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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.
