PROBLEM 4.1
For the beam and loading shown, determine (a) the reaction at A,
(b) the tension in cable BC.
SOLUTION
Free–Body Diagram:
PROBLEM 4.2
A 3200–lb forklift truck is used to lift a 1700–lb crate. Determine
the reaction at each of the two (a) front wheels A, (b) rear wheels B.
SOLUTION
Free–Body Diagram:
(a) Front wheels:
0: (1700 lb)(52 in.) (3200 lb)(12 in.) 2 (36 in.) 0
B
MAΣ= + − =
1761.11lbA= +
1761lb=A
(b) Rear wheels:
0: 1700 lb 3200 lb 2(1761.11lb) 2 0
y
FBΣ= − − + + =
688.89 lbB= +
689 lb=B
PROBLEM 4.3
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:
0: (2 )(1m) (60 N)(0.15 m) (250 N)(0.3 m) 0
A
MFΣ= − − =
42.0 N=F
PROBLEM 4.4
A load of lumber weighing
25 kNW=
is being raised as
shown by a mobile crane. Knowing that the tension is
25 kN in all portions of cable AEF and that the weight of
boom ABC is 3 kN, determine (a) the tension in rod CD,
(b) the reaction at pin B.
SOLUTION
Free–Body Diagram: (boom)
(a)
( )( ) ( )( ) ( )( ) ( )
0: 25 kN 2.6 m + 3 kN 0.6 m 25 kN 0.4 m 0.7 m 0
B CD
MTΣ= − − =
81.143 kN
CD
T=
or
81.1 kN
CD
T=
(b)
0: 0
xx
FBΣ= =
so that
y
BB=
( )
0: 25 3 25 81.143 kN 0
y
FBΣ = − −− − + =
134.143 kNB=
or
134.1 kN=B
consent of McGraw–Hill Education.
PROBLEM 4.5
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Neglecting the weight of the beam,
determine the range of values of Q for which neither cable becomes
slack when
0.P=
SOLUTION
Free–Body Diagram:
For
min, 0
D
QT=
( )( ) ( )
min
0: 7.5 kN 0.5 m 3 m 0
B
MQΣ= − =
min
1.250 kNQ=
For
max
, 0
B
QT=
( )( ) ( )
max
0: 7.5 kN 2.75 m 0.75 m 0
D
MQΣ= − =
max 27.5 kN
Q=
Therefore:
1.250 kN 27.5 kNQ≤≤
consent of McGraw–Hill Education.
PROBLEM 4.6
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Knowing that the maximum allowable
tension in each cable is 12 kN and neglecting the weight of the
beam, determine the range of values of Q for which the loading is
safe when
0.P=
SOLUTION
Free–Body Diagram:
( )( ) ( ) ( )
0: 7.5 kN 2.75 m 2.25 m 0.75 m 0
DB
M TQΣ= − − =
27.5 3 B
QT= −
(1)
( )( ) ( ) ( )
0: 7.5 kN 0.5 m + 2.25 m 3 m 0
BD
M TQΣ= − =
( )
1.25 0.75 D
QT= +
(2)
For the loading to be safe, cables must not be slack and tension must not exceed 12 kN.
consent of McGraw–Hill Education.
PROBLEM 4.7
For the beam and loading shown, determine the range of
the distance a for which the reaction at B does not exceed
100 lb downward or 200 lb upward.
SOLUTION
Assume B is positive when directed .
Sketch showing distance from D to forces.
0: (300 lb)(8 in. ) (300 lb)( 2 in.) (50 lb)(4 in.) 16 0
D
M aa BΣ = −− − − + =
600 2800 16 0aB− + +=
(2800 16 )
600
B
a+
=
(1)
PROBLEM 4.8
For the beam of Sample Prob. 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 25 kips and that the
reaction at A must be directed upward.
SOLUTION
Free–Body Diagram:
( ) ( ) ( )( )
0: (3 ft) 9 ft (6 kips) 11ft 6 kips 13 ft 0
Ay
M PBΣ= − + − − =
( )
3B 48 kips
y
Q= −
(1)
( ) ( ) ( ) ( )
0: 9 ft 6 ft (6 kips) 2 ft (6 kips) 4 ft 0
By
M APΣ=−+−−=
( )
1.5 +6 kips
y
PA=
(2)
For the loading to meet the design criteria, the reactions must not exceed 25 kips.
