13–1.
The 6-lb particle is subjected to the action of
its weight and forces
and where tis in
seconds. Determine the distance the ball is from the origin
2safter being released from rest.
F3=5–2ti6lb,5t2i–4tj–1k6lb,
F2=F1=52i+6j–2tk6lb,
SOLUTION
Equating components:
Since , integrating from ,,yields
Since , integrating from , yields
When then, ,
Thus,
Ans.s=(14.31)2+(35.78)2+(–89.44)2=97.4 ft
sz=-89.44 ftsy=35.78 ftsx=14.31 ftt=2s
¢
6
32.2
≤
sx=t4
12 –t3
3+t2
¢
6
32.2
≤
sy=-
2t3
3+3t2
¢
6
32.2
≤
sz=-
t3
3–7t2
2
t=0s=0ds =vdt
¢
6
32.2
≤
vx=t3
3–t2+2t
¢
6
32.2
≤
vy=-2t2+6t
¢
6
32.2
≤
vz=-t2–7t
t=0n=0dv =adt
¢
6
32.2
≤
ax=t2–2t+2
¢
6
32.2
≤
ay=-4t+6
¢
6
32.2
≤
az=-2t–7
©F=ma;(2i+6j–2tk)+(t2i–4tj–1k)–2ti–6k=
¢
6
32.2
≤
(axi+ayj+azk)
z
y
x
F1
F3
F2
246
13–2.
C
5°
AB
The two boxcars A and B have a weight of 20000lb and
30 000 lb, respectively. If they are freely coasting down
the incline when the brakes are applied to all the wheels
of car A, determine the force in the coupling C between
the two cars. The coeffi cient of kinetic friction between
the wheels of A and the tracks is mk=0.5. The wheels of
car B are free to roll. Neglect their mass in the calculation.
Suggestion: Solve the problem by representing single
resultant normal forces acting on A and B, respectively.
SOLUTION
13–3.
If the coefficient of kinetic friction between the 50-kg crate
and the ground is , determine the distance the
crate travels and its velocity when The crate starts
from rest, and .P=200 N
t=3 s.
mk =0.3
SOLUTION
the motion of the crate which is to the right, Fig.a.
Equations of Motion: Here,.Thus,
;
;
Kinematics: Since the acceleration aof the crate is constant,
Ans.
and
Ans.s=0+0+1
2
(1.121)
A
32
B
=5.04 m
s=s0+v0t+1
2 act2
A
:
+
B
v=0+1.121(3) =3.36 m>s
v=v0+act
A
:
+
B
a=1.121 m>s2
200 cos 30° –0.3(390.5) =50a©Fx=max
:
+
N=390.5 N
N–50(9.81) +200 sin 30° =0+c©Fy=0
ay=0
30
P
*13–4.
If the 50-kg crate starts from rest and achieves a velocity of
when it travels a distance of 5 m to the right,
determine the magnitude of force Pacting on the crate.
The coefficient of kinetic friction between the crate and the
ground is .mk=0.3
v=4 m>s
SOLUTION
Free-Body Diagram: Here, the kinetic friction is required to be
directed to the left to oppose the motion of the crate which is to the right, Fig.a.
Equations of Motion:
;
Using the results of Nand a,
;
Ans.P=224 N
P cos 30° –0.3(490.5 –0.5P)=50(1.60):
+©Fx=max
N=490.5 –0.5P
N+P sin 30° –50(9.81) =50(0)+c©Fy=may
Ff=mkN=0.3N
a=1.60 m>s2 :
42=02+2a(5 –0)
v2=v0 2+2ac(s–s0)( :
30
P
13–5.
SOLUTION
are required to act up the plane to oppose the motion of
the blocks which are down the plane. Since the blocks are connected, they have a
common acceleration a.
