acting in the same direction
as the 3.0 kN and 0.5 kN forces.
Now taking clockwise moments as positive and choosing C as the moment center, we get
 M C  1.537.5  0.512.5
 56.25  6.25  50.5 kN.cm (+ve)
where the positive sign indicates that the resultant must act in such a manner as to produce a
clockwise moment about C. Then the length of the moment arm r of R is obtained from R r  50
r  50 2  25 cm
[Note: 1. that the line of action of R is 25 cm from C, 2. that R acts downward, and 3. that R must
produce a clockwise moment about C.]
Problem-2.21 Determine the magnitude and direction of the couple moment.
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Figure for Problem-2.21
Solution:
Soln. Fig. 2.21
The couple forces are resolved. Now there are two couples.
M C x  4 sin 30 o 4  8 kN.m (counterclockwise)
MC y


 4 cos 30 4  13.86 kN.m (counterclockwise)
o
 MC  MCx  MC y
 8.00  13.86  21.86 kN.m (counterclockwise)
Problem-2.22 A couple acts at the end of the beam shown. Replace it by an equivalent one having a
pair of forces that act through (a) points A and B; (b) points D and E.
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Figure for Problem-2.22
Solution:
The couple has a magnitude of M C  F  d  4000.2  80 N.m. Here the forces tend to rotate
clockwise.
(a) To preserve the direction of M C horizontal forces acting through points A and B must be
directed as shown.
Soln. Fig. 2.22a
The magnitude of each force is M C  F  d
 80  F  0.25
 F  320 N
(b) To generate the required clockwise rotation, forces acting through points D and E must be
vertical and directed as shown.
Soln. Fig. 2.22b
The magnitude of each force is M C  P  d
 80  P  0.1
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 P  800 N
Problem-2.23 Replace the force F = 300 Ib in the figure with a force through the axis and a couple.
Figure for Problem-2.23
Solution:
Introduce at the axis two equal, opposite and collinear forces F  300 lb, shown dotted, Soln. Fig.
2.23(a). We now have a couple F d  30014.1  4230 lb-in and a force F  300 lb acting
through the axis downward toward the left.
Soln. Fig. 2.23a
Soln. Fig. 2.23b
As suggested by Soln. Fig. 2.23(b), we may say that the original force F produced a force at the shaft
of 300 lb and a turning moment of C  4030 lb-in.
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Problem-2.24 Determine the horizontal and vertical components of reaction for the beam loaded as
shown in the figure. Neglect the weight of the beam in the calculations.
Figure for Problem- 2.24
Solution:
Soln. Fig. 2.24
F
x
 0 yields 600 cos 45 o  B x  0
 Bx  424.26 N
F
y
 0 yields Ay  By  600 sin 45o  100  0
 Ay  B y  524.26
M
A
...

...
...
...
...
...
(a)

 0 yields  600 sin sin 45o  2  100  5  By  7  0
 B y  192.65 N
Substitute this value of B y in Eq. a
Ay  524.26  192.65  331.61 N
Problem-2.25 Determine the reactions at the fixed support A for the loaded frame shown in the
figure.
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Figure for Problem- 2.25
Solution:
Soln. Fig. 2.25
F
y
 0 yields Ay  200  400  0
 Ay  600 N
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M
A
 0 yields M A  400 1  200  2  0
 M A  800 N-m
Problem-2.26 The uniform beam shown in the figure is subjected to a force and couple. If the beam
is supported at A by a smooth wall and at B and C either at the top or bottom by smooth contacts,
determine the reactions at these supports. Neglect the weight of the beam.
Figure for Problem- 2.26
Solution:
Soln. Fig. 2.26
F
x
 0 yields  FA cos 30 o  300 sin 30 o  0
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 FA  173.20 N
F
y
 0 yields FC  FB  FA sin 30 o  300 cos 30 o  0
 FC  FB  0.5FA  259.81
 FC  FB  0.5  173.20  259.81
...
...
 FC  FB  346.41
M
A
 0 yields


...
...

...
...
(a)
...
(b)

4000  FB  2  FA sin 30 o  4  FC  2  300 cos 30 o  4  0
 2 FB  2 FC  4000  173.20 sin 30 o  4  1039.23  0
...
...
...
...
 FB  FC  2346.41
...
Add Eq. a and b to get
2FB  2000
 FB  1000 N
Put this value of FB in Eq. a
 1000  FC  346.41
 FC  1346.41 N
Problem-2.27 The beam shown in the figure is pin-connected at A and rests against a roller at B.
Compute the horizontal and vertical components of reaction at the pin A.
Figure for Problem- 2.27
Solution:
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Soln. Fig. 2.27
F
x
 0 yields Ax  FB sin 30 o  0
 Ax  0.5FB  0
F
y
...
...
...
...
(a)
...
...
...
...
(b)
 0 yields Ay  60  FB cos 30o  0
 Ay  0.866 FB  60
M
...
A
...
 0 gives FB  0.75  60 1  60  0
 FB  160 N
Put this value in Eq. a
Ax  0.5  160  0
 Ax  80 N
Now put FB  160 N in Eq. b
Ay  0.866  160  60
 Ay  198.56 N
Problem-2.28 The man has a weight of 150 lb and stands at the end of the beam shown. Determine
the magnitude and direction of the reaction at the pin A and the tension in the cable.
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Figure for Problem- 2.28
Solution:
Soln. Fig. 2.28
3
 36.87 o
4
 Fx  0 yields T cos 36.87 o  Ax  0
  tan 1
 0.8T  Ax  0
F
y
...
...
...
...
...
(a)
...
...
...
(b)
 0 yields  Ay  T sin 36.87 o  150  0
 0.6T  Ay  150
...
...
 0 gives 150  10  T sin 36.87 o  2  T cos 36.87 o  3  0
 150  1.2 T  2.4 T  0
 3.6 T  1500
 T  416.67 lb
M
A
Put this value of T in Eq. a
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0.8  416.67  Ax  0
 Ax  333.33 lb
Now put T  416.67 lb in Eq. b
0.6  416.67  Ay  150
 Ay  100.00 lb
FA  Ax2  Ay2
 333.332  100.00 2  348.00 lb
Direction of FA ,
Ay
100.00
  tan 1
 tan 1
 16.70 o
Ax
333.33
Problem-2.29 The 100-kg uniform beam AB shown in the figure is supported at A by a pin and at B
and C by a continuous cable which wraps around a frictionless pulley located at D. If a maximum
tension force of 800 N can be developed in the cable before it breaks, determine the greatest length b
of a uniform 2.5 kN/m distributed load that can be placed on the beam. What are the horizontal and
vertical components of reaction at A just before the cable breaks?
Figure for Problem- 2.29
Solution:
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Soln. Fig. 2.29
Referring to Soln. Fig. 2.29
 Fx  0 yields  Ax  800 cos 60 o  0
 Ax  400 N
F
y
 0 yields Ay  2500 b  981  800 sin 60o  800  0
 Ay  2500 b  511.82  0
...
...
...
...
...
(a)
and,
b
 0 gives 80010  800 sin 60 o 8  9815  2500 b    0
2
2
 800  5542.56  4905  1250 b  0
 b 2  6.91
 b  2.63 m
M
A
Put this value in Eq. (a) to get
Ay  25002.63  511.82  6063.18 N
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