PROBLEM 4.79
Determine the smallest value of P required to (a) start the block up the
incline, (b) keep it moving up.
SOLUTION
(a) To start block up the incline:
1
0.40
tan 0.40 21.80
s
s
µ
φ
−
=
= = °
From force triangle:
50 lb
sin 51.80 sin 28.20
P=
°°
83.2 lbP=
(b) To keep block moving up:
1
0.30
tan 0.30 16.70
k
k
µ
φ
−
=
= = °
SOLUTION Continued
From force triangle:
50 lb
sin 46.70 sin 33.30°
P=
°
66.3 lbP=
PROBLEM 4.80
The 80–lb block is attached to link AB and rests on a moving belt.
Knowing that
0.25
s
µ
=
and
0.20,
k
µ
=
determine the magnitude of
the horizontal force P that should be applied to the belt to maintain its
motion (a) to the right, (b) to the left.
SOLUTION
We note that link AB is a two-force member, since there is motion between belt and block
0.20
k
µ
=
and
1
tan 0.20 11.31
k
φ
−
= = °
PROBLEM 4.81
The 50–lb block A and the 25–lb block B are supported by an incline
that is held in the position shown. Knowing that the coefficient of
static friction is 0.15 between the two blocks and zero between block
B and the incline, determine the value of
θ
for which motion is
impending.
SOLUTION
Since motion impends,
s
FN
µ
=
between A+B
Free body: Block A
PROBLEM 4.82
The 50–lb block A and the 25–lb block B are supported by an incline
that is held in the position shown. Knowing that the coefficient of
static friction is 0.15 between all surfaces of contact, determine the
value of
θ
for which motion is impending.
SOLUTION
Since motion impends,
s
FN
µ
=
between A+B
Free body: Block A
PROBLEM 4.83
The coefficients of friction between the block and the rail are
0.30
s
µ
=
and
0.25.
k
µ
=
Knowing that
65 ,
θ
= °
determine the smallest value of P required
(a) to start the block moving up the rail, (b) to keep it from moving down.
SOLUTION
(a) To start block up the rail:
1
0.30
tan 0.30 16.70
s
s
µ
φ
−
=
= = °
Force triangle:
500 N
sin 51.70 sin (180° 25 51.70 )
P=
° − °− °
403 NP=
(b) To prevent block from moving down:
Force triangle:
500 N
sin 18.30 sin (180° 25° 18.30°)
P=
° −−
229 NP=
consent of McGraw–Hill Education.
PROBLEM 4.84
Knowing that P = 100 N, determine the range of values of θ for which equilibrium
of the 7.5-kg block is maintained.
SOLUTION
Free-body diagram of block and force triangle:
For motion impending downward,
( )
( )
2
7.5 kg 9.81 m/s 73.575 NW= =
( )
( )
( )
11
tan tan 0.45 24.23
100 N 73.575 N
sin
sin 65.77
sin 24.23 0.67093
24.23 42.14
ss
s
φµ
θφ
θ
θ
−−
= = = °
=+
+=
+=
17.91
θ
=
For motion impending upward,
consent of McGraw–Hill Education.
SOLUTION Continued
( )
( )
100 N 73.575 N
sin
sin 114.23
sin 24.23 0.67093
24.23 42.14
s
θφ
θ
θ
=−
−=
−=
66.37
θ
=
17.91 66.4
θ
≤≤
consent of McGraw–Hill Education.
PROBLEM 4.85
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. If
32 in.,h=
determine the magnitude of the force P required to move
the cabinet to the right (a) if all casters are locked, (b) if the casters at B are
locked and the casters at A are free to rotate, (c) if the casters at A are locked
and the casters at B are free to rotate.
SOLUTION
FBD cabinet: Note: for tipping,
0
AA
NF= =
tip
0: (12 in.) (32 in.) 0
B
M WPΣ= − =
tip 2.66667P=
(a) All casters locked. Impending slip:
A sA
B sB
FN
FN
µ
µ
=
=
0: 0
y AB
F NNWΣ= + −=
AB
NNW+=
120 lbW=
So
AB s
FF W
µ
+=
0.3
s
µ
=
0: 0
x AB
AB s
F PF F
PF F W
µ
Σ= − − =
=+=
0.3(120 lb)P=
or 36.0 lb=P
tip
( 0.3 OK)P WP= <
(b) Casters at A free, so
0
A
F=
Impending slip:
B sB
FN
µ
=
0: 0
xB
F PFΣ= − =
B sB B s
P
PF N N
µ
µ
= = =
consent of McGraw–Hill Education.
