PROBLEM 4.86
A 120–lb cabinet is mounted on casters that can be locked to prevent their
rotation. The coefficient of static friction between the floor and each caster is
0.30. Assuming that the casters at both A and B are locked, determine (a) the
force P required to move the cabinet to the right, (b) the largest allowable value
of h if the cabinet is not to tip over.
SOLUTION
FBD cabinet:
(a)
0: 0
y AB
AB
F NNW
NNW
Σ= + −=
+=
Impending slip:
A sA
B sB
FN
FN
µ
µ
=
=
So
AB s
FF W
µ
+=
0: 0
x AB
AB s
F PF F
PF F W
µ
Σ= − − =
=+=
120 lb
0.3
s
W
µ
=
=
0.3(120 lb) 141.26 NP= =
36.0 lb=P
(b) For tipping,
0
AA
NF= =
0: (12 in.) 0
B
M hP W
Σ= − =
max 1 12 in.
(12 in.) (12 in.) 0.3
s
W
hP
µ
= = =
max
40.0 in.h=
PROBLEM 4.87
A 40–kg packing crate must be moved to the left along the floor without
tipping. Knowing that the coefficient of static friction between the crate
and the floor is 0.35, determine (a) the largest allowable value of α, (b) the
corresponding magnitude of the force P.
SOLUTION
(a) Free-body diagram
If the crate is about to tip about C, contact between crate and ground is only at C and the reaction R is
applied at C. As the crate is about to slide, R must form with the vertical an angle
11
tan tan
0.35 19.29
ss
φµ
−−
= =
= °
Since the crate is a 3-force body. P must pass through E where R and W intersect.
0.4 m 1.1429 m
tan 0.35
1.1429 0.5 0.6429 m
0.6429 m
tan 0.4 m
s
CF
EF
EH EF HF
EH
HB
θ
a
= = =
=− = −=
= =
58.11
a
= °
58.1
a
= °
(b) Force Triangle
0.424
sin19.29 sin128.82
PW
PW= =
°°
2
0.424(40 kg)(9.81 m/s ),P=
166.4 NP=
Note: After the crate starts moving,
s
µ
should be replaced by the lower value
.
k
µ
This will
yield a larger value of
.
a
consent of McGraw–Hill Education.
PROBLEM 4.88
A 40–kg packing crate is pulled by a rope as shown. The coefficient of static
friction between the crate and the floor is 0.35. If α = 40°, determine (a) the
magnitude of the force P required to move the crate, (b) whether the crate will
slide or tip.
SOLUTION
Force P for which sliding is impending
(We assume that crate does not tip)
2
(40 kg)(9.81 m/s ) 392.4 NW= =
0: sin 40 0
y
F NW PΣ = − + °=
sin 40NWP=−°
(1)
0: 0.35 cos40 0
x
F NP
Σ = − °=
Substitute for
N
from Eq. (1):
0.35( sin 40 ) cos40 0WP P− ° − °=
0.35
0.35sin 40 cos40
W
P=°+ °
0.3532PW=
Force P for which crate rotates about C
(We assume that crate does not slide)
0: ( sin 40 )(0.8 m) ( cos40 )(0.5 m)
C
MP PΣ= ° + °
(0.4 m) 0W−=
0.4 0.4458
0.8sin 40 0.5cos 40
W
PW= =
°+ °
(b) Crate will first slide
(a)
0.3532(392.4 N)P=
138.6 NP=
consent of McGraw–Hill Education.
PROBLEM 4.89
The coefficients of friction are
0.40
s
µ
=
and
0.30
k
µ
=
between all
surfaces of contact. Determine the smallest force P required to start the
30–kg block moving if cable AB (a) is attached as shown, (b) is removed.
SOLUTION
(a) Free body: 20-kg block
2
1
11
(20 kg)(9.81 m/s ) 196.2 N
0.4(196.2 N) 78.48 N
s
W
FN
µ
= =
= = =
11
0: 0 78.48 NF TF TFΣ= − = = =
Free body: 30-kg block
12
0: 0F PF F TΣ= − − −=
78.48 N 196.2 N 78.48 N 353.2 NP=++=
353 N=P
(b) Free body: Both blocks
Blocks move together
2
(50 kg)(9.81 m/s ) 490.5 NW= =
0: 0F PFΣ= − =
0.4(490.5 N) 196.2 N
s
PN
µ
= = =
196.2 N=P
consent of McGraw–Hill Education.
