Chapter 04

CHAPTER 3CHAPTER 7CHAPTER 7CHAPTER 3CHAPTER 4CHAPTER 4

3
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PROBLEM 4.1
A tractor of mass 950 kg is used to lift gravel weighing 4 kN.
Determine the reaction at each of the two (a) rear wheels A,
(b) front wheels B.
SOLUTION
(a) Rear wheels + S - -M AB = + =0 9319 5 1000 4000 1250 2 1500 0: ( . )( ) ( )( ) ( )N mm N mm mm
A = +1439 83. N A =1440 N ≠ 
(b) Front wheels + SM BA: ( . )( ) ( )( ) ( )- - + =9319 5 500 4000 2750 2 1500 0N mm N mm mm
B = +5219 92. N B = 5220 N ≠ b
Check: + = + - - =SF A By 0 2 2 9320 4000: N N 0
2 1440 2 5220 9320 4000 0( ) ( )+ - - =
0 0= ( )Checks

4
PROBLEM 4.2
A gardener uses a 60-N wheelbarrow to transport a 250-N bag of
fertilizer. What force must she exert on each handle?
SOLUTION
Free-Body Diagram:
+ SM FA = - - =0 2 1 60 0 15 250 0 3 0: ( )( ) ( )( . ) ( )( . )m N m N m
F = 42 0. N ≠ 

5
PROBLEM 4.3
The gardener of Problem 4.2 wishes to transport a second 250-N
bag of fertilizer at the same time as the first one. Determine the
maximum allowable horizontal distance from the axle A of the
wheelbarrow to the center of gravity of the second bag if she can
hold only 75 N with each arm.
PROBLEM 4.2 A gardener uses a 60-N wheelbarrow to transport
a 250-N bag of fertilizer. What force must she exert on each handle?
SOLUTION
Free-Body Diagram:
+ SM
x
A = -
- - =
0 2 75 1 60 0 15
250 0 3 250 0
: ( )( ) ( )( . )
( )( . ) ( )
N m N m
N m N

x = 0 264. m 

6
PROBLEM 4.4
For the beam and loading shown, determine (a) the reaction at A, (b) the
tension in cable BC.
SOLUTION
Free-Body Diagram:
(a) Reaction at A SF Ax x= =0 0:
+ SMB = + +
+
0 75 700 100 550 175 350
100 150
: ( )( ) ( )( ) ( )( )
( )(
N mm N mm N mm
N mmm mm) ( )- =Ay 150 0
Ay = +1225 N A =1225 N ≠ 
(b) Tension in BC + SMA = + +
-
0 75 550 100 400 175 200
75 150
: ( )( ) ( )( ) ( )( )
( )(
N mm N mm N mm
N mm)) ( )- =FBC 150 0mm
FBC = +700 N FBC = 700 N 
Check: + SF A Fy BC= - - - - - + - =
- + - =
0 75 100 175 100 75 0
525 1225 700 0
: N N N N N
0 0= ( )Checks

7
PROBLEM 4.5
Two crates, each of mass 350 kg, are placed as shown in the bed
of a 1400-kg pickup truck. Determine the reactions at each of
the two (a) rear wheels A, (b) front wheels B.
SOLUTION
Free-Body Diagram:

W
Wt
= =
= =
( )( . ) .
( )( . ) .
350 9 81 3 434
1400 9 81 13 734
kg m/s kN
kg m/s kN
2
2

(a) Rear wheels + SM W W W AB t= + + + - =0 1 7 2 05 1 2 2 3 0: ( . ) ( . ) ( . ) ( )m 2.05 m m m m

( . )( . ) ( . )( . )
( . )( . ) ( )
3 434 3 75 3 434 2 05
13 734 1 2 2 3
kN m kN m
kN m m
+
+ - =A 00

A = +6 0663. kN A = 6 07. kN ≠ 
(b) Front wheels + SF W W W A By t= - - - + + =0 2 2 0:
- - - + + =3 434 3 434 13 734 0. . .kN kN kN 2(6.0663 kN) 2B
B = +4 2347. kN B = 4 23. kN ≠ 

8
PROBLEM 4.6
Solve Problem 4.5, assuming that crate D is removed and that
the position of crate C is unchanged.
PROBLEM 4.5 Two crates, each of mass 350 kg, are placed
as shown in the bed of a 1400-kg pickup truck. Determine the
reactions at each of the two (a) rear wheels A, (b) front wheels B.
SOLUTION
Free-Body Diagram:

W
Wt
= =
= =
( )( . ) .
( )( . ) .
350 9 81 3 434
1400 9 81 13 734
kg m/s kN
kg m/s kN
2
2

(a) Rear wheels + SM W W AB t= + + - =0 1 7 1 2 2 3 0: ( . ) ( . ) ( )m 2.05 m m m
( . )( . ) ( . )( . ) ( )3 434 3 75 13 734 1 2 2 3 0kN m kN m m+ - =A
A = +4 893. kN A = 4 89. kN ≠ 
(b) Front wheels + SM W W A By t= - - + + =0 2 2 0:
- - + + =3 434 13 734 0. .kN kN 2(4.893 kN) 2B
B = +3 691. kN B = 3 69. kN ≠ 

9
PROBLEM 4.7
A T-shaped bracket supports the four loads shown. Determine the
reactions at A and B (a) if a = 250mm, (b) if a =175 mm. Neglect the
weight of the bracket.
SOLUTION
Free-Body Diagram:
+
SF Bx x= =0 0:
+ SM a a AB = - - + + =0 200 150 150 50 200 300 0: ( )( ) ( ) ( )( ) ( )N mm N N mm mm
A
a
=
-( )200 20000
300
(1)
+ SM a
a
A = - - - +
-
0 200 150 250 300 150 300
50
: ( )( ) ( )( ) ( )( )
( )(
N mm N mm N mm
N ++ + =500 300 0mm mm) ( )By
B
a
y =
+( )175000 200
300

Since B B
a
x = =
+
0
175000 200
300
( )
(2)
(a) For mma = 250
Eq. (1): A =
¥ -
= +
( )200 250 20000
300
100 N A =100 0. N Ø 
Eq. (2): B =
+ ¥
= +
( )175000 200 250
300
750 N B = 750 N Ø 
(b) For mma =175
Eq. (1): A =
¥ -
= +
( )200 175 20000
300
50 N A = 50 0. N Ø 
Eq. (2): B =
+ ¥
= +
( )175000 200 175
300
700 N B = 700 N ≠ 

10
PROBLEM 4.8
For the bracket and loading of Problem 4.7, determine the smallest
distance a if the bracket is not to move.
PROBLEM 4.7 A T-shaped bracket supports the four loads shown.
Determine the reactions at A and B (a) if a = 250 mm, (b) if a =175 mm.
Neglect the weight of the bracket.
SOLUTION
Free-Body Diagram:
For no motion, reaction at A must be downward or zero; smallest distance a for no motion corresponds to
A = 0.
+ SM a a AB = - - + + =0 200 150 150 50 200 300 0: ( )( ) ( ) ( )( ) ( )N mm N N mm mm
A
a
=
-( )200 20000
300

A a= - =0 200 20000 0: ( ) a =100 0. mm 

11
PROBLEM 4.9
The maximum allowable value of each of the reactions is 180 N.
Neglecting the weight of the beam, determine the range of the
distance d for which the beam is safe.
SOLUTION
SF Bx x= =0 0:
B By=
+ SM d d d B dA = - - - - + - =0 50 100 0 45 150 0 9 0 9 0: ( ) ( )( . ) ( )( . ) ( . )N N m N m m
50 45 100 135 150 0 9d d d B Bd- + - + + -.
d
B
A B
=
◊ -
-
180 0 9
300
N m m( . )
(1)
+ SM A dB = - - + =0 50 0 9 0 9 100 0: ( )( . ) ( . ) (N m m N)(0.45 m)
45 0 9 45 0- + + =. A Ad
d
A
A
=
- ◊( . )0 9 90m N m
(2)
Since B Յ180 N, Eq. (1) yields
d Ն
180 0 9 180
300 180
18
120
0 15
-
-
= =
( . )
. m d Ն150 0. mm v
Since AՅ180 N, Eq. (2) yields
d Յ
( . )
.
0 9 180 90
180
72
180
0 40
-
= = m d Յ 400 mm v
Range: 150 0. mm 400 mmՅ Յd 

12
PROBLEM 4.10
Solve Problem 4.9 if the 50-N load is replaced by an 80-N load.
PROBLEM 4.9 The maximum allowable value of each of the reactions
is 180 N. Neglecting the weight of the beam, determine the range of
the distance d for which the beam is safe.
SOLUTION
SF Bx x= =0 0:
B By=
+ SM d d d B dA = - - - - + - =0 80 100 0 45 150 0 9 0 9 0: ( ) ( )( . ) ( )( . ) ( . )N N m N m m
80 45 100 135 150 0 9 0d d d B Bd- + - + + - =.
d
B
B
=
◊ -
-
180 0 9
330
N m
N
.
(1)
+ SM A dB = - - + =0 80 0 9 0 9 100 0: ( )( . ) ( . ) (N m m N)(0.45 m)
d
A
A
=
-0 9 117.
(2)
Since B Յ180 N, Eq. (1) yields
d Ն ( . ) ( ) .180 0 9 180 330 180
18
150
0 12- ¥ - = =/ m d =120 0. mm v
Since AՅ180 N, Eq. (2) yields
d Յ ( . ) .0 9 180 112
45
180
0 25¥ - = =/180 m d = 250 mm v
Range: 120 0. mm 250 mmՅ Յd 