Equations of Motion: By referring to Figs. (a) and (b),
(1)
and
(2)
Solving Eqs. (1) and (2) yields
Ans.F=6.37 N
a=3.42 m>s2
F+14.14 =6a
R+©Fx¿=max¿;F+6(9.81) sin 30° –0.3(50.97) =6a
NB=50.97 N
+Q©Fy¿=may¿;NB–6(9.81) cos 30° =6(0)
40.55 –F=10a
R+©Fx¿=max¿; 10(9.81) sin 30° –0.1(84.96) –F=10a
NA=84.96 N
+Q©Fy¿=may¿;NA–10(9.81) cos 30° =10(0)
(Ff)B=mBNB=0.3NB
If blocks Aand Bof mass 10 kg and 6 kg, respectively, are
placed on the inclined plane and released, determine the
force developed in the link.The coefficients of kinetic
friction between the blocks and the inclined plane are
and . Neglect the mass of the link.mB=0.3mA=0.1
A
B
30
13–6.
The 10-lb block has a speed of 4 ft
>
s when the force of
F
=
(8t2) lb is applied. Determine the velocity of the block
when t
=
2 s. The coefficient of kinetic friction at the surface
is m
k=0.2.
SOLUTION
v
4 ft
/
s
F (8t2) lb
13–7.
The 10-lb block has a speed of 4 ft
>
s when the force of
F
=
(8t2) lb is applied. Determine the velocity of the block
when it moves s
=
30 ft. The coefficient of kinetic friction at
the surface is m
s=0.2.
SOLUTION
v
4 ft
/
s
F (8t2) lb
*13–8.
SOLUTION
we have
For ,,.Applying equation
v={t+50} ft>s
v–60
t–10 =80 –60
30 –10
10 6t…30 s
a=dv
dt =6ft>s2
The speed of the 3500-lb sports car is plotted over the 30-s
time period. Plot the variation of the traction force Fneeded
to cause the motion.
v(ft/s)
80
60
F
13–9.
The conveyor belt is moving at 4 m>s. If the
coeffi cient of static friction between the conveyor and
the 10-kg package B is ms=0.2, determine the shortest time
the belt can stop so that the package does not slide on the belt.
SOLUTION
+ΣFx=max; 0.2(98.1) =10 a
B
254
13–10.
The conveyor belt is designed to transport packages of
various weights. Each 10-kg package has a coefficient of
kinetic friction m
k=0.15.
If the speed of the conveyor is
5 m
>
s, and then it suddenly stops, determine the distance
the package will slide on the belt before coming to rest.
SOLUTION
S
+ΣF
x
=ma
x
;
0.15 m(9.81) =ma
a
=
1.4715 m
>
s
2
(
S
+)
v
2
=v
2
0+2ac(s–s0)
0=(5)2+2(–1.4715)(s–0)
s=8.49 m
Ans.
Ans:
s=8.49 m
B
255
13–11.
Determine the time needed to pull the cord at B down 4 ft
starting from rest when a force of 10 lb is applied to the
cord. Block A weighs 20 lb. Neglect the mass of the pulleys
and cords.
SOLUTION
+
c
Σ
Fy
=
may; 40 –20 =
20
32.2
aA Ans.
a
A=
32.2 ft
>
s
2
sB+2sC=l;
aB=–2aC
2sA–sC=l′;
2aA= aC
aB=–4aA
a
B=
128.8 ft
>
s
2
(+ T)
s=s0+v0t+
1
2
act2
4=0+0+
1
2
(128.8) t2
t=0.249 s
Ans.
Ans:
t=0.249 s
A
10 lb
B
C
256
*13–12.
Cylinder B has a mass m and is hoisted using the cord and
pulley system shown. Determine the magnitude of force F
as a function of the block’s vertical position y so that when
F is applied the block rises with a constant acceleration aB.
Neglect the mass of the cord and pulleys.
SOLUTION
+
c
Σ
Fy=may;
2F cos
u
–mg =maB
where cos u=
y
2
y2+
1
d
2
2
2
2F
ay
2
y2
+
1
d
2
2
2
b
–mg =maB
F=
m(aB+g)
2
4y2+d2
4y Ans.
aBB
F
y
d
13–13.
Block A has a weight of 8 lb and block B has a weight of 6 lb.
They rest on a surface for which the coefficient of kinetic
friction is m
k=0.2.
If the spring has a stiffness of k
=
20 lb
>
ft,
and it is compressed 0.2 ft, determine the acceleration of each
block just after they are released.
k
BA
SOLUTION
13–14.
