13–21.
The conveyor belt delivers each 12-kg crate to the ramp at
A such that the crate’s speed is vA
=2.5 m>s,
directed down
along the ramp. If the coefficient of kinetic friction between
each crate and the ramp is m
k=0.3,
determine the smallest
incline
u
of the ramp so that the crates will slide off and fall
into the cart.
vA 2.5 m/s
3 m
A
B
u
13–22.
The 50-kg block A is released from rest. Determine the
velocity of the 15-kg block B in 2 s.
Kinematics. As shown in Fig. a, the position of block B and point A are specified by
sB and sA respectively. Here the pulley system has only one cable which gives
sA+ sB+2(sB–a)=l
sA+ 3sB=l+2a
(1)
Taking the time derivative of Eq. (1) twice,
aA+3aB=0
(2)
Equations of Motion. The FBD of blocks B and A are shown in Fig. b and c. To be
consistent to those in Eq. (2), aA and aB are assumed to be directed towards the
positive sense of their respective position coordinates sA and sB. For block B,
+
c
ΣF
y=
ma
y
;
3T–15(9.81) =15(–aB)
(3)
For block A,
+
c
ΣF
y=
ma
y
;
T–50(9.81) =50(–aA)
(4)
Solving Eqs. (2), (3) and (4),
aB=–2.848 m>s2=2.848 m>s2
c
aA=8.554 m>s2
T=63.29 N
The negative sign indicates that aB acts in the sense opposite to that shown in FBD.
The velocity of block B can be determined using
+
c
v
B=(
v
A)0+aBt;
vB=0+2.848(2)
v
B=5.696 m>s=5.70 m>s
c
Ans.
A
D
C
E
13–23.
If the supplied force F
=
150 N, determine the velocity of
the 50-kg block A when it has risen 3 m, starting from rest.
SOLUTION
Equations of Motion. Since the pulleys are smooth, the tension is constant
throughout each entire cable. Referring to the FBD of pulley C, Fig. a, of which its
mass is negligible.
+
c
ΣF
y=
0;
150 +150 –T=0
T=300 N
Subsequently, considered the FBD of block A shown in Fig. b,
+
c
ΣF
y=
ma
y
;
300 +300 –50(9.81) =50a
a=2.19 m>s2
c
Kinematics. Using the result of a,
(+
c
)
v
2=
v
2
0+2acs;
v
2=02+2(2.19)(3)
v=3.6249 m>s=3.62 m>s
Ans.
A
B
C
F
SOLUTION
*13–24.
A 60-kg suitcase slides from rest 5 m down the smooth ramp.
Determine the distance R where it strikes the ground at B.
How long does it take to go from A to B?
5 m
2.5 m
30
C
A
SOLUTION
13–25.
Solve Prob. 13–24 if the suitcase has an initial velocity down
the ramp of vA
=2 m>s
, and the coefficient of kinetic
friction along AC is m
k =0.2.
5 m
2.5 m
30
C
A
13–26.
The 1.5 Mg sports car has a tractive force of
F=4.5 kN
. If
it produces the velocity described by v–t graph shown, plot
the air resistance R versus t for this time period.
SOLUTION
Equation of Motion. Referring to the FBD of the car shown in Fig. a,
(d
+)ΣFx=max;
4500 –R=1500(–0.1t+3)
R=5150t6N
The plot of R vs t is shown in Fig. b
v (m/s)
45
F
R
v (–0.05t2 + 3t) m/s
13–27.
The conveyor belt is moving downward at 4 m
>
s. If the
coefficient of static friction between the conveyor and the
15-kg package B is m
s =0.8,
determine the shortest time
the belt can stop so that the package does not slide on
thebelt.
SOLUTION
B
30
4 m/s
*13–28.
At the instant shown the 100-lb block A is moving down the
plane at 5 ft
>
s while being attached to the 50-lb block B. If
the coefficient of kinetic friction between the block and the
incline is m
k=0.2,
determine the acceleration of A and
thedistance A slides before it stops. Neglect the mass of the
pulleys and cables.
SOLUTION
C
B
D
3
A4
5
13–29.