13
PROBLEM 4.11
For the beam of Sample Problem 4.2, determine the range of values
of P for which the beam will be safe, knowing that the maximum
allowable value of each of the reactions is 135 kN and that the reaction
at A must be directed upward.
SOLUTION
SF Bx x= =0 0:
B = By ≠
+ SM P BA = - + - - =0 0 9 2 7 27 3 3 27: ( . ) ( . ) ( )( . ) (m m kN m kN)(3.9 m) 0
P B= -3 216 kN (1)
+ SM A PB = - + - - =0 2 7 1 8 27 0 6 27 0: ( . ) ( . ) ( )( . ) (m m kN m kN)(1.2 m)
P A= +1 5 27. kN (2)
Since B Յ135 kN, Eq. (1) yields
P Յ ( )( )3 135 216- kN P Յ189 0. kN v
Since 0 135Յ ՅA kN, Eq. (2) yields
0 27 1 5 135 27+ +kN Յ ՅP ( . )( )
27 229 5kN kNՅ ՅP . v
Range of values of P for which beam will be safe:
27 0. kN 189.0 kNՅ ՅP 

14
PROBLEM 4.12
The 10-m beam AB rests upon, but is not attached to, supports at C
and D. Neglecting the weight of the beam, determine the range of values
of P for which the beam will remain in equilibrium.
SOLUTION
Free-Body Diagram:
+ SM P DC = - - + =0 2 4 3 20 8 6: ( ) ( )( ) ( )( ) (m kN m kN m m) 0
P D= -86 3kN (1)
+ SM P CD = + - - =0 8 4 3 20 2 6 0: ( ) ( )( ) ( )( ) (m kN m kN m m)
P C= +3 5 0 75. .kN (2)
For no motion C Ն 0 and D Ն 0
For C Ն 0 from (2) P Յ 3 50. kN
For D Ն 0 from (1) P Յ86 0. kN
Range of P for no motion:
3 50. kN 86.0 kNՅ ՅP 

15
PROBLEM 4.13
The maximum allowable value of each of the reactions is 50 kN, and
each reaction must be directed upward. Neglecting the weight of the
beam, determine the range of values of P for which the beam is safe.
SOLUTION
Free-Body Diagram:
+ SM P DC = - - + =0 2 4 3 20 8 6: ( ) ( )( ) ( )( ) (m kN m kN m m) 0
P D= -86 3kN (1)
+ SM P CD = + - - =0 8 4 3 20 2 6 0: ( ) ( )( ) ( )( ) (m kN m kN m m)
P C= +3 5 0 75. .kN (2)
For C Ն 0, from (2): P Ն 3 50. kN v
For D Ն 0, from (1): P Յ86 0. kN v
For C Յ 50 kN, from (2):

P
P
Յ
Յ
3 5 0 75 50
41 0
. . ( )
.
kN kN
kN
+
v
For D Յ 50 kN, from (1):

P
P
Ն
Ն
86 3 50
64 0
kN kN
kN
-
-
( )
. v
Comparing the four criteria, we find
3 50. kN 41.0 kNՅ ՅP 

16
PROBLEM 4.14
For the beam and loading shown, determine the range of the
distance a for which the reaction at B does not exceed 450-N
downward or 900-N upward.
SOLUTION
Assume B is positive when directed ≠
Sketch showing distance from D to forces.
+ SM a a BD = - - - - + =0 1350 200 1350 50 225 100 400: ( )( ) ( )( ) ( )( ) ( )N mm N N mm 00
- + + =2700 315000 400 0a B
a
B
=
+( )315000 400
2700
(1)
For B = 450 N Ø = -450 N, Eq. (1) yields:
a Ն
[ ( )]315000 400 450
2700
+ -
a Ն 50 0. mm v
For B = 900 N ≠ = +900 N, Eq. (1) yields:
a Յ
[ ( )]315000 400 900
2700
+
a Յ 250 mm v
Required range: 50 0 250. mm mmՅ Յa 

17
PROBLEM 4.15
Two links AB and DE are connected by a bell crank as
shown. Knowing that the tension in link AB is 720 N,
determine (a) the tension in link DE, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
+ SM F FC AB DE= - =0 100 120 0: ( ) ( )mm mm
F FDE AB=
5
6
(1)
(a) For FAB = 720 N
FDE =
5
6
720( )N FDE = 600 N 
(b)
+
SF Cx x= - + =0
3
5
720 0: ( )N
Cx = +432 N
+ SF C
C
y y
y
= - + - =
= +
0
4
5
720 600
1176
: ( )N N 0
N

C =
= ∞
1252 84
69 829
.
.
N
a

C =1253 N 69.8° 

18
PROBLEM 4.16
Two links AB and DE are connected by a bell crank as
shown. Determine the maximum force that may be safely
exerted by link AB on the bell crank if the maximum
allowable value for the reaction at C is 1600 N.
SOLUTION
See solution to Problem 4.15 for F. B. D. and derivation of Eq. (1)
F FDE AB=
5
6
(1)
+
SF F C C Fx AB x x AB= - + = =0
3
5
0
3
5
:
+ SF F C Fy AB y DE= - + - =0
4
5
0:

- + - =
=
4
5
5
6
0
49
30
F C F
C F
AB y AB
y AB

C C C
F
C F
x y
AB
AB
= +
= +
=
2 2
2 21
30
49 18
1 74005
( ) ( )
.

For C FAB= =1600 1 74005N, 1600 N . FAB = 920 N 

19
PROBLEM 4.17
The required tension in cable AB is 900 N. Determine (a) the vertical
force P that must be applied to the pedal, (b) the corresponding reaction
at C.
SOLUTION
Free-Body Diagram:
BC =175 mm
(a) + SM PC = - =0 375 900 151 5544 0: ( ) ( )( . )mm N mm
P = 363 731. N P = 364 N Ø 
(b) + SF Cx x= - =0 900: N 0 Cx = 900 N Æ
+ SF C P Cy y y= - = - =0 0 363 731: . N 0 Cy = 364 N ≠
a = ∞
=
22 0
970 722
.
.C N

C = 971 N 22.0∞ 

20
PROBLEM 4.18
Determine the maximum tension that can be developed in cable AB if
the maximum allowable value of the reaction at C is 1125 N.
SOLUTION
Free-Body Diagram:
BC =175 mm
+ SM P T P TC = - = =0 375 151 5544 0 0 40415: ( ) ( . ) .mm mm
+ SF P C T Cy y y= - + = - + =0 0 0 40415: . 0
C Ty = 0 40415.

+ SF T Cx x= - + =0: 0 C Tx =

C C C T T
C T
x y= + = +
=
2 2 2 2
0 40415
1 0786
( . )
.

For C =1125 N
1125 1 0786 1043 02N N= ﬁ =. .T T
T =1043 N 

21
PROBLEM 4.19
The bracket BCD is hinged at C and attached to a control cable
at B. For the loading shown, determine (a) the tension in the
cable, (b) the reaction at C.
SOLUTION
At B:
T
T
y
x
=
0 18
0 24
.
.
m
m

T Ty x=
3
4
(1)
(a) + SM TC x= - - =0 0 18 240 0 4 240 0 8 0: ( . ) ( )( . ) ( )( . )m N m N m
Tx = +1600 N
Eq. (1) Ty = =
3
4
1600 1200( )N N
T T Tx y= + = + =2 2 2 2
1600 1200 2000 N T = 2 00. kN 
(b) + SF C Tx x x= - =0: 0
C Cx x- = = +1600 1600N 0 N Cx =1600 N Æ
+ SF C Ty y y= - - - =0 240 240 0: N N
Cy - - =1200 480 0N N
Cy = +1680 N Cy =1680 N ≠

a = ∞
=
46 4
2320
.
C N C = 2 32. kN 46.4° 

22
PROBLEM 4.20
Solve Problem 4.19, assuming that a = 0.32 m.
PROBLEM 4.19 The bracket BCD is hinged at C and attached
to a control cable at B. For the loading shown, determine
(a) the tension in the cable, (b) the reaction at C.
SOLUTION
At B:
T
T
T T
y
x
y x
=
=
0 32
0 24
4
3
.
.
m
m

+ SM TC x= - - =0 0 32 240 0 4 240 0 8 0: ( . ) ( )( . ) ( )( . )m N m N m
Tx = 900 N
Eq. (1) Ty = =
4
3
900 1200( )N N
T T Tx y= + = + =2 2 2 2
900 1200 1500 N T =1 500. kN 
+ SF C Tx x x= - =0: 0
C Cx x- = = +900 900N 0 N Cx = 900 N Æ
+ SF C Ty y y= - - - =0 240 240 0: N N
Cy - - =1200 480 0N N
Cy = +1680 N Cy =1680 N ≠

a = ∞
=
61 8
1906
.
C N C =1 906. kN 61.8° 

23
PROBLEM 4.21
DeterminethereactionsatAandBwhen(a) h = 0, (b) h = 200 mm.
SOLUTION
Free-Body Diagram:
+ SM B B hA = ∞ - ∞ - =0 60 0 5 60 150 0 25 0: ( cos )( . ) ( sin ) ( )( . )m N m
B
h
=
-
37 5
0 25 0 866
.
. .
(1)
(a) When h = 0:
Eq. (1): B = =
37 5
0 25
150
.
.
N B =150 0. N 30.0° 
+ SF A By x= - ∞ =0 60 0: sin
Ax = ∞ =( )sin .150 60 129 9 N Ax =129 9. N Æ
+ SF A By y= - + ∞ =0 150 60 0: cos
Ay = - ∞ =150 150 60 75( )cos N Ay = 75 N ≠

a = ∞
=
30
150 0A . N A =150 0. N 30.0° 
(b) When h = 200 mm = 0.2 m
Eq. (1): B =
-
=
37 5
0 25 0 866 0 2
488 3
.
. . ( . )
. N B = 488 N 30.0° 
+ SF A Bx x= - ∞ =0 60 0: sin
Ax = ∞ =( . )sin .488 3 60 422 88 N Ax = 422 88. N Æ
+ SF A By y= - + ∞ =0 150 60 0: cos
Ay = - ∞ = -150 488 3 60 94 15( . )cos . N Ay = 94 15. N Ø

a = ∞
=
12 55
433 2
.
.A N A = 433 N 12.6° 

24
PROBLEM 4.22
For the frame and loading shown, determine the reactions at A and E
when (a) a = ∞30 , (b) a = ∞45 .
SOLUTION
Free-Body Diagram:
+ SM E EA = +
- -
0 200 125
90 250 90 75
: ( sin )( ) ( cos )( )
( )( ) ( )(
a amm mm
N mm N mmm) = 0

