Solution Manual for Engineering Mechanics Dynamics, 13th Edition

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R1–1.
An automobile is traveling with a constant speed along a
horizontal circular curve that has a radius If the
magnitude of acceleration is determine the
speed at which the automobile is traveling.
a = 8 ft>s2 ,
r = 750 ft.
SOLUTION
Ans.v = 77.5 ft>s
8 = v2
750
a = an = 8 = v2
r

R1–2.
SOLUTION
a) The maximum friction force between blocks A and B is
Thus, both blocks move together.
Ans.
b) In this case
Block A:
Ans.
Block B:
Ans.a B = 3.86 ft>s2
:+ a Fx = max; 20(0.3) = 50
32.2 aB
aA = 70.8 ft>s2
:+ a Fx = max; 20(0.3) = 20
32.2 aA
8 lb 6 F = 50 lb
aB = aA = a = 2.76 ft>s2
:+ a Fx = max; 6 = 70
32.2 a
Fmax = 0.4(20) = 8 lb 7 6 lb
Block B rests on a smooth surface. If the coefficients of
friction between A and B are and
determine the acceleration of each block if (a)
and (b) F = 50 lb.
F = 6 lb,
mk = 0.3,ms = 0.4
20 lb
A
B 50 lb
F

R1–3.
The small 2-lb collar starting from rest at A slides down
along the smooth rod. During the motion, the collar is acted
erehwecrofaybnopu x, y, z are
in feet. Determine the collar’s speed when it strikes the wall
at B.
F = 510i + 6yj + 2zk6 lb,
y
x
z
8 ft
10 ft
F
1 ft
B
4 ft
A
SOLUTION
Ans.vB = 47.8 ft>s
0 + 2(10-1) + L
0
4
10dx + L
8
0
6y dy + L
1
10
2z dz = 1
2 a 2
32.2bv2
B
T1 + © LFds = T2
rAB = rB - rA = -4i + 8j - 9k

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*R1–4.
SOLUTION
Since
Since
Ans.a = 2(-0.217)2 + (0) 2 + (0)2 = 0.217 m>s2
a z = z
$ = 0
au = r
# u
$ + 2 r u = 10(0) + 2(0)(0.1473) = 0
a r = r - r
# u2 = 0 - 10(0.1473) 2 = -0.217
u = 0
vu = r u
# = 1.473 u = 1.473
10 = 0.1473
r = 10 r
# = 0 r = 0
vz = -1.5 sin 10.81° = -0.2814 m>s
vu = 1.5 cos 10.81° = 1.473 m>s
vr = 0
v = 1.5 m>s
f = tan-1 a 12
2p(10) b = 10.81°
The automobile travels from a parking deck down along a
cylindrical spiral ramp at a constant speed of . If
the ramp descends a distance of 12 m for every full
revolution, rad, determine the magnitude of the car’s
acceleration as it moves along the ramp, . Hint: For
part of the solution, note that the tangent to the ramp at any
point is at an angle of
from the horizontal. Use this to determine the velocity
components and , which in turn are used to determine
and .z
# u
#
vzvu
f = tan- 1 (12>32p(10)4) = 10.81°
r = 10 m
u = 2p
v = 1.5 m>s
10 m
12 m

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R1–7.
The van is traveling at 20 km h when the coupling of the
trailer at A fails. If the trailer has a mass of 250 kg and
coasts 45 m before coming to rest, determine the constant
horizontal force F created by rolling friction which causes
the trailer to stop.
>
A
20 km/h
F
SOLUTION
Ans.:+ ©Fx = max ; F = 250(0.3429) = 85.7 N
a = -0.3429 m>s2 = 0.3429 m>s2 :
0 = 5.556 2 + 2(a)(45 - 0)
a ;+ b y2 = y2
0 + 2ac (s - s0)
20 km>h = 20(103
)
3600 = 5.556 m>s

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*R1–8.
SOLUTION
Ans.
Ans.ymax = 3(10)2 - 18(10) + 15 = 135 ft>s
ymax occurs when t = 10 s
amax = 6(10) - 18 = 42 ft>s2
amax occurs at t = 10 s,
a = dy
dt = 6t - 18
y = ds
dt = 3t2 - 18t + 15
s = t3 - 9t2 + 15t
The position of a particle along a straight line is given by
where t is in seconds. Determine its
maximum acceleration and maximum velocity during the
time interval 0 … t … 10 s.
s = 1t3 - 9t2 + 15t2 ft,