PROBLEM 4.90
The coefficients of friction are
0.40
s
µ
=
and
0.30
k
µ
=
between all
surfaces of contact. Determine the smallest force P required to start
the 30–kg block moving if cable AB (a) is attached as shown, (b) is
removed.
SOLUTION
(a) Free body: 20-kg block
2
1
11
(20 kg)(9.81 m/s ) 196.2 N
0.4(196.2 N) 78.48 N
s
W
FN
µ
= =
= = =
11
0: 0 78.48 NF TF TFΣ= − = = =
Free body: 30-kg block
2
2
2
22
(30 kg)(9.81 m/s ) 294.3 N
196.2 N 294.3 N 490.5 N
0.4(490.5 N) 196.2 N
s
W
N
FN
µ
= =
=+=
= = =
12
0: 0
F PF FΣ= − − =
78.48 N 196.2 N 274.7 NP=+=
275 N=P
(b) Free body: Both blocks
Blocks move together
2
(50 kg)(9.81 m/s )
490.5 N
W=
=
0: 0
0.4(490.5 N) 196.2 N
s
F PF
PN
µ
Σ= − =
= = =
196.2 N=P
consent of McGraw–Hill Education.
PROBLEM 4.91
A 6.5–m ladder AB leans against a wall as shown. Assuming that the coefficient
of static friction
s
µ
is zero at B, determine the smallest value of
s
µ
at A for
which equilibrium is maintained.
SOLUTION
Free body: Ladder
Three-force body.
Line of action of A must pass through D, where W and B intersect.
At A:
1.25 m
tan 0.2083
6m
ss
µφ
= = =
0.208
s
µ
=
consent of McGraw–Hill Education.
PROBLEM 4.92
A 6.5–m ladder AB leans against a wall as shown. Assuming that the coefficient
of static friction
s
µ
is the same at A and B, determine the smallest value of
s
µ
for which equilibrium is maintained.
SOLUTION
Free body: Ladder
Motion impending:
A sA
B sB
FN
FN
µ
µ
=
=
0: (1.25 m) (6 m) (2.5 m) 0
A B sB
MW N N
µ
Σ= − − =
1.25
6 2.5
Bs
W
N
µ
=+
(1)
0: 0
y A sB
F N NW
µ
Σ= + −=
1.25
6 2.5
A sB
s
As
NW N
W
NW
µ
µ
µ
= −
= − +
(2)
0: 0
x sA B
F NN
µ
Σ= − =
Substitute for
A
N
and
B
N
from Eqs. (1) and (2):
2
22
2
1.25 1.25
6 2.5 6 2.5
6 2.5 1.25 1.25
1.25 6 1.25 0
0.2
s
sss
ss s
ss
s
WW
W
µ
µµµ
µµ µ
µµ
µ
−=
++
+− =
+−=
=
and
5 (Discard)
s
µ
= −
0.200
s
µ
=
consent of McGraw–Hill Education.
PROBLEM 4.93
End A of a slender, uniform rod of length L and weight W bears on a surface as
shown, while end B is supported by a cord BC. Knowing that the coefficients
of friction are
0.40
s
µ
=
and
0.30,
k
µ
=
determine (a) the largest value of
θ
for which motion is impending, (b) the corresponding value of the tension in the
cord.
SOLUTION
Free-body diagram
Rod AB is a three-force body. Thus, line of action of R must pass through D,
Where W and T intersect.
Triangle ABC bisects the angle
θ
.
(a) Thus,
1
2
s
φθ
=
Since motion impends,
1
tan 0.40 21.80
2 2(21.8 )
s
s
φ
θφ
−
= = °
= = °
43.6
θ
= °
(b) Force triangle:
This is a right triangle.
sin
sin 21.8
s
TW
W
φ
=
= °
0.371TW=
consent of McGraw–Hill Education.
PROBLEM 4.94
End A of a slender, uniform rod of length L and weight W bears on a
surface as shown, while end B is supported by a cord BC. Knowing that
the coefficients of friction are
0.40
s
µ
=
and
0.30,
k
µ
=
determine (a) the
largest value of
θ
for which motion is impending, (b) the corresponding
value of the tension in the cord.
SOLUTION
Free-body diagram
Three-force body. Line of action of R must pass through D, where T and R intersect.
Motion impends:
tan 0.4
21.80
s
s
φ
φ
=
= °
90
2
21.8 90
2
s
θφ
θ
+=°
+ °= °
136.4
θ
= °
(b) Force triangle (right triangle):
cos 21.8TW= °
0.928TW
=
consent of McGraw–Hill Education.