E =
+
29250
200 125sin cosa a

(a) When a = 30°: E =
∞ + ∞
=
29250
200 30 125 30
140 454
sin cos
. N E =140 5. N 60.0° 
+ SF Ax x= - + ∞ =0 90 30 0: sinN (140.454 N)
Ax = +19 773. N Ax =19 77. N Æ
+ SF Ay y= - + ∞ =0 90 140 454 30 0: ( . )cosN N
Ay = -31 637. N Ay = 31 7. N Ø
A = 37 3. N 58.0° 

25
PROBLEM 4.22 (Continued)
(b) When a = ∞45 :
E =
∞ +
=
29250
200 45 125
127 279
sin cos
.
a
N E =127 3. N 45.0° 
+ SF Ax x= - + ∞ =0 90 127 279 45 0: ( . )sinN N
Ax = 0 Ax = 0
+ SF Ay y= - + ∞ =0 90 127 279 45 0: ( . )cosN
Ay = 0 Ay = 0 A = 0 

26
PROBLEM 4.23
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a) Free-Body Diagram:
+ SM BA = - - =0 500 250 100 200 250 0: ( ) ( )( ) ( )( )mm N mm N mm
B = +150 N B =150 0. N ≠
+
SF Ax x= + =0 200 0: N
Ax = -200 N Ax = 200 N ¨
+ SF A By y= + - =0 250: N 0
Ay + - =150 250 0N N
Ay = +100 N Ay =100 0. N ≠

a = ∞
=
26 56
223 607
.
.A N A = 224 N 26.6° 

27
(b) Free-Body Diagram:
+ SM BA = ∞ - - =0 30 500 200 250 250 100 0: ( cos )( ) ( )( ) ( )( )mm N mm N mm
B =173 205. N B =173 2. N 60.0° 
+ SF A Bx x= - ∞ +0 30 200: sin N
Ax - ∞ + =( . )sin173 205 30 200 0N N
Ax = -113 3975. N Ax =113 4. N ¨
+ SF A By y= + ∞ - =0 30 250: cos N 0
Ay + ∞ - =( . )cos173 205 30 250 0N
Ay = +100 N Ay =100 0. N ≠

a =
=
41 4
151 192
.
.
∞
A N A =151 2. N 41.4° 

28
PROBLEM 4.24
For each of the plates and loadings shown, determine the reactions at A and B.
SOLUTION
(a) Free-Body Diagram:
+ SM AB = - + - =0 500 250 400 200 250 0: ( ) ( )( ) ( )( )mm N mm N mm
A = +100 N A =100 0. N ≠ 
+
SF Bx x= + =0 200 0: N
Bx = -200 N Bx = 200 N ¨
+ SF A By y= + - =0 250: N 0
100 250 0N N+ - =By
By = +150 N By =150 0. N ≠

a = ∞
=
36 87
250
.
B N B = 250 N 36.9° 

29
(b)
+ SM AA = - ∞ - + =0 30 500 200 250 250 400 0: ( cos )( ) ( )( ) ( )( )mm N mm N mm
A =115 47. N A =115 5. N 60.0° 
+ SF A Bx x= ∞ + + =0 30 200 0: sin N
( . )sin115 47 30 200 0N N∞ + + =Bx
Bx = -257 735. N Bx = 258 N ¨
+ SF A By y= ∞ + - =0 30 250: cos N 0
( . )cos115 47 30 250N N 0∞ + - =By
By = +150 N By =150 0. N ≠

a = ∞
=
30 2
298 207
.
.B N B = 298 N 30.2° 

30
PROBLEM 4.25
Determine the reactions at A and B when (a) a = 0, (b) a = 90 ,∞
(c) a = 30 .∞
SOLUTION
(a) a = 0∞
+ SM BA = - ◊ =0 0 5 100 0: ( . )m N m
B = 200 N
+
SF Ax x= =0 0:
+ SF Ay y= + =0 200 0:
Ay = -200 N
A = 200 N Ø B = 200 N ≠ 
(b) a = ∞90
+ SM BA = - ◊ =0 0 3 100: ( . )m N m 0
B = =
1000
3
333 33N N.
+
SF A AA x x= - = =0 333 33 0 333 33: . , .N N
+ SF Ay y= =0 0:
A = 333 N Æ B = 333 N ¨ 
(c) a = ∞30
+ SM B BA = ∞ + ∞ - ◊ =0 30 0 5 30 0 3 100: ( cos )( . ) ( sin )( . )m m N m 0
B =171 523. N
+
SF Ax x= - ∞ =0 171 523 30 0: ( . )sinN
Ax = 85 762. N
+ SF A Ay y y= + ∞ = = -0 171 523 0 148 543: ( . cos .N) 30 N
A = = ∞171 523. N 60.0a
A =171 5. N 60.0° B =171 5. N 60.0° 

31
PROBLEM 4.26
A rod AB hinged at A and attached at B to cable BD supports
the loads shown. Knowing that d = 200 mm, determine (a) the
tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a) Move T along BD until it acts at Point D.
+ SM TA = ∞ + + =0 45 0 2 90 0 1 90 0 2 0: ( sin )( . ) ( )( . ) ( )( . )m N m N m
T =190 92. N T =190 9. N 
(b) + SF Ax x= - ∞ =0 190 92 45 0: ( . )cosN
Ax = +135 N Ax =135 0. N Æ
+ SF Ay y= - - + ∞ =0 90 90 0: sinN N (190.92 N) 45
Ay = +45 N Ay = 45 0. N ≠
A =142 3. N 18.43°

32
PROBLEM 4.27
A rod AB hinged at A and attached at B to cable BD supports
the loads shown. Knowing that d =150 mm, determine (a) the
tension in cable BD, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
tan ;a a= = ∞
10
15
33.69
(a) Move T along BD until it acts at Point D.
+ SM TA = ∞ - - =0 33 69 0 15 90 0 1 90 0 2 0: ( sin . )( . ) ( )( . ) ( )( . )m N m N m
T = 324 5. N T = 324 N 
(b) + SF Ax x= - ∞ =0 324 99 33 69 0: ( . )cos .N
Ax = +270 N Ax = 270 N Æ
+ SF Ay y= - - + ∞ =0 90 90 33 69 0: sin .N N (324.5 N)
Ay = 0 Ay = 0 A = 270 N Æ 

33
PROBLEM 4.28
A lever AB is hinged at C and attached to a control cable at A. If the
lever is subjected to a 500-N horizontal force at B, determine (a) the
tension in the cable, (b) the reaction at C.
SOLUTION
Triangle ACD is isosceles with S C = ∞ + ∞ = ∞90 30 120 S SA D= = ∞ - ∞ = ∞
1
2
180 120 30( )
Thus DA forms angle of 60° with horizontal.
(a) We resolve FAD into components along AB and perpendicular to AB.
+ SM FC AD= ∞ - =0 30 250 500 100 0: ( sin )( ) ( )( )mm N mm FAD = 400 N 
(b) +
SF Cx x= - ∞ + - =0 400 60 500 0: ( )cosN N Cx = +300 N
+ SF Cy y= - ∞ + =0 400 0: ( sinN) 60 Cy = +346 4. N
C = 458 N 49.1° 

34
PROBLEM 4.29
A force P of magnitude 1260 N is applied to member ABCD, which is
supported by a frictionless pin at A and by the cable CED. Since the cable
passes over a small pulley at E, the tension may be assumed to be the
same in portions CE and ED of the cable. For the case when a = 75 mm,
determine (a) the tension in the cable, (b) the reaction at A.
SOLUTION
Free-Body Diagram:
(a) + SM
T T
T
A = -
-
- =
0 1260 75
300
7
25
300
24
25
200 0
: ( )( )
( ) ( )
( )
N mm
mm mm
mm

24 94500
3937 5
T
T
=
= . N T = 3940 N 
(b) + SF Ax x= + + =0
7
25
3937 5 3937 5 0: ( . ) .
Ax = -5040 N Ax = 5040 N ¨
+ SF Ay y= - - =0 1260
24
25
3937 5 0: ( . )N N
Ay = +5040 N Ay = 5040 N ≠
A = 7128 N 45.0° 

35
PROBLEM 4.30
Neglecting friction, determine the tension in cable ABD and the reaction
at support C.
SOLUTION
Free-Body Diagram:
+ SM T TC = - - =0 0 25 0 1 120 0 1 0: ( . ) ( . ) ( )( . )m m N m T = 80 0. N 
+ SF C Cx x x= - = = +0 80 80: N 0 N Cx = 80 0. N Æ
+ SF C Cy y y= - + = = +0 120 40: N 80 N 0 N Cy = 40 0. N ≠
C = 89 4. N 26.6° 

36
PROBLEM 4.31
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the q = 30°,
determine the reaction (a) at B, (b) at C.
SOLUTION
Free-Body Diagram: + SM C R P RD x= - =0 0: ( ) ( )
C Px = +
+ SF C Bx x= - =0 0: sinq