The 2-Mg truck is traveling at 15 ms when the brakes on all its
wheels are applied, causing it to skid for a distance of 10 m
before coming to rest. Determine the constant horizontal force
developed in the coupling C,and the frictional force
developed between the tires of the truck and the road during
this time.The total mass of the boat and trailer is 1 Mg.
>
Free-Body Diagram:The free-body diagram of the truck and trailer are shown in
Figs. (a) and (b), respectively. Here, Frepresentes the frictional force developed
when the truck skids, while the force developed in coupling Cis represented by T.
Equations of Motion:Using the result of aand referrning to Fig. (a),
Ans.
Using the results of aand Tand referring to Fig. (b),
Ans.F=33 750 N =33.75 kN
+c©Fx=max;11 250 –F=2000(–11.25)
T=11 250 N =11.25 kN
:
+©Fx=max;–T=1000(–11.25)
a=-11.25 m>s2=11.25 m>s2;
0=152+2a(10 –0)
a:
C
13–15.
The motor lifts the 50-kg crate with an acceleration of
6 m
>
s2. Determine the components of force reaction and
the couple moment at the fixed support A.
SOLUTION
4 m
y
x
B
A
6 m/s2
30
260
*13–16.
The 75-kg man pushes on the 150-kg crate with a horizontal
force F. If the coefficients of static and kinetic friction
between the crate and the surface are m
s=0.3
and
m
k=0.2,
and the coefficient of static friction between the
man’s shoes and the surface is m
s=0.8
, show that the man
is able to move the crate. What is the greatest acceleration
the man can give the crate?
F
SOLUTION
Ans:
a
=
1.96 m
>
s
2
261
13–17.
Determine the acceleration of the blocks when the system is
released. The coefficient of kinetic friction is
mk,
and the
mass of each block is m. Neglect the mass of the pulleys
andcord.
SOLUTION
2
2
A
B
2
13–18.
SOLUTION
Ans.
R=5.30 ft
tBC =0.2413 s
4 =0+25.38 sin 30° tBC +1
2(32.2)(tBC)2
(+T) sy=(sy)0+(vy)0 t+1
2act2
R=0+25.38 cos 30°(tBC)
( :
+)sx=(sx)0+(vx)0 t
tAB =1.576 s
25.38 =0+16.1 tAB
(+R) v=v0+ac t ;
vB=25.38 ft>s
v2
B=0+2(16.1)(20)
(+R)v2=v2
0+2 ac(s–s0);
a=16.1 ft>s2
+R ©F
x=m ax ;
40 sin 30° =40
32.2a
A 40-lb suitcase slides from rest 20 ft down the smooth
ramp. Determine the point where it strikes the ground at C.
How long does it take to go from Ato C?
20 ft
4 ft
30
C
A
B
13–19.
Solve Prob. 13–18 if the suitcase has an initial velocity down
the ramp of and the coefficient of kinetic
friction along AB is .mk =0.2
vA =10 ft>s
SOLUTION
;
;
Ans.
Total time Ans.
=
t
AB +
t
BC =
1.48 s
R=5.08 ft
tBC =0.2572 s
4=0+22.82 sin 30° tBC +1
2(32.2)(tBC)2
(+T) sy=(sy)0+(vy)0 t+1
2 ac t2
R=0+22.82 cos 30° (tBC)
(:
+) sx=(sx)0+(vx)0 t
tAB =1.219 s
22.82 =10 +10.52 tAB
(+R) v=v0+ac t
vB=22.82 ft>s
vB
2=(10)2+2(10.52)(20)
(+R) v2=v0
2+2 ac (s–s0)
a=10.52 ft>s2
20 ft
4 ft
30
C
A
B
264
*13–20.
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
speed at which each crate slides off the ramp at B. Assume
that no tipping occurs. Take
u=30°.
vA 2.5 m/s
3 m
A
B
u
SOLUTION
Q + ΣF
y
=ma
y
;
NC–12(9.81) cos 30°=0
NC=101.95 N
+R ΣFx=max;
12(9.81)
sin 30°–0.3(101.95) =12 aC
aC=2.356 m>s2
(+R)
v
2
B=
v
2
A+
2 a
C
(s
B–
s
A
)
v
2
B=
(2.5)
2+
2(2.356)(3
–
0)
vB=4.5152 =4.52 m>s
Ans.
Ans:
vB=4.52 m>s