Thus, making
25 kips
y
B≤
in (1), we have
P
≤
27.0 kips (3)
And making 0
≤
Ay
≤
25 kips in (2), we have
6.00
≤
P
≤
43.5 kips (4)
(3) and (4) now give: 6.00 kips
≤
P
≤
27.0 kips
consent of McGraw–Hill Education.
PROBLEM 4.9
The 40–ft boom AB weighs 2 kips; the distance from the
axle A to the center of gravity G of the boom is 20 ft. For
the position shown, determine (a) the tension T in the cable,
(b) the reaction at A.
SOLUTION
(a)
( )sin10 (40 ft)sin10
6.9459 ft
AD AB
AD
= =
=
AA
0: ( ) 2(20cos30 ) 5(40cos30 ) 0
A
M T ADΣ= − − =
AA
29.924 kipsT=
29.9 kipsT=
(b)
0: (29.924)cos20 0
xx
FAΣ= − =
A
28.119 kips
x
A= +
0: (29.924)sin 20 2 5 0
yy
FAΣ = − −−=
A
17.2346 kips
y
A=
22
28.119 17.2346 32.980 kips= +=A
117.2346
tan 28.119
31.5
a
−
=
= °
33.0 kips=A
31.5°
consent of McGraw–Hill Education.
PROBLEM 4.10
The lever BCD is hinged at C and attached to a control rod at B. If
P = 100 lb, determine (a) the tension in rod AB, (b) the reaction at C.
SOLUTION
Free–Body Diagram:
(a)
0: (5 in.) (100 lb)(7.5 in.) 0
C
MT
Σ= − =
150.0 lbT=
(b)
3
0: 100 lb (150.0 lb) 0
5
xx
FCΣ= + + =
190 lb
x
C= −
190 lb
x
=C
4
0: lb) 0
5
yy
FCΣ = + (150.0 =
120 lb
y
C= −
120 lb
y=C
32.3
α
= °
225 lbC=
225 lb=C
32.3°
consent of McGraw–Hill Education.
PROBLEM 4.2
A 3200–lb forklift truck is used to lift a 1700–lb crate. Determine
the reaction at each of the two (a) front wheels A, (b) rear wheels B.
SOLUTION
Free–Body Diagram:
(a) Front wheels:
0: (1700 lb)(52 in.) (3200 lb)(12 in.) 2 (36 in.) 0
B
MAΣ= + − =
1761.11lbA= +
1761lb=A
(b) Rear wheels:
0: 1700 lb 3200 lb 2(1761.11lb) 2 0
y
FBΣ= − − + + =
688.89 lbB= +
689 lb=B
PROBLEM 4.3
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:
0: (2 )(1m) (60 N)(0.15 m) (250 N)(0.3 m) 0
A
MFΣ= − − =
42.0 N=F
PROBLEM 4.4
A load of lumber weighing
25 kNW=
is being raised as
shown by a mobile crane. Knowing that the tension is
25 kN in all portions of cable AEF and that the weight of
boom ABC is 3 kN, determine (a) the tension in rod CD,
(b) the reaction at pin B.
SOLUTION
Free–Body Diagram: (boom)
(a)
( )( ) ( )( ) ( )( ) ( )
0: 25 kN 2.6 m + 3 kN 0.6 m 25 kN 0.4 m 0.7 m 0
B CD
MTΣ= − − =
81.143 kN
CD
T=
or
81.1 kN
CD
T=
(b)
0: 0
xx
FBΣ= =
so that
y
BB=
( )
0: 25 3 25 81.143 kN 0
y
FBΣ = − −− − + =
134.143 kNB=
or
134.1 kN=B
consent of McGraw–Hill Education.
PROBLEM 4.5
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Neglecting the weight of the beam,
determine the range of values of Q for which neither cable becomes
slack when
0.P=
SOLUTION
Free–Body Diagram:
For
min, 0
D
QT=
( )( ) ( )
min
0: 7.5 kN 0.5 m 3 m 0
B
MQΣ= − =
min
1.250 kNQ=
For
max
, 0
B
QT=
( )( ) ( )
max
0: 7.5 kN 2.75 m 0.75 m 0
D
MQΣ= − =
max 27.5 kN
Q=
Therefore:
1.250 kN 27.5 kNQ≤≤
consent of McGraw–Hill Education.