The force exerted by the motor on the cable is shown in the
graph. Determine the velocity of the 200-lb crate when
.t=2.5 s
SOLUTION
Equation of Motion: For ,.By referring to Fig.a,
;
Kinematics: The velocity of the crate can be obtained by integrating the kinematic
equation, .For ,at will be used as the lower
integration limit.Thus,
When ,
Ans.v
=
8.05(2.52)
–
32.2(2.5)
+
32.2
=
2.01 ft
>
s
t=2.5 s
=
A
8.05t2–32.2t+32.2
B
ft>s
v=
A
8.05t2–32.2t
B
2t
2 s
Lv
0
dv =Lt
2 s
(16.1t–32.2)dt
Ldv =Ladt(+c)
t=2 sv=02 s…t62.5 sdv =adt
a=(16.1t–32.2) ft>s2
100t–200 =200
32.2 a+c©Fy=may
F=250
2.5 t=(100
t) lb2 s6t62.5 s
M
250 lb
2.5
F (lb)
t (s)
13–30.
The force of the motor Mon the cable is shown in the graph.
Determine the velocity of the 400-kg crate Awhen .t=2 s
SOLUTION
Equations of Motion: . By referring to Fig.a,
;
Kinematics: The velocity of the crate can be obtained by integrating the kinematic
equation, .For , at will be used as the
lower integration limit. Thus,
When ,
Ans.v=1.0417(23)–9.81(2) +11.587 =0.301 m>s
t=2 s
=
A
1.0417t3–9.81t+11.587
B
m>s
v=
A
1.0417t3–9.81t
B
2t
1.772 s
Lv
0
dv =Lt
1.772 s
A
3.125t2–9.81
B
dt
Ldv =Ladt(+c)
t=1.772 sv=01.772 s…t62 sdv =adt
a=(3.125t2–9.81) m>s2
2
A
625t2
B
–400(9.81) =400a+c©Fy=may
F=
A
625t2
B
N
t=1.772 s
M
F (N)
F 625 t2
2500
2t (s)
13–31.
The tractor is used to lift the 150-kg load Bwith the 24-m-
long rope, boom, and pulley system. If the tractor travels to
the right at a constant speed of 4 ms, determine the tension
in the rope when .When ,.sB=0sA=0sA=5m >
sB
B
12 m
SOLUTION
+c©Fy=may;T–150(9.81) =150(1.0487)
aB=-C(5)2(4)2
((5)2+144)3
2
–(4)2+0
((5)2+144)1
2S=1.0487 m>s2
s
$
B=-Cs2
As
#2
A
A
s2
A+144
B
3
2
–s
#2
A+sAs
$
A
A
s2
A+144
B
1
2S
–s
$
B–
A
s2
A+144
B
–3
2asAs
#
Ab2
+
A
s2
A+144
B
–1
2as
#2
Ab+
A
s2
A+144
B
–1
2asAs
$
Ab=0
–sB+
A
s2
A+144
B
Ab=0
12 –sB+2s2
A+(12)2=24
*13–32.
The tractor is used to lift the 150-kg load Bwith the 24-m-
long rope, boom, and pulley system. If the tractor travels to
the right with an acceleration of and has a velocity of
4m s at the instant , determine the tension in the
rope at this instant. When ,.sB=0sA=0
sA=5m>3m>s2
SOLUTION
Ans.T=1.80 kN
+c©Fy=may;T–150(9.81) =150(2.2025)
aB=-C(5)2(4)2
((5)2+144)3
2
–(4)2+(5)(3)
((5)2+144)1
2S=2.2025 m>s2
s
$
B=-Cs2
As
#2
A
A
s2
A+144
B
3
2
–s
#2
A+sAs
$
A
A
s2
A+144
B
1
2S
–s
$
B–
A
s2
A+144
B
–3
2asAs
#
Ab2
+
A
s2
A+144
B
–1
2as
#2
Ab+
A
s2
A+144
B
–1
2asAs
$
Ab=0
–s
#
B+1
2
A
s2
A+144
B
–3
2a2sAs
#
Ab=0
A+(12)2=24
sA
sB
B
12 m
277
13–33.