P B
B P
- =
=
sin
/sin
q
q
0

B =
P
sinq
q
+ SF C B Py y= + - =0 0: cosq
C P P
C P
y
y
+ - =
= -
Ê
ËÁ
ˆ
¯˜
( /sin )cos
tan
q q
q
0
1
1

For q = ∞30 :
(a) B P P= ∞ =/sin30 2 B = 2P 60.0° 
(b) C P C Px x= + = Æ
C P Py = - ∞ = -( /tan ) . /1 1 30 0 732 Cy P= 0 7321. Ø
C =1 239. P 36.2° 

37
PROBLEM 4.32
Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the q = 60°,
determine the reaction (a) at B, (b) at C.
SOLUTION
See the solution to Problem 4.31 for the free-body diagram and analysis leading to the following expressions:
C P
C P
B
P
x
y
= +
= -
Ê
ËÁ
ˆ
¯˜
=
1
1
tan
sin
q
q

For q = ∞60 :
(a) B P P= ∞ =/sin .60 1 1547 B =1 155. P 30.0° 
(b) C P C Px x= + = Æ
C P Py = - ∞ = +( /tan ) .1 1 60 0 4226 Cy P= 0 4226. Ø
C =1 086. P 22.9° 

38
PROBLEM 4.33
at C when q = 60°.
SOLUTION
+ SM T a a Ta PaC = + - + =0 2 0: ( cos )q
T
P
=
+1 cosq
(1)

+
SF C Tx x= - =0 0: sinq
C T
P
x = =
+
sin
sin
cos
q
q
q1

+ SF C T T Py y= + + - =0 0: cosq
C P T P P
C
y
y
= - + = -
+
+
=
( cos )
cos
cos
1
1
1
0
q
q
q

Since C C Cy x= =0, C =
+
P
sin
cos
q
q1
Æ (2)
For q = ∞60 :
Eq. (1): T
P P
=
+ ∞
=
+1 60 1 1
2cos
T P=
2
3

Eq. (2): C P P=
∞
+ ∞
=
+
sin
cos
.60
1 60
0 866
1 1
2
C = 0 577. P Æ 

39
PROBLEM 4.34
at C when q = 45°.
SOLUTION
Free-Body Diagram:
Equilibrium for bracket:
+ SM T a P a T a
T a a
C = - - + ∞ ∞
+ ∞ + ∞
0 45 2 45
45 2 45
: ( ) ( ) ( sin )( sin )
( cos )( cos ) == 0

T = 0 58579. or T P= 0 586. 
+ SF C Px x= + ∞ =0 0 58579 45 0: ( . )sin
C Px = 0 41422.
+ SF C P P Py y= + - + ∞ =0 0 58579 0 58579 45 0: . ( . )cos
Cy = 0 or C = 0 414. P Æ 

40
PROBLEM 4.35
A light rod AD is supported by frictionless pegs at B and C and rests
against a frictionless wall at A. A vertical 600 N force is applied
at D. Determine the reactions at A, B, and C.
SOLUTION
Free-Body Diagram:
SF Ax = ∞ - ∞ =0 30 600 60 0: cos ( )cosN
A = 346 4102. N A = 346 N Æ 
+ SM CB = - ∞
+
0 200 600 400 30
346 4102 200 30
: ( ) ( )( )cos
( . )( )sin
mm N mm
N mm ∞∞ = 0

C = 866 0253. N C = 866 N 60.0° 
+ SM BC = - ∞
+
0 200 600 200 30
346 4102 400 30
: ( ) ( )( )cos
( . )( )sin
mm N mm
N mm ∞∞ = 0

B =173 205. N B =173 2. N 60.0° 

41
PROBLEM 4.36
A light bar AD is suspended from a cable BE and supports a 250-N
block at C. The ends A and D of the bar are in contact with frictionless
vertical walls. Determine the tension in cable BE and the reactions
at A and D.
SOLUTION
Free-Body Diagram:
SF A Dx = =0:
SFy = 0: TBE = 250 N 
We note that the forces shown form two couples.
+ SM A= - =0 200 250 75 0: ( ) ( )( )mm N mm
A = 93 75. N
A = 93 75. N Æ D = 93 75. N ¨ 

42
PROBLEM 4.37
Bar AC supports two 400-N loads as shown. Rollers at A and C
rest against frictionless surfaces and a cable BD is attached at
B. Determine (a) the tension in cable BD, (b) the reaction at A,
(c) the reaction at C.
SOLUTION
Similar triangles: ABE and ACD

AE
AD
BE
CD
BE
BE= = =;
.
. .
; .
0 15
0 5 0 25
0 075
m
m m
m
(a) + SM T TA x x= -
Ê
ËÁ
ˆ
¯˜ - -0 0 25
0 075
0 35
0 5 400 0 1 400: ( . )
.
.
( . ) ( )( . ) (m m N m N))( . )0 4 0m =
Tx =1400 N

Ty =
=
0 075
0 35
1400
300
.
.
( )N
N T =1432 lb 

43
(b) + SF Ay = - - - =0 300 400 400 0: N N N
A = +1100 N A =1100 N ≠ 
(c) + SF Cx = - + =0 1400 0: N
C = +1400 N C =1400 N ¨ 

44
PROBLEM 4.38
Determine the tension in each cable and the reaction at D.
SOLUTION

tan
.
.
tan
.
.
a
a
b
b
=
= ∞
=
= ∞
0 08
21 80
0 08
38 66
m
0.2 m
m
0.1 m

+ SM TB CF= - ∞ =0 600 0 1 38 66 0 1 0: ( )( . ) ( sin . )( . )N m m
TCF = 960 47. N TCF = 96 0. N 
+ SM TC BE= - ∞ =0 600 0 2 21 80 0 1 0: ( )( . ) ( sin . )( . )N m m
TBE = 3231 1. N TBE = 3230 N 
+ SF T T Dy BE CF= + - =0 0: cos cosa b
( . )cos . ( . )cos .3231 1 21 80 960 47 38 66 0N N∞ + ∞ - =D
D = 3750 03. N D = 3750 N ¨ 

45
PROBLEM 4.39
Two slots have been cut in plate DEF, and the plate has
been placed so that the slots fit two fixed, frictionless pins A
and B. Knowing that P = 75 N, determine (a) the force each
pin exerts on the plate, (b) the reaction at F.
SOLUTION
Free-Body Diagram:
+ SF Bx = - ∞ =0 75 30 0: sinN B =150 N 60.0° 
+ SM B B FA = - + ∞ + ∞ -0 150 100 30 75 30 275 325: ( )( ) sin ( ) cos ( ) ( )N mm mm mm mm == 0
- + + - =15000 5625 35723 55 325 0. ( )F
F = +81 0725. N F = 81 1. N Ø 
+ SF A B Fy = - + ∞ - =0 150 30 0: cosN
A - + ∞ - =150 150 30 81 0725 0N N N( )cos .
A = +101 169. N A =101 2. N ≠ 

46
PROBLEM 4.40
For the plate of Problem 4.39 the reaction at F must be
directed downward, and its maximum allowable value
is 100 N. Neglecting friction at the pins, determine the
required range of values of P.
PROBLEM 4.39 Two slots have been cut in plate DEF,
and the plate has been placed so that the slots fit two fixed,
frictionless pins A and B. Knowing that P = 75 N, determine
(a) the force each pin exerts on the plate, (b) the reaction at F.
SOLUTION
Free-Body Diagram:
+
SF P Bx = - ∞ =0 30 0: sin B = 2P 60°
+ SM B B FA = - + ∞ + ∞ -0 150 100 30 75 30 275 325: ( )( ) sin ( ) cos ( ) ( )N mm mm mm mm == 0

- ◊ ∞ + ∞ - =
-
15000 2 30 75 2 30 275 325 0
15000
N mm+ mm mm mmP P Fsin ( ) cos ( ) ( )
++ + - =75 476 314 325 0P P F.
P
F
=
+325 15000
551 314. (1)
For F = 0: P = =
15000
551 314
27 208
.
. N
For P =100 N: P =
+
=
32500 15000
551 314
86 158
.
. N
For 0 100Յ ՅF N 27 2. N 86.2 NՅ ՅP 

47
PROBLEM 4.41
Bar AD is attached at A and C to collars that can move freely on
the rods shown. If the cord BE is vertical (a = 0), determine the
tension in the cord and the reactions at A and C.
SOLUTION
Free-Body Diagram:
+ SF Ty = - ∞ + ∞ =0 30 80 30 0: cos ( )cosN
T = 80 N T = 80 0. N 
+ SM AC = ∞ - - =0 30 0 4 80 0 2 80 0 2 0: ( sin )( . ) ( )( . ) ( )( . )m N m N m
A = +160 N A =160 0. N 30.0° 
+ SM CA = - + ∞ =0 80 0 2 80 0 6 30 0 4 0: ( )( . ) ( )( . ) ( sin )( . )N m N m m
C = +160 N C =160 0. N 30.0° 

48
PROBLEM 4.42
Solve Problem 4.41 if the cord BE is parallel to the rods
(a = 30°).
PROBLEM 4.41 Bar AD is attached at A and C to collars that
can move freely on the rods shown. If the cord BE is vertical
(a = 0), determine the tension in the cord and the reactions at
A and C.
SOLUTION
Free-Body Diagram:
+
SF Ty = - + ∞ =0 80 30 0: ( )cosN
T = 69 282. N T = 69 3. N 
+ SM
A
C = - ∞
- + ∞ =
0 69 282 30 0 2
80 0 2 30 0 4 0
: ( . )cos ( . )
( )( . ) ( sin )( . )
N m
N m m

A = +140 000. N A =140 0. N 30.0° 
+ SM
C
A = + ∞
- + ∞ =
0 69 282 30 0 2
80 0 6 30 0 4 0
: ( . )cos ( . )
( )( . ) ( sin )( . )
N m
N m m