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R1–9.
The spool, which has a mass of 4 kg, slides along the rotating
rod. At the instant shown, the angular rate of rotation of the
rod is and this rotation is increasing at
At this same instant, the spool has a velocity
of 3 m/s and an acceleration of both measured
relative to the rod and directed away from the center O
when Determine the radial frictional force and
the normal force, both exerted by the rod on the spool at
this instant.
r = 0.5 m.
1 m>s2 ,
u
$ = 2 rad>s2
.
u
# = 6 rad >s
SOLUTION
Ans.
Ans.N = 2(148)2 + (39.24)2 = 153 N
Fr = -68 N
Nz = 39.24 N
©F z = maz ; N z - 4(9.81) = 0
©Fu = mau ; Nu = 4(37) = 148 N
©Fr = mar ; F r = 4(-17) = -68 N
au = ru
$ + 2r
# u
# = 0.5(2) + 2(3)(6) = 37
ar = r
$ - ru
# 2 = 1 - 0.5(6)2 = -17
r
$ = 1 m>s 2 u
$ = 2 rad>s
r
# = 3 m>s u
# = 6 rad>s
r = 0.5 m
θ
••
= 2 rad/s2
θ
•
= 6 rad/s
vs = 3 m/s
as = 1 m/s2
r = 0.5 m
O

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R1–10.
Packages having a mass of 6 kg slide down a smooth chute
and land horizontally with a speed of 3 m/ s on the surface
of a conveyor belt. If the coefficient of kinetic friction
between the belt and a package is determine the
time needed to bring the package to rest on the belt if the
belt is moving in the same direction as the package with a
speed v = 1 m/s.
mk = 0.2,
SOLUTION
Ans.t = 1.02 s
)3(6 - 0.2(58.86)(t) = 6(1)
( :+ ) m(v1 )x + © LFx dt = m(v2 )x
Np = 58.86 N
0 + Np(t) - 58.86(t) = 0
(+c) m(v1 )y + © LFy dt = m(v2 )y
1 m/s
3 m/s

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R1–11.
SOLUTION
The crate starts moving when
From the graph since
The time needed for the crate to start moving is
Hence, the impulse due to F is equal to the area under the curve from
Ans.v2 = 31.7 m>s
384.436 = 20v2
(04 1
2 t2 ) 2 5
2.943
+ 200(10 - 5) - 692.292 = 20v2
0 + L
5
2.943
200
5 t dt + L
10
5
200 dt - (0.5)196.2(10 - 2.943) = 20v2
:+ m(vx)1 + © LF dt = m(vx)2
2.943 s … t … 10 s
t = 5
200 (117.72) = 2.943 s
F = 200
5 t, 0 … t … 5 s
F = Fr = 0.6(196.2) = 117.72 N
A 20-kg block is originally at rest on a horizontal surface
for which the coefficient of static friction is and
the coefficient of kinetic friction is If a horizontal
force F is applied such that it varies with time as shown,
determine the speed of the block in 10 s. Hint: First
determine the time needed to overcome friction and start
the block moving.
mk = 0.5.
ms = 0.6 F
F (N)
t (s)
200
5 10
x

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R1–14.
The rocket sled has a mass of 4 Mg and travels from rest
along the smooth horizontal track such that it maintains a
constant power output of . Neglect the loss of fuel
mass and air resistance, and determine how far it must
travel to reach a speed of .v = 60 m>s
450 kW
SOLUTION
Ans.s = 4 (103 )(60)3
3(450)(103
) = 640 m
s = m v3
3 P
Ps = m v3
3
P L
s
0
ds = m L
v
0
v2 dv
LP ds = m Lv2 dv
P = Fv = ma v2 dv
ds b
F = m a = ma v dv
ds b©F x = m ax;+:
v
T

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■R1–15.
SOLUTION
at
Using the Runge–Kutta method:
t = 0s0 = 0
n = 3 (8 e- 1 + t)1>2
A projectile, initially at the origin, moves vertically
downward along a straight-line path through a fluid medium
such that its velocity is defined as
where t is in seconds. Plot the position s of the projectile
during the first 2 s. Use the Runge-Kutta method to evaluate
s with incremental values of h = 0.25 s.
v = 318e-t + t21>2 m>s,
ts
0 0
2.01 0.25
3.83 0.50
5.49 0.75
7.03 1.00
8.48 1.25
9.87 1.50
11.2 1.75
12.5 2.00