PROBLEM 4.86
A 120–lb cabinet is mounted on casters that can be locked to prevent their
rotation. The coefficient of static friction between the floor and each caster is
0.30. Assuming that the casters at both A and B are locked, determine (a) the
force P required to move the cabinet to the right, (b) the largest allowable value
of h if the cabinet is not to tip over.
SOLUTION
FBD cabinet:
(a)
0: 0
y AB
AB
F NNW
NNW
Σ= + −=
+=
Impending slip:
A sA
B sB
FN
FN
µ
µ
=
=
So
AB s
FF W
µ
+=
0: 0
x AB
AB s
F PF F
PF F W
µ
Σ= − − =
=+=
120 lb
0.3
s
W
µ
=
=
0.3(120 lb) 141.26 NP= =
36.0 lb=P
(b) For tipping,
0
AA
NF= =
0: (12 in.) 0
B
M hP W
Σ= − =
max 1 12 in.
(12 in.) (12 in.) 0.3
s
W
hP
µ
= = =
max
40.0 in.h=
PROBLEM 4.87
A 40–kg packing crate must be moved to the left along the floor without
tipping. Knowing that the coefficient of static friction between the crate
and the floor is 0.35, determine (a) the largest allowable value of α, (b) the
corresponding magnitude of the force P.
SOLUTION
(a) Free-body diagram
If the crate is about to tip about C, contact between crate and ground is only at C and the reaction R is
applied at C. As the crate is about to slide, R must form with the vertical an angle
11
tan tan
0.35 19.29
ss
φµ
−−
= =
= °
Since the crate is a 3-force body. P must pass through E where R and W intersect.
0.4 m 1.1429 m
tan 0.35
1.1429 0.5 0.6429 m
0.6429 m
tan 0.4 m
s
CF
EF
EH EF HF
EH
HB
θ
a
= = =
=− = −=
= =
58.11
a
= °
58.1
a
= °
(b) Force Triangle
0.424
sin19.29 sin128.82
PW
PW= =
°°
2
0.424(40 kg)(9.81 m/s ),P=
166.4 NP=
Note: After the crate starts moving,
s
µ
should be replaced by the lower value
.
k
µ
This will
yield a larger value of
.
a
consent of McGraw–Hill Education.
PROBLEM 4.88
A 40–kg packing crate is pulled by a rope as shown. The coefficient of static
friction between the crate and the floor is 0.35. If α = 40°, determine (a) the
magnitude of the force P required to move the crate, (b) whether the crate will
slide or tip.
SOLUTION
Force P for which sliding is impending
(We assume that crate does not tip)
2
(40 kg)(9.81 m/s ) 392.4 NW= =
0: sin 40 0
y
F NW PΣ = − + °=
sin 40NWP=−°
(1)
0: 0.35 cos40 0
x
F NP
Σ = − °=
Substitute for
N
from Eq. (1):
0.35( sin 40 ) cos40 0WP P− ° − °=
0.35
0.35sin 40 cos40
W
P=°+ °
0.3532PW=
Force P for which crate rotates about C
(We assume that crate does not slide)
0: ( sin 40 )(0.8 m) ( cos40 )(0.5 m)
C
MP PΣ= ° + °
(0.4 m) 0W−=
0.4 0.4458
0.8sin 40 0.5cos 40
W
PW= =
°+ °
(b) Crate will first slide
(a)
0.3532(392.4 N)P=
138.6 NP=
consent of McGraw–Hill Education.
PROBLEM 4.89
The coefficients of friction are
0.40
s
µ
=
and
0.30
k
µ
=
between all
surfaces of contact. Determine the smallest force P required to start the
30–kg block moving if cable AB (a) is attached as shown, (b) is removed.
SOLUTION
(a) Free body: 20-kg block
2
1
11
(20 kg)(9.81 m/s ) 196.2 N
0.4(196.2 N) 78.48 N
s
W
FN
µ
= =
= = =
11
0: 0 78.48 NF TF TFΣ= − = = =
Free body: 30-kg block
12
0: 0F PF F TΣ= − − −=
78.48 N 196.2 N 78.48 N 353.2 NP=++=
353 N=P
(b) Free body: Both blocks
Blocks move together
2
(50 kg)(9.81 m/s ) 490.5 NW= =
0: 0F PFΣ= − =
0.4(490.5 N) 196.2 N
s
PN
µ
= = =
196.2 N=P
consent of McGraw–Hill Education.