C = +180 000. N C =180 0. N 30.0° 

49
PROBLEM 4.43
An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm
radius, determine the reaction at A in each case.
SOLUTION
W mg= = =( )( . ) .8 9 81 78 48kg m/s N2

(a) SF Ax x= =0 0:
+ SF A Wy y= - =0 0: Ay = 78 48. N ≠
+ SM M WA A= - =0 1 6 0: ( . )m
MA = +( . )( . )78 48 1 6N m MA = ◊125 56. N m
A = 78 5. N ≠ MA = ◊125 6. N m 
(b) + SF A Wx x= - =0 0: Ax = 78 48. ≠
+ SF A Wy y= - =0 0: Ay = 78 48. ¨
A = =( . ) .78 48 2 110 99N N 45°
+ SM M WA A= - =0 1 6 0: ( . )m
MA A= + = ◊( . )( . ) .78 48 1 6 125 56N m N mM
A =111 0. N 45° MA = ◊125 6. N m 
(c) SF Ax x= =0 0:
+ SF A Wy y= - =0 2 0:
A Wy = = =2 2 78 48 156 96( . ) .N N ≠
+ SM M WA A= - =0 2 1 6 0: ( . )m
MA = +2 78 48 1 6( . )( . )N m MA = ◊125 1. N m
A =157 0. N ≠ MA = ◊125 N m 

50
PROBLEM 4.44
A tension of 22.5 N is maintained in a tape as it passes through
the support system shown. Knowing that the radius of each
pulley is 10 mm, determine the reaction at C.
SOLUTION
From f.b.d. of system
+ SF Cx x= + =0 22 5 0: ( . )N
Cx = -22 5. N
+ SF Cy y= - =0 22 5 0: ( . )N
Cy = 22 5. N
Then C C Cx y= +
= +
=
( ) ( )
( . ) ( . )
.
2 2
2 2
22 5 22 5
31 8198 N

and q =
+
-
Ê
ËÁ
ˆ
¯˜ = - ∞-
tan
.
.
1 22 5
22 5
45 or C = 31 8. N 45.0∞ 
+ SM MC C= + + =0 22 5 160 22 5 60 0: ( . )( ) ( . )( )N mm N mm
MC = - ◊4950 N mm or MC = ◊4950 N mm 

51
PROBLEM 4.45
Solve Problem 4.44, assuming that 15 mm radius pulleys are
used.
PROBLEM 4.44 A tension of 22.5 N is maintained in a tape
as it passes through the support system shown. Knowing that
the radius of each pulley is 10 mm, determine the reaction at C.
SOLUTION
From f.b.d. of system
+ SF Cx x= + =0 22 5 0: N( . )
Cx = -22 5. N
+ SF Cy y= - =0 22 5 0: N( . )
Cy = 22 5. N
Then C C Cx y= +
= +
=
( ) ( )
( . ) ( . )
.
2 2
2 2
22 5 22 5
31 8198 N

and q =
-
Ê
ËÁ
ˆ
¯˜ = - ∞-
tan
.
.
.1 22 5
22 5
45 0 or C = 31 8. N 45 0. ∞ 
+ SM MC C= + + =0 22 5 165 22 5 65 0: N mm N mm( . )( ) ( . )( )
MC = - ◊5175 N mm or MC = ◊5180 N mm 

52
PROBLEM 4.46
The frame shown supports part of the roof of a small building.
Knowing that the tension in the cable is 150 kN, determine the
reaction at the fixed end E.
SOLUTION
Free-Body Diagram:
4.5m
1.8m
20kN
D
C
6m
150kN
ME
Ex
My
E F
BA
20kN20kN20kN
1.8m1.8m1.8m
Equilibrium Equations DF = + =( . ( .4 5 6 7 52 2
m) m) m,
+
SF Ex x= + =0
4 5
7 5
150 0:
.
.
( )kN
Ex = -90 0. kN Ex = 90 0. kN ¬ 
+ SF Ey y= - - =0 4 20
6
7 5
150 0: (
.
( )kN) kN
Ey = +200 kN Ey = 200 kN  
+ SME = + +
+
0 20 20 5 4 20 3 6
20
: ( ( )( . ) ( )( . )
(
kN)(7.2m) kN m kN m
kN)(1.8m) --
6
7 5
150 0
.
( kN)(4.5m)+ ME =

ME = + ◊180 0. kN m ME = ◊180 0. kN m 

53
PROBLEM 4.47
Beam AD carries the two 200-N loads shown. The beam is held by a fixed
support at D and by the cable BE that is attached to the counterweight W.
Determine the reaction at D when (a) W = 500 N, (b) W = 450 N.
SOLUTION
(a) W = 500 N
From f.b.d. of beam AD
+ SF Dx x= =0 0:
+ SF Dy y= - - + =0 200 200 500: N N N 0
Dy = -100 N or D =100 0. N Ø 
+ SM MD D= - +
+ =
0 500 1 5 200 2 4
200
: ( )( . ) ( )( . )
(
N m N m
N)(1.2 m) 0

MD = ◊30 N m or MD = ◊30 0. N m 
(b) W = 450 N
From f.b.d. of beam AD
+
SF Dx x= =0 0:
+ SF Dy y= + - - =0 450 200 200: N N N 0
Dy = -50 N or D = 50 0. N Ø 
+ SM MD D= - +
+ =
0 450 1 5 200 2 4
200 1 2 0
: ( )( . ) ( )( . )
( )( . )
N m N m
N m

MD = - ◊45 N m or MD = - ◊45 0. N m 

54
PROBLEM 4.48
For the beam and loading shown, determine the range of values of W for
which the magnitude of the couple at D does not exceed 60 N · m.
SOLUTION
For Wmin , MD = - ◊60 N m
From f.b.d. of beam AD + SM WD = -
+ - ◊ =
0 200 2 4 1 5
200 1 2 60 0
: ( )( . ) ( . )
( )( . )
minN m m
N m N m

Wmin N= 440
For Wmax, MD = ◊60 N m
From f.b.d. of beam AD + SM WD = -
+ + ◊ =
0 200 2 4 1 5
200 1 2 60 0
: ( )( . ) ( . )
( )( . )
maxN m m
N m N m

Wmax = 520 N or N 520 N440 Յ ՅW 

55
PROBLEM 4.49
Knowing that the tension in wire BD is 1300 N, determine the
reaction at the fixed support C of the frame shown.
SOLUTION

T
T T
T T
x
y
=
=
=
=
=
1300
5
13
500
12
13
1200
N
N
N
+ SM C Cx x x= - + = = -0 450 500 50: N N 0 N Cx = 50 N ¨
+ SF C Cy y y= - - = = +0 750 120 1950: N 0 N 0 N Cy =1950 N ≠
C =1951 N 88.5° 
+ SM MC C= + +
- =
0 750 0 5 4 50 0 4
1200 0 4 0
: ( )( . ) ( . )( . )
( )( . )
N m N m
N m

MC = - ◊75 0. N m MC = ◊75 0. N m 

56
PROBLEM 4.50
Determine the range of allowable values of the tension in wire BD if
the magnitude of the couple at the fixed support C is not to exceed
100 N m.◊
SOLUTION
+ SM T
T
C = + -
Ê
ËÁ
ˆ
¯˜
-
Ê
ËÁ
0 750 450 0 4
5
13
0 6
12
13
: ( ( )( . ) ( . )N)(0.5 m) N m m
ˆˆ
¯˜ + =( .0 15 0m) MC

375 180
4 8
13
0N m N m m◊ + ◊ -
Ê
ËÁ
ˆ
¯˜ + =
.
T MC
T MC= +
13
4 8
555
.
( )
For M TC = - ◊ = - =100
13
4 8
555 100 1232N m: N
.
( )
For M TC = + ◊ = + =100
13
4 8
555 100 1774N m: N
.
( )
For | | :MC Յ100 N m◊ 1 232. kN 1.774 kNՅ ՅT 

57
PROBLEM 4.51
A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the
rod, express the angle q corresponding to the equilibrium position in terms of
P, l, and the counterweight W. (b) Determine the value of q corresponding to
equilibrium if P W= 2 .
SOLUTION
(a) Triangle ABC is isosceles.
We have CD BC l= =( )cos cos
q q
2 2
+ SM W l P lC =
Ê
ËÁ
ˆ
¯˜ - =0
2
0: cos ( sin )
q
q
Setting sin sin cos : cos sin cosq
q q q q q
= - =2
2 2 2
2
2 2
0Wl Pl
W P- =2
2
0sin
q
q =
Ê
ËÁ
ˆ
¯˜
-
2
2
1
sin
W
P
b
(b) For P W= 2 : sin .
q
2 2 4
0 25= = =
W
P
W
W

q
2
14 5= ∞. q = ∞29 0. b
or
q
q
2
165 5 331= ∞ = ∞. ( )discard

58
PROBLEM 4.52
A slender rod AB, of weight W, is attached to blocks A and B, which
move freely in the guides shown. The blocks are connected by an
elastic cord that passes over a pulley at C. (a) Express the tension in
the cord in terms of W and q. (b) Determine the value of q for which
the tension in the cord is equal to 3W.
SOLUTION
(a) From f.b.d. of rod AB
+ SM T l W T lC = +
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙ - =0
1
2
0: ( sin ) cos ( cos )q q q
T
W
=
-
cos
(cos sin )
q
q q2
Dividing both numerator and denominator by cos q,
T
W
=
-
Ê
ËÁ
ˆ
¯˜
2
1
1 tanq
or T
W
=
( )
-
2
1( tan )q
b
(b) For T W= 3 , 3
1
1
1
6
2
W
W
=
( )
-
- =
( tan )
tan
q
q
or q =
Ê
ËÁ
ˆ
¯˜ = ∞-
tan .1 5
6
39 806 or q = ∞39 8. b