Block A and B each have a mass m. Determine the largest
horizontal force P which can be applied to B so that it will
not slide on A. Also, what is the corresponding acceleration?
The coefficient of static friction between A and B is
ms.
Neglect any friction between A and the horizontal surface.
SOLUTION
cos u
–
m
s
sin u
P
A
B
cos u
–
m
s
sin u
13–34.
The 4-kg smooth cylinder is supported by the spring having
a stiffness of kAB
=
120 N
>
m. Determine the velocity of the
cylinder when it moves downward s
=
0.2 m from its
equilibrium position, which is caused by the application of
the force F
=
60 N.
SOLUTION
s
kAB 120 N/
m
F
60 N
B
13–35.
The coefficient of static friction between the 200-kg crate
and the flat bed of the truck is Determine the
shortest time for the truck to reach a speed of 60 km h,
starting from rest with constant acceleration, so that the
crate does not slip.
>
ms =0.3.
SOLUTION
280
*13–36.
SOLUTION
Ans.v=14.6 ft>s
–
C
2s2–431+s2
D
1
0=1
32.2
A
v2–152
B
–L1
0
¢
4sds–4sds
21+s2
≤
=Lv
15 a2
32.2 bvdv
:
+©Fx=max;–4
A
21+s2–1
B
¢
s
21+s2
≤
=a2
32.2 bavdv
ds b
Fs=kx;Fs=4
A
21+s2–1
B
The 2-lb collar Cfits loosely on the smooth shaft. If the
spring is unstretched when and the collar is given a
velocity of 15 ft s, determine the velocity of the collar when
.s=1ft>s=0
C
1ft
k4lb/ft
15 ft/s
s
Ans:
v
=14.6 ft>s
13–37.
The 10-kg block A rests on the 50-kg plate B in the position
shown. Neglecting the mass of the rope and pulley, and
using the coefficients of kinetic friction indicated, determine
the time needed for block A to slide 0.5 m on the plate when
the system is released from rest.
SOLUTION
C
A
B
mAB 0.2
0.5 m
mBC 0.1
13–38.
The 300-kg bar B, originally at rest, is being towed over a
series of small rollers. Determine the force in the cable
when t
=
5 s, if the motor M is drawing in the cable for a
short time at a rate of v
=
(0.4t
2
) m
>
s, where t is in seconds
(0 …t…6 s).
How far does the bar move in 5 s? Neglect
the mass of the cable, pulley, and the rollers.
SOLUTION
M
=
283
13–39.
An electron of mass mis discharged with an initial
horizontal velocity of v0. If it is subjected to two fields of
force for which and , where is
constant, determine the equation of the path, and the speed
of the electron at any time t.
F0
Fy=0.3F0
Fx=F0
SOLUTION
;
;
Thus,
Ans.
Ans.x=y
0.3
+v0a
B
2m
0.3F0
by 1
2
x=F0
2ma2m
0.3F0by+v0aB2m
0.3F0by 1
2
t=aB2m
0.3F0by 1
2
y=0.3F0 t2
2m
Ly
0
dy =
Lt
0
0.3F0
m t dt
x=F0 t2
2m+v0 t
Lx
0
dx =Lt
0aF0
m t+v0b dt
=1
m21.09F0
2t2+2F0tmv0+m2v0
2
v=
C
aF0
m t+v0b2
+a0.3F0
m tb2
vy=0.3F0
m t
Lvy
0
dvy=Lt
0
0.3F0
m dt
vx=F0
m t+v0
Lvx
v0
dvx=Lt
0
F0
m dt
0.3 F0=may
+c©Fy=may
F0=max
:
+©Fx=max
y
v0
+ + + + + + + + + + + + + +
+ + + + + + + + + + + + + +
x=
0.3 +v0
a
A
0.3F
0b
y1
*13–40.
The 400-lb cylinder at A is hoisted using the motor and the
pulley system shown. If the speed of point B on thecable is
increased at a constant rate from zero to vB
=10 ft>s
in
t=5 s
, determine the tension in the cable at B to cause
themotion.
SOLUTION
B