PROBLEM 4.79
Determine the smallest value of P required to (a) start the block up the
incline, (b) keep it moving up.
SOLUTION
(a) To start block up the incline:
1
0.40
tan 0.40 21.80
s
s
µ
φ
−
=
= = °
From force triangle:
50 lb
sin 51.80 sin 28.20
P=
°°
83.2 lbP=
(b) To keep block moving up:
1
0.30
tan 0.30 16.70
k
k
µ
φ
−
=
= = °
SOLUTION Continued
From force triangle:
50 lb
sin 46.70 sin 33.30°
P=
°
66.3 lbP=
PROBLEM 4.80
The 80–lb block is attached to link AB and rests on a moving belt.
Knowing that
0.25
s
µ
=
and
0.20,
k
µ
=
determine the magnitude of
the horizontal force P that should be applied to the belt to maintain its
motion (a) to the right, (b) to the left.
SOLUTION
We note that link AB is a two-force member, since there is motion between belt and block
0.20
k
µ
=
and
1
tan 0.20 11.31
k
φ
−
= = °
PROBLEM 4.81
The 50–lb block A and the 25–lb block B are supported by an incline
that is held in the position shown. Knowing that the coefficient of
static friction is 0.15 between the two blocks and zero between block
B and the incline, determine the value of
θ
for which motion is
impending.
SOLUTION
Since motion impends,
s
FN
µ
=
between A+B
Free body: Block A
PROBLEM 4.82
The 50–lb block A and the 25–lb block B are supported by an incline
that is held in the position shown. Knowing that the coefficient of
static friction is 0.15 between all surfaces of contact, determine the
value of
θ
for which motion is impending.
SOLUTION
Since motion impends,
s
FN
µ
=
between A+B
Free body: Block A
PROBLEM 4.83
The coefficients of friction between the block and the rail are
0.30
s
µ
=
and
0.25.
k
µ
=
Knowing that
65 ,
θ
= °
determine the smallest value of P required
(a) to start the block moving up the rail, (b) to keep it from moving down.
SOLUTION
(a) To start block up the rail:
1
0.30
tan 0.30 16.70
s
s
µ
φ
−
=
= = °
Force triangle:
500 N
sin 51.70 sin (180° 25 51.70 )
P=
° − °− °
403 NP=
(b) To prevent block from moving down:
Force triangle:
500 N
sin 18.30 sin (180° 25° 18.30°)
P=
° −−
229 NP=
consent of McGraw–Hill Education.
PROBLEM 4.84
Knowing that P = 100 N, determine the range of values of θ for which equilibrium
of the 7.5-kg block is maintained.
SOLUTION
Free-body diagram of block and force triangle:
For motion impending downward,
( )
( )
2
7.5 kg 9.81 m/s 73.575 NW= =
( )
( )
( )
11
tan tan 0.45 24.23
100 N 73.575 N
sin
sin 65.77
sin 24.23 0.67093
24.23 42.14
ss
s
φµ
θφ
θ
θ
−−
= = = °
=+
+=
+=
17.91
θ
=
For motion impending upward,
consent of McGraw–Hill Education.
SOLUTION Continued
( )
( )
100 N 73.575 N
sin
sin 114.23
sin 24.23 0.67093
24.23 42.14
s
θφ
θ
θ
=−
−=
−=
66.37
θ
=
17.91 66.4
θ
≤≤
consent of McGraw–Hill Education.
PROBLEM 4.85
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. If
32 in.,h=
determine the magnitude of the force P required to move
the cabinet to the right (a) if all casters are locked, (b) if the casters at B are
locked and the casters at A are free to rotate, (c) if the casters at A are locked
and the casters at B are free to rotate.
SOLUTION
FBD cabinet: Note: for tipping,
0
AA
NF= =
tip
0: (12 in.) (32 in.) 0
B
M WPΣ= − =
tip 2.66667P=
(a) All casters locked. Impending slip:
A sA
B sB
FN
FN
µ
µ
=
=
0: 0
y AB
F NNWΣ= + −=
AB
NNW+=
120 lbW=
So
AB s
FF W
µ
+=
0.3
s
µ
=
0: 0
x AB
AB s
F PF F
PF F W
µ
Σ= − − =
=+=
0.3(120 lb)P=
or 36.0 lb=P
tip
( 0.3 OK)P WP= <
(b) Casters at A free, so
0
A
F=
Impending slip:
B sB
FN
µ
=
0: 0
xB
F PFΣ= − =
B sB B s
P
PF N N
µ
µ
= = =
consent of McGraw–Hill Education.