59
PROBLEM 4.53
Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive
an equation in q, P, M, and l that must be satisfied when the rod is in equilibrium.
(b) Determine the value of q corresponding to equilibrium when M =1500 N m,и
P = 200 N, and l = 600 mm.
SOLUTION
Free-Body Diagram: (a) From free-body diagram of rod AB
+ SM P l P l MC = + - =0 0: ( cos ) ( sin )q q
or sin cosq q+ =
M
Pl
b
(b) For M P l= ◊ = =
+ =
◊
=
1500 600
1500
N m, 200 N, and mm
N m
(200 N)(600 mm)
sin cosq q
55
4
1 25= .
Using identity sin cos
sin ( sin ) .
( sin ) . sin
2 2
2 1 2
2 1 2
1
1 1 25
1 1 25
1
q q
q q
q q
+ =
+ - =
- = -
/
/
-- = - +sin . . sin sin2 2
1 5625 2 5q q q
2 2 5 0 5625 02
sin . sin .q q- + =
Using quadratic formula
sin
( . ) ( ) ( )( . )
( )
. .
q =
- - ± -
=
±
2 5 625 4 2 0 5625
2 2
2 5 1 75
4
or sin . sin .
. .
q q
q q
= =
= ∞ = ∞
0 95572 0 29428
72 886 17 1144
and
and
or andq q= ∞ = ∞17 11 72 9. . b

60
PROBLEM 4.54
Rod AB is attached to a collar at A and rests against a small roller at C.
(a) Neglecting the weight of rod AB, derive an equation in P, Q, a, l, and q
that must be satisfied when the rod is in equilibrium. (b) Determine the value
of q corresponding to equilibrium when P = 80 N, Q = 60 N, l = 500 mm, and
a =125 mm.
SOLUTION
Free-Body Diagram:

+ SF C P Qy = - - =0 0: cosq
C
P Q
=
+
cosq
(a) + SM C
a
PlA = - =0 0:
cos
cos
q
q

P Q a
Pl
+
◊ - =
cos cos
cos
q q
q 0 cos
( )3
q =
+a P Q
Pl
b
(b) For P Q l a= = = =80 60 500 125N, N, mm and mm,

cos
( )( )
( )( )
.
3 125 80 60
80 500
0 4375
q =
+
=
mm N N
N mm
cos .q = 0 75915 q = ∞40 6. b

61
PROBLEM 4.55
A collar B of weight W can move freely along the vertical rod shown. The
constant of the spring is k, and the spring is unstretched when q = 0. (a) Derive an
equation in q, W, k, and l that must be satisfied when the collar is in equilibrium.
(b) Knowing that W = 300 N, l = 500 mm, and k = 800 N/m, determine the
value of q corresponding to equilibrium.
SOLUTION
First note T ks=
where k
s
l
l
l
=
=
= -
= -
spring constant
elongation of spring
cos
cos
( co
q
q
1 ss )
cos
( cos )
q
q
qT
kl
= -1
(a) From f.b.d. of collar B + Â = - =F T Wy 0 0: sinq
or
kl
W
cos
( cos )sin
q
q q1 0- - = or tan sinq q- =
W
kl
b
(b) For W
l
k
l
=
=
=
= =
- =
300
500
800
500
0 5
300
N
mm
N m
mm
2400 N/m
m
N
800
/
.
tan sin
(
q q
NN/m m)( . )
.
0 5
0 75=
Solving numerically, q = ∞57 957. or q = ∞58.0 b

62
PROBLEM 4.56
A vertical load P is applied at end B of rod BC. The constant of the spring is
k, and the spring is unstretched when q = ∞90 . (a) Neglecting the weight of the
rod, express the angle q corresponding to equilibrium in terms of P, k, and l.
(b) Determine the value of q corresponding to equilibrium when P kl= 1
4
.
SOLUTION
First note T ks= =tension in spring
where s
AB AB
l l
=
= -
=
Ê
ËÁ
ˆ
¯˜ -
= ∞
elongation of spring
( ) ( )
sin sin
q q
q
90
2
2
2
990
2
2
2
1
2
∞Ê
ËÁ
ˆ
¯˜
=
Ê
ËÁ
ˆ
¯˜ -
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙l sin
q
T kl=
Ê
ËÁ
ˆ
¯˜ -
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙2
2
1
2
sin
q

(1)
(a) From f.b.d. of rod BC + SM T l P lC =
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙ - =0
2
0: cos ( sin )
q
q
Substituting T from Equation (1)
2
2
1
2 2
0kl l P lsin cos sin
q q
q
Ê
ËÁ
ˆ
¯˜ -
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙ - ( ) =
22
2
1
2 2
2
2
2
kl Plsin cos sin c
q q qÊ
ËÁ
ˆ
¯˜ -
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙
Ê
ËÁ
ˆ
¯˜ -
Ê
ËÁ
ˆ
¯˜ oos
q
2
0
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙ =
Factoring out 2
2
l cos ,
qÊ
ËÁ
ˆ
¯˜ leaves
kl Psin sin
q q
2
1
2 2
0
Ê
ËÁ
ˆ
¯˜ -
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙ -
Ê
ËÁ
ˆ
¯˜ =
or sin
q
2
1
2
Ê
ËÁ
ˆ
¯˜ =
-
Ê
ËÁ
ˆ
¯˜
kl
kl P
q =
-
È
Î
Í
˘
˚
˙
-
2
2
1
sin
( )
kl
kl P
b

63
(b) P
kl
kl
kl
kl
kl
kl
=
=
-( )
È
Î
Í
˘
˚
˙
=
Ê
ËÁ
ˆ
¯˜
È
Î
Í
˘
˚
˙
=
-
-
4
2
2
2
2
4
3
2
1
4
1
q sin
sin
ssin- Ê
ËÁ
ˆ
¯˜
1 4
3 2

=
= ∞
-
2 0 94281
141 058
1
sin ( . )
. or 141.1q = ∞ b

64
PROBLEM 4.57
Solve Sample Problem 4.5, assuming that the spring is unstretched when q = ∞90 .
SOLUTION
First note T ks= =tension in spring
where s
r
F kr
=
=
=
deformation of spring
b
b
From f.b.d. of assembly + SM W l F r0 0 0= - =: ( cos ) ( )b
or Wl kr
kr
Wl
cos
cos
b b
b b
- =
=
2
2
0
For k
r
l
W
=
=
=
=
=
45
75
200
1800
45 75
1800 20
2
N mm
mm
mm
N
N/mm mm
N
/
cos
( )( )
( )(
b
00 mm)
b
or cos .b b= 0 703125
Solving numerically, b = 0 89245. rad
or b = ∞51 134.
Then q = ∞ + ∞ = ∞90 51 134 141 134. . or 141.1q = ∞ b

65
PROBLEM 4.58
A slender rod AB, of weight W, is attached to blocks A and B that
move freely in the guides shown. The constant of the spring is k, and
the spring is unstretched when q = 0. (a) Neglecting the weight of
the blocks, derive an equation in W, k, l, and q that must be satisfied
when the rod is in equilibrium. (b) Determine the value of q when
W = 375 N, l = 750 mm, and k = 600 N/m.
SOLUTION
Free-Body Diagram:
Spring force: F ks k l l kls = = - = -( cos ) ( cos )q q1
(a) + SM F l W
l
D s= -
Ê
ËÁ
ˆ
¯˜ =0
2
0: ( sin ) cosq q
kl l
W
l( cos ) sin cos1
2
0- - =q q q
kl
W
( cos )tan1
2
0- - =q q or ( cos )tan1
2
- =q q
W
kl
b
(b) For given values of W
l
k
=
= =
=
- = -
=
375
750 0 75
600
1
375
2 60
N
mm m
N/m
N
.
( cos )tan tan sin
(
q q q q
00 0 75
0 41667
N/m m)( . )
.=
Solving numerically: q = ∞49 710. or q = ∞49 7. b

66
PROBLEM 4.59
Eight identical 500 750-mm¥ rectangular plates, each of mass m = 40 kg, are held in a vertical plane as
shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether
(a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or
indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible,
compute the reactions.
SOLUTION
1. Three non-concurrent, non-parallel reactions
(a) Plate: completely constrained
(b) Reactions: determinate
(c) Equilibrium maintained
A C= =196 2. N ≠
B C D= = =0 196 2, . N ≠
3. Four non-concurrent, non-parallel reactions
(b) Reactions: indeterminate
Ax = 294 N Æ, Dx = 294 N ¨
(A Dy y+ = 392 N ≠ )
4. Three concurrent reactions (through D)
(a) Plate: improperly constrained
(c) No equilibrium ( )SMD π 0

67
5. Two reactions
(a) Plate: partial constraint
C R= =196 2. N ≠
B = 294 N Æ, D = 491 N 53.1°
7. Two reactions
(a) Plate: improperly constrained
(b) Reactions determined by dynamics
(c) No equilibrium ( )SFy π 0
8. For non-concurrent, non-parallel reactions
B D= =y 196 2. N ≠
( )C D+ =x 0

68
PROBLEM 4.60
The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins,
rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible,
compute the reactions, assuming that the magnitude of the force P is 500 N.
SOLUTION
(a) Bracket: complete constraint
A = 601N 56.3°, B = 333 N ¨
2. Four concurrent, reactions (through A)
(a) Bracket: improper constraint
(c) No equilibrium ( )SMA π 0
3. Two reactions
(a) Bracket: partial constraint
A = 250 N ≠, C = 250 N ≠
A = 250 N ≠, B = 417 N 36.9°, C = 333N Æ