PROBLEM 4.6
Three loads are applied as shown to a light beam supported by
cables attached at B and D. Knowing that the maximum allowable
tension in each cable is 12 kN and neglecting the weight of the
beam, determine the range of values of Q for which the loading is
safe when
0.P=
SOLUTION
Free–Body Diagram:
( )( ) ( ) ( )
0: 7.5 kN 2.75 m 2.25 m 0.75 m 0
DB
M TQΣ= − − =
27.5 3 B
QT= −
(1)
( )( ) ( ) ( )
0: 7.5 kN 0.5 m + 2.25 m 3 m 0
BD
M TQΣ= − =
( )
1.25 0.75 D
QT= +
(2)
For the loading to be safe, cables must not be slack and tension must not exceed 12 kN.
consent of McGraw–Hill Education.
PROBLEM 4.7
For the beam and loading shown, determine the range of
the distance a for which the reaction at B does not exceed
100 lb downward or 200 lb upward.
SOLUTION
Assume B is positive when directed .
Sketch showing distance from D to forces.
0: (300 lb)(8 in. ) (300 lb)( 2 in.) (50 lb)(4 in.) 16 0
D
M aa BΣ = −− − − + =
600 2800 16 0aB− + +=
(2800 16 )
600
B
a+
=
(1)
PROBLEM 4.8
For the beam of Sample Prob. 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 25 kips and that the
reaction at A must be directed upward.
SOLUTION
Free–Body Diagram:
( ) ( ) ( )( )
0: (3 ft) 9 ft (6 kips) 11ft 6 kips 13 ft 0
Ay
M PBΣ= − + − − =
( )
3B 48 kips
y
Q= −
(1)
( ) ( ) ( ) ( )
0: 9 ft 6 ft (6 kips) 2 ft (6 kips) 4 ft 0
By
M APΣ=−+−−=
( )
1.5 +6 kips
y
PA=
(2)
For the loading to meet the design criteria, the reactions must not exceed 25 kips.
Thus, making
25 kips
y
B≤
in (1), we have
P
≤
27.0 kips (3)
And making 0
≤
Ay
≤
25 kips in (2), we have
6.00
≤
P
≤
43.5 kips (4)
(3) and (4) now give: 6.00 kips
≤
P
≤
27.0 kips
consent of McGraw–Hill Education.
PROBLEM 4.9
The 40–ft boom AB weighs 2 kips; the distance from the
axle A to the center of gravity G of the boom is 20 ft. For
the position shown, determine (a) the tension T in the cable,
(b) the reaction at A.
SOLUTION
(a)
( )sin10 (40 ft)sin10
6.9459 ft
AD AB
AD
= =
=
AA
0: ( ) 2(20cos30 ) 5(40cos30 ) 0
A
M T ADΣ= − − =
AA
29.924 kipsT=
29.9 kipsT=
(b)
0: (29.924)cos20 0
xx
FAΣ= − =
A
28.119 kips
x
A= +
0: (29.924)sin 20 2 5 0
yy
FAΣ = − −−=
A
17.2346 kips
y
A=
22
28.119 17.2346 32.980 kips= +=A
117.2346
tan 28.119
31.5
a
−
=
= °
33.0 kips=A
31.5°
consent of McGraw–Hill Education.
PROBLEM 4.10
The lever BCD is hinged at C and attached to a control rod at B. If
P = 100 lb, determine (a) the tension in rod AB, (b) the reaction at C.
SOLUTION
Free–Body Diagram:
(a)
0: (5 in.) (100 lb)(7.5 in.) 0
C
MT
Σ= − =
150.0 lbT=
(b)
3
0: 100 lb (150.0 lb) 0
5
xx
FCΣ= + + =
190 lb
x
C= −
190 lb
x
=C
4
0: lb) 0
5
yy
FCΣ = + (150.0 =
120 lb
y
C= −
120 lb
y=C
32.3
α
= °
225 lbC=
225 lb=C
32.3°
consent of McGraw–Hill Education.