PROBLEM 4.90
The coefficients of friction are
0.40
s
µ
=
and
0.30
k
µ
=
between all
surfaces of contact. Determine the smallest force P required to start
the 30–kg block moving if cable AB (a) is attached as shown, (b) is
removed.
SOLUTION
(a) Free body: 20-kg block
2
1
11
(20 kg)(9.81 m/s ) 196.2 N
0.4(196.2 N) 78.48 N
s
W
FN
µ
= =
= = =
11
0: 0 78.48 NF TF TFΣ= − = = =
Free body: 30-kg block
2
2
2
22
(30 kg)(9.81 m/s ) 294.3 N
196.2 N 294.3 N 490.5 N
0.4(490.5 N) 196.2 N
s
W
N
FN
µ
= =
=+=
= = =
12
0: 0
F PF FΣ= − − =
78.48 N 196.2 N 274.7 NP=+=
275 N=P
(b) Free body: Both blocks
Blocks move together
2
(50 kg)(9.81 m/s )
490.5 N
W=
=
0: 0
0.4(490.5 N) 196.2 N
s
F PF
PN
µ
Σ= − =
= = =
196.2 N=P
consent of McGraw–Hill Education.
PROBLEM 4.91
A 6.5–m ladder AB leans against a wall as shown. Assuming that the coefficient
of static friction
s
µ
is zero at B, determine the smallest value of
s
µ
at A for
which equilibrium is maintained.
SOLUTION
Free body: Ladder
Three-force body.
Line of action of A must pass through D, where W and B intersect.
At A:
1.25 m
tan 0.2083
6m
ss
µφ
= = =
0.208
s
µ
=
consent of McGraw–Hill Education.
PROBLEM 4.92
A 6.5–m ladder AB leans against a wall as shown. Assuming that the coefficient
of static friction
s
µ
is the same at A and B, determine the smallest value of
s
µ
for which equilibrium is maintained.
SOLUTION
Free body: Ladder
Motion impending:
A sA
B sB
FN
FN
µ
µ
=
=
0: (1.25 m) (6 m) (2.5 m) 0
A B sB
MW N N
µ
Σ= − − =
1.25
6 2.5
Bs
W
N
µ
=+
(1)
0: 0
y A sB
F N NW
µ
Σ= + −=
1.25
6 2.5
A sB
s
As
NW N
W
NW
µ
µ
µ
= −
= − +
(2)
0: 0
x sA B
F NN
µ
Σ= − =
Substitute for
A
N
and
B
N
from Eqs. (1) and (2):
2
22
2
1.25 1.25
6 2.5 6 2.5
6 2.5 1.25 1.25
1.25 6 1.25 0
0.2
s
sss
ss s
ss
s
WW
W
µ
µµµ
µµ µ
µµ
µ
−=
++
+− =
+−=
=
and
5 (Discard)
s
µ
= −
0.200
s
µ
=
consent of McGraw–Hill Education.
PROBLEM 4.93
End A of a slender, uniform rod of length L and weight W bears on a surface as
shown, while end B is supported by a cord BC. Knowing that the coefficients
of friction are
0.40
s
µ
=
and
0.30,
k
µ
=
determine (a) the largest value of
θ
for which motion is impending, (b) the corresponding value of the tension in the
cord.
SOLUTION
Free-body diagram
Rod AB is a three-force body. Thus, line of action of R must pass through D,
Where W and T intersect.
Triangle ABC bisects the angle
θ
.
(a) Thus,
1
2
s
φθ
=
Since motion impends,
1
tan 0.40 21.80
2 2(21.8 )
s
s
φ
θφ
−
= = °
= = °
43.6
θ
= °
(b) Force triangle:
This is a right triangle.
sin
sin 21.8
s
TW
W
φ
=
= °
0.371TW=
consent of McGraw–Hill Education.
PROBLEM 4.94
End A of a slender, uniform rod of length L and weight W bears on a
surface as shown, while end B is supported by a cord BC. Knowing that
the coefficients of friction are
0.40
s
µ
=
and
0.30,
k
µ
=
determine (a) the
largest value of
θ
for which motion is impending, (b) the corresponding
value of the tension in the cord.
SOLUTION
Free-body diagram
Three-force body. Line of action of R must pass through D, where T and R intersect.
Motion impends:
tan 0.4
21.80
s
s
φ
φ
=
= °
90
2
21.8 90
2
s
θφ
θ
+=°
+ °= °
136.4
θ
= °
(b) Force triangle (right triangle):
cos 21.8TW= °
0.928TW
=
consent of McGraw–Hill Education.