69
(c) Equilibrium maintained ( )SMC y= =0 250 NA ≠
Ax = 333 N ®, B = 333 N ¨
(A By y+ = 500 N ≠ )
A C= = 250 N ≠
8. Three concurrent, reactions (through A)
(a) Bracket: improper constraint
(c) No equilibrium ( )SMA π 0

70
PROBLEM 4.61
Determine the reactions at A and B when a =180 mm.
SOLUTION
Reaction at B must pass through D where A and 300-N load intersect. Free-Body Diagram:
(Three-force member)
DBCD: tan b
b
=
= ∞
240
180
53 13.
Force triangle

A = ∞
=
( )tan .300 53 13
400
N
N A = 400 N ≠ b

B =
=
300
500
N
cos 53.13
N
∞
B = 500 N 53 1. ∞ b

71
PROBLEM 4.62
For the bracket and loading shown, determine the range of values of the
distance a for which the magnitude of the reaction at B does not exceed 600 N.
SOLUTION
Reaction at B must pass through D where A and 300-N load intersect.
Free-Body Diagram:
(Three-force member)
a =
240 mm
tanb
(1)
Force Triangle
(with N)B = 600

cos .
.
b
b
= =
= ∞
300
0 5
60 0
N
600 N
Eq. (1) a =
=
240
138 56
mm
tan 60.0
mm
∞
.
For B Յ 600 138 6N mma ≥ . b

72
PROBLEM 4.63
Using the method of Section 4.7, solve Problem 4.17.
PROBLEM 4.17 The required tension in cable AB is 900 N. Determine
(a) the vertical force P that must be applied to the pedal, (b) the
corresponding reaction at C.
SOLUTION
Free-Body Diagram:
(Three-Force body)
Reaction at C must pass through E, where P and 900 N force
intersect.
tan
.
.
b
b
=
=
151 554
375
22 006
m
mm
∞
Force triangle
(a) P = ( )900 N tan 22.006∞
P = 363 7. P = 364 N Ø b
(b) C =
∞
=
900
22 006
970 7
N
N
cos .
. C = 971 N 22 0. ∞ b

73
PROBLEM 4.64
PROBLEM 4.18 Determine the maximum tension that can be
developed in cable AB if the maximum allowable value of the reaction
at C is 1125 N.
SOLUTION
Free-Body Diagram:
(Three-Force body)
Reaction at C must pass through E, where D and the force T intersect.

tan
.
.
b
b
=
= ∞
151 554
375
22 006
mm
mm
Force triangle

T
T
= ∞
=
( )cos .
.
1125 22 006
1043 04
N
N T =1043 N b

74
PROBLEM 4.65
The spanner shown is used to rotate a shaft. A pin fits in a hole
at A, while a flat, frictionless surface rests against the shaft at B. If
a 300-N force P is exerted on the spanner at D, find the reactions
at A and B.
SOLUTION
Free-Body Diagram:
(Three-Force body)
The line of action of A must pass through D, where B and P intersect.

tan
sin
cos
.
.
a
a
=
∞
∞ +
=
= ∞
75 50
75 50 375
0 135756
7 7310
Force triangle A
B
=
=
=
=
300
2230 11
300
2209 842
N
sin 7.7310
N
N
tan 7.7310
N
∞
∞
.
.
A = 2230 N 7 73. ∞ b
B = 2210 N Æ b

75
PROBLEM 4.66
Determine the reactions at B and D when b = 60 mm.
SOLUTION
Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D.
Free-Body Diagram:
(Three-Force body)
Reaction at B must pass through E, where the reaction at D and 80 N force intersect.
tan
.
b
b
=
= ∞
220
41 348
mm
250 mm
Force triangle
Law of sines

80
45 131 348
888 0
942 8
N
sin 3.652
N
N
∞ ∞ ∞
= =
=
=
B D
B
D
sin sin .
.
. B = 888 N 41 3. ∞ D = 943 N 45 0. ∞ b

76
PROBLEM 4.67
Determine the reactions at B and D when b =120 mm.
SOLUTION
Since CD is a two-force member, line of action of reaction at D must pass through C and D.
Free-Body Diagram:
(Three-Force body)
Reaction at B must pass through E, where the reaction at D and 80-N force intersect.
tan
.
b
b
=
= ∞
280
48 24
mm
250 mm
Force triangle
Law of sines
80
135 41 76
N
sin 3.24∞
=
∞
=
∞
B D
sin sin .

B
D
=
=
1000 9
942 8
.
.
N
N B =1001 N 48 2 943. ∞ =D N 45 0. ∞ b

77
PROBLEM 4.68
Determine the reactions at B and C when a = 37 5. mm.
SOLUTION
Since CD is a two-force member, the force it exerts on member ABD is directed along DC.
Free-Body Diagram of ABD: (Three-Force member)
The reaction at B must pass through E, where D and the 250 N load intersect.
Triangle CFD: tan .
.
a
a
= =
= ∞
75
125
0 6
30 964
Triangle EAD: AE
GE AE AG
= =
= - = - =
250 150
150 37 5 112 5
tan
. .
a mm
mm
Triangle EGB: tan
.
.
b
b
= =
= ∞
GB
GE
75
112 5
33 690
Force triangle
B D
sin . sin .120 964 33 690
250
∞
=
∞
=
∞
N
sin 25.346

B
D
=
=
500 773
323 943
.
.
N
N B = 501 N 56 3. ∞ b
C D= = 324 N 31 0. ∞ b

78
PROBLEM 4.69
A 50-kg crate is attached to the trolley-beam system shown.
Knowing that a =1 5. m, determine (a) the tension in cable CD,
(b) the reaction at B.
SOLUTION
Free-Body Diagram:
Three-Force body: W and TCD intersect at E.
tan
.
.
.
b
b
=
= ∞
0 7497
1 5
26 56
m
m
Force triangle 3 forces intersect at E.
W =
=
( )
.
50
490 5
kg 9.81 m/s
N
2
Law of sines
490 5
63 44 55
498 9
456 9
.
sin . sin
.
.
N
sin 61.56
N
N
∞
=
∞
=
∞
=
=
T B
T
B
CD
CD
(a) TCD = 499 N b
(b) B = 457 N 26 6. ∞ b

79
PROBLEM 4.70
Solve Problem 4.69, assuming that a = 3 m.
PROBLEM 4.69 A 50 kg crate is attached to the trolley-beam
system shown. Knowing that a =1 5. m, determine (a) the tension
in cable CD, (b) the reaction at B.
SOLUTION
W and TCD intersect at E
Free-Body Diagram:
Three-Force body:
tan
.
.
b
b
= =
= ∞
AE
AB
0 301
5 722
m
3 m
Force Triangle (Three forces intersect at E.)

W =
=
( )50
490
kg 9.81 m/s
.5 N
2
Law of sines 490 5
95 722 55
997 99
821
.
sin . sin
.
.
N
sin 29.278
N
∞
=
∞
=
∞
=
=
T B
T
B
CD
CD
559 N
(a) TCD = 998 N b
(b) B = 822 N 5 72. ∞ b

80
PROBLEM 4.71
One end of rod AB rests in the corner A and the other end is attached to cord
BD. If the rod supports a 200-N load at its midpoint C, find the reaction at A
and the tension in the cord.
SOLUTION
Free-Body Diagram: (Three-Force body)
The line of action of reaction at A must pass through E, where T and the 200 N load intersect.

tan
.
tan
.
a
a
b
b
= =
= ∞
= =
= ∞
EF
AF
EH
DH
575
300
62 447
125
300
22 620
Force triangle A T
sin . sin .67 380 27 553
200
∞
=
∞
=
∞
N
sin 85.067
A =185 3. N 62 4. ∞ b
T = 92 9. N b

81
PROBLEM 4.72
Determine the reactions at A and D when b = ∞30 .
SOLUTION
From f.b.d. of frame ABCD
+ SM AD = - + ∞
+ ∞
0 0 18 150 30 0 10
150 30 0 28
: ( . [( sin ]( .
[( cos ]( .
m) N) m)
N) m)) = 0
A = 243 74. N
or A = 244 N Æ b
+ SF Dx x= + ∞ + =0 243 74 150 30 0: ( . ( sinN) N)
Dx = -318 74. N
+ SF Dy y= - ∞ =0 150 30 0: ( cosN)
Dy =129 904. N
Then D D Dx x= +
= +
=
( )
( . ) ( . )
.
2 2
2 2
318 74 129 904
344 19 N
and q =
Ê
ËÁ
ˆ
¯˜
=
-
Ê
ËÁ
ˆ
¯˜
-
-
tan
tan
.
.
1
1 129 904
318 74
D
D
y
x
= - ∞22 174. or D = 344 N 22 2. ∞ b

82
PROBLEM 4.73
Determine the reactions at A and D when b = ∞60 .
SOLUTION
From f.b.d. of frame ABCD
+ SM AD = - + ∞
+ ∞
0 0 18 150 60 0 10
150 60 0 28
: ( . [( sin ]( .
[( cos ]( .
m) N) m)
N) m)) = 0
A =188 835. N
or A =188 8. N Æ b
+ SF Dx x= + ∞ + =0 188 835 150 60 0: ( . ) ( )sinN N
Dx = -318 74. N
+ SF Dy y= - ∞ =0 150 60 0: ( )cosN
Dy = 75 0. N
Then D D Dx y= +
= +
=
( ) ( )
( . ) ( . )
.
2 2
2 2
318 74 75 0
327 44 N
and q =
Ê
ËÁ
ˆ
¯˜
-
tan 1
D
D
y
x

=
-
Ê
ËÁ
ˆ
¯˜
= - ∞
-
tan
.
.
.
1 75 0
318 74
13 2409 or D = 327 N 13 24. ∞ b

83
PROBLEM 4.74
A 20-kg roller, of diameter 200 mm, which is to be used on a tile floor, is
resting directly on the subflooring as shown. Knowing that the thickness of
each tile is 8 mm, determine the force P required to move the roller onto the
tiles if the roller is (a) pushed to the left, (b) pulled to the right.
SOLUTION
(a) Based on the roller having impending motion to the left, the only
contact between the roller and floor will be at the edge of the tile.
First note W mg= = =( )( . ) .20 9 81 196 2kg m/s N2
From the geometry of the three forces acting on the roller
a =
Ê
ËÁ
ˆ
¯˜ =-
cos .1 92
100
23 074
mm
mm
∞
and q a= - - = - =90 30 60 23 074 36 926∞ ∞ ∞ ∞ ∞. .
Applying the law of sines to the force triangle,
W P
sin sinq a
=
or 196 2
36 926 23 074
.
sin . sin .
N
∞
=
∞
P
=P 127 991. N
or NP =128 0. 30∞ b
(b) Based on the roller having impending motion to the right, the only
contact between the roller and floor will be at the edge of the tile.
First note W mg= = ( )( . )20 9 81kg m/s2
=196 2. N
From the geometry of the three forces acting on the roller
a =
Ê
ËÁ
ˆ
¯˜ =-
cos .1 92
100
23 074
mm
mm
∞
and q a= + + = - =90 30 120 23 074 96 926∞ ∞ ∞ ∞ ∞. .
Applying the law of sines to the force triangle,
W P
sin sinq a
=
or
196 2
96 926 23 074
.
sin . sin .
N
∞
=
P
=P 77 460. N
or NP = 77 5. 30∞ b
Free-Body Diagram:
Free-Body Diagram:
100 mm
f.b.d
30°
92 mm
N
P
W
120°
W
N
P
100 mm P
W
N
α
G
30°
92 mm
N
W
P
60°
θ
α

84
PROBLEM 4.75
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions
AD and CD of the cord. For the loading shown and neglecting the size
of the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Reaction at B must pass through D. Free-Body Diagram:

tan
.
tan
.
a
a
b
b
=
= ∞
=
= ∞
175
30 256
175
16 26
mm
300 mm
mm
600mm
Force triangle
Law of sines T T B
T T
sin . sin . sin .
(sin . ) (
59 744
360
13 996 106 26
13 996
∞
=
-
∞
=
∞ = -
N
∞
3360
0 24185 360 0 86378
N)(sin 59.744 )∞
T T( . ) ( )( . )= -
T = 500 N T = 500 N b

B =
=
(
sin .
.
500
59 744
555 695
N)
sin 106.26
N
∞
B = 556 N 30 3. ∞ b

85
PROBLEM 4.76
Member ABC is supported by a pin and bracket at B and by an
inextensible cord attached at A and C and passing over a frictionless
pulley at D. The tension may be assumed to be the same in portions
AD and CD of the cord. For the loading shown and neglecting the size
of the pulley, determine the tension in the cord and the reaction at B.
SOLUTION
Free-Body Diagram:
Force triangle

Reaction at B must pass through D.
tan ; .a a= = ∞
120
160
36 9
T T B
4
75
3 5
=
-
=
lb

3 4 300 300
5
4
5
4
300 375
T T T
B T
= - =
= = =
;
( )
lb
lb lb B = 375 lb 36.9°

86
PROBLEM 4.77
Rod AB is supported by a pin and bracket at A and rests against a frictionless peg
at C. Determine the reactions at A and C when a 170-N vertical force is applied
at B.
SOLUTION
Free-Body Diagram:
(Three-Force body)
The reaction at A must pass through D where C and 170-N force intersect.

tan
.
a
a
=
= ∞
160
28 07
mm
300 mm
We note that triangle ABD is isosceles (since AC = BC ) and, therefore
SCAD = = ∞a 28 07.
Also, since CD CB^ , reaction C forms angle a = ∞28 07. with horizontal.
Force triangle
We note that A forms angle 2a with vertical. Thus A and C form angle
180 90 2 90∞ - ∞ - - = ∞ -( )a a a
Force triangle is isosceles and we have

A
C
=
=
=
170
2 170
160 0
N
N
N
( )sin
.
a
A =170 0. N 33.9° C =160 0. N 28.1° b

87
PROBLEM 4.78
Solve Problem 4.77, assuming that the 170-N force applied at B is horizontal
and directed to the left.
PROBLEM 4.77 Rod AB is supported by a pin and bracket at A and rests
against a frictionless peg at C. Determine the reactions at A and C when a 170-N
vertical force is applied at B.
SOLUTION
Free-Body Diagram: (Three-Force body)
The reaction at A must pass through D, where C and the 170-N force intersect.

tan
.
a
a
=
= ∞
160
28 07
mm
300 mm
We note that triangle ADB is isosceles (since AC = BC). Therefore S SA B= = ∞ -90 a.
Also S ADB = 2a
Force triangle
The angle between A and C must be 2a a a- =
Thus, force triangle is isosceles and
a = ∞28 07.

A
C
=
= =
170 0
2 170 300
.
( )cos
N
N Na
A =170 0. N 56.1° C = 300 N 28.1° b

88
PROBLEM 4.79
PROBLEM 4.21 Determine the reactions at A and B when
(a) h = 0, (b) h = 200 mm.
SOLUTION
Free-Body Diagram: (a) h = 0
Reaction A must pass through C where 150-N weight and
B interect.
Force triangle is equilateral
A =150 0. N 30.0° b
B =150 0. N 30.0° b
(b) h = 200 mm
tan
.
.
b
b
=
= ∞
55 662
250
12 552
Force triangle
Law of sines

150
60 102 552
433 247
488 31
N
sin17.448
N
N
∞
=
∞
=
∞
=
=
A B
A
B
sin sin .
.
.
A = 433 N 12.55° b
B = 488 N 30.0° b

89
PROBLEM 4.80
PROBLEM 4.28 A lever AB is hinged at C and attached to a control
cable at A. If the lever is subjected to a 500-N horizontal force at B,
determine (a) the tension in the cable, (b) the reaction at C.
SOLUTION
Free-Body Diagram:
(Three-Force body)
Reaction at C must pass through E, where FAD and 500-N force intersect.
Since AC CD= = 250 mm, triangle ACD is isosceles.
We have SC = ∞ + ∞ = ∞90 30 120
and S SA D= = ∞ - ∞ = ∞
1
2
180 120 30( ) Dimensions in mm
On the other hand, from triangle BCF:

CF BC
FD CD CF
= ∞ = ∞ =
= - = - =
( )sin30 200 30 100
250 100 150
sin mm
mm
From triangle EFD, and since SD = ∞30 :
EF FD= ∞ = ∞ =( )tan tan .30 150 30 86 60 mm
From triangle EFC: tan
.
a
a
= =
= ∞
CF
EF
100
49 11
mm
86.60 mm
Force triangle
Law of sines
F C
F C
AD
AD
sin . sin49 11 60
500
400 458
∞
=
∞
=
= =
N
sin 70.89
N, N
∞
(a) FAD = 400 N b
(b) C = 458 N 49.1° b

90
PROBLEM 4.81
Knowing that q = 30°, determine the reaction (a) at B, (b) at C.
SOLUTION
Free-Body Diagram:
(Three-Force body)
Reaction at C must pass through D where force P and reaction at B intersect.
In DCDE: tan
( )
.
b
b
=
-
= -
= ∞
3 1
3 1
36 2
R
R
Force triangle
Law of sines
P B C
B P
C P
sin . sin . sin
.
.
23 8 126 2 30
2 00
1 239
∞
=
∞
=
∞
=
=
(a) B = 2P 60.0° b
(b) C =1 239. P 36.2° b

91
PROBLEM 4.82
Knowing that q = 60°, determine the reaction (a) at B, (b) at C.
SOLUTION
Reaction at C must pass through D where force P and reaction at B intersect.
In DCDE: tan
.
b
b
=
-
= -
= ∞
R
R
R
3
1
1
3
22 9
Force triangle

Law of sines
P B C
B P
C P
sin . sin . sin
.
.
52 9 67 1 60
1 155
1 086
∞
=
∞
=
∞
=
=
(a) B =1 155. P 30.0° b
(b) C =1 086. P 22.9° b
Free-Body Diagram:
(Three-Force body)

92
PROBLEM 4.83
Rod AB is bent into the shape of an arc of circle and is lodged between two pegs
D and E. It supports a load P at end B. Neglecting friction and the weight of the
rod, determine the distance c corresponding to equilibrium when a = 20 mm and
R = 100 mm.
SOLUTION
Since y x aED ED= = , Free-Body Diagram:
Slope of ED is 45∞
slope of isHC 45∞
Also DE a= 2
and DH HE DE
a
= =
Ê
ËÁ
ˆ
¯˜ =
1
2 2
For triangles DHC and EHC sinb = =
a
R
a
R
2
2
Now c R= ∞ -sin( )45 b
For a R= =
=
=
= ∞
20 100
2 100
0 141421
8 1301
mm and mm
20 mm
mm
sin
( )
.
.
b
b
and c = ∞ - ∞( )sin( . )100 mm 45 8 1301
= 60 00. mm or c = 60 0. mm b

93
PROBLEM 4.84
A slender rod of length L is attached to collars that can slide freely along the
guides shown. Knowing that the rod is in equilibrium, derive an expression for
the angle q in terms of the angle b.
SOLUTION
As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force
geometry
Free-Body Diagram:
tanb =
x
y
GB
AB
where
y LAB = cosq
and
x LGB =
1
2
sinq

tan
sin
cos
tan
b
q
q
q
=
=
1
2
1
2
L
L
or tan tanq b= 2 b

Chapter 04

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