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*R1–16.
The chain has a mass of 3 kg/m. If the coefficient of kinetic
friction between the chain and the plane is
determine the velocity at which the end A will pass point
B when the chain is released from rest.
mk = 0.2,
SOLUTION
Ans.
Also,
Ans.v = 4.38 m>s
0 + 2(3)(9.81)(2 sin 40°) - 0.2(45.09)(2) = 1
2 (2)(3)(v2
)
T1 + ©U1-2 = T2
v2 = 4.38 m>s
v2
2 = 0 + 2(4.80)(2)
+ b v2
2 = v2
1 + 2as
a = 4.80 m>s2
+ b©Fx = max ; 2(3)(9.81) sin 40° - 0.2(45.09) = 2(3) a
NC = 45.09 N
+ a©Fy = may ; - 2(3)(9.81) cos 40° + NC = 0
2 m
A
B
40°

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R1–17.
The motor M pulls in its attached rope with an acceleration
Determine the towing force exerted by M on
the rope in order to move the 50-kg crate up the inclined
plane. The coefficient of kinetic friction between the crate
and the plane is Neglect the mass of the pulleys
and rope.
mk = 0.3.
a p = 6 m/s2
.
SOLUTION
(1)
Kinematics,
Taking two time derivatives, yields
Thus,
Substituting into Eq. (1) and solving,
Ans.T = 158 N
aC = 6
3 = 2
3aC = ap
2sC + (sC - sp) = l
Q+ ©Fx = max ; 3T - 0.3(424.79) - 50(9.81) sin 30° = 50aC
N C = 424.79
a+ ©Fy = may ; N C - 50(9.81) cos 30° = 0
aP = 6 m/s
2
P
M
θ = 30°

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R1–18.
The drinking fountain is designed such that the nozzle is
located from the edge of the basin as shown. Determine the
maximum and minimum speed at which water can be
ejected from the nozzle so that it does not splash over the
sides of the basin at B and C.
SOLUTION
(1)
(2)
Substituting Eq. (1) into (2) yields:
At point B, .
Ans.
At point C, .
Ans.vA = A
4.905 sin 40°(0.35)2
sin2 40°(0.35 cos 40° + 0.05 sin 40°) = 1.76 m>s
R = 0.35 m
vA = A
4.905 sin 40°(0.1)2
sin2 40°(0.1 cos 40° + 0.05 sin 40°) = 0.838 m>s
R = 0.1 m
vA = A
4.905 sin 40°R2
sin2 40°(R cos 40° + 0.05 sin 40°)
-0.05 = vA cos 40°¢ R
vA sin 40° ≤ + 1
2(-9.81)¢ R
vA sin 40° ≤
2
-0.05 = 0 + vA cos 40°t + 1
2 (-9.81)t2
(+c) sy = Asy B0 + vy t + 1
2 ac t2
R = vA sin 40°t t = R
vA sin 40°
a :+ b sx = vx t
vA
50 mm
A
B C
40°
250 mm
100 mm

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R1–19.
The 100-kg crate is subjected to the action of two forces,
and as shown. If it is originally at
rest, determine the distance it slides in order to attain a
speed of 6 m/s. The coefficient of kinetic friction between
the crate and the surface is mk = 0.2.
F2 = 1.5 kN,F1 = 800 N
SOLUTION
Ans.s = 0.933 m
s(1928.7) = 1800
0 + 800 cos 30°(s) - 0.2(867.97)(s) + 1500 cos 20° (s) = 1
2 (100)(6)2
T1 + a U1 - 2 = T2
N C = 867.97 N
+ c a Fy = 0; N C - 800 sin 30° - 100(9.81) + 1500 sin 20° = 0
F2
F1
30° 20°

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R1–21.
SOLUTION
Ans.h = 0.390 m
h = 0 + 8.3328(0.048112) - 1
2 (9.81)(0.048112)2
(+ c) s = s0 + v0 t + 1
2 a c t2
t = 0.048112 s
0.75 = 0 + 15.5885t
( :+ ) s = s0 + v0 t
(vy)2 = 8.3328 m>s
(+ c) e = 0.8 = (vy)2
10.4160
(vy)1 = 10.4160 m>s
(vy)1 = 18 sin 30° + 9.81(0.14434)
(+ T) v = v0 + a c t
(vx)1 = (vx)2 = 18 cos 30° = 15.5885 m>s
t = 0.14434 s
2.25 = 0 + 18 cos 30°t
( :+ ) s = s0 + v0 t
The ping-pong ball has a mass of 2 g. If it is struck with the
velocity shown, determine how high h it rises above the end
of the smooth table after the rebound. Take e = 0.8.
h
2.25 m
18 m/s
30
0.75 m

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R1–22.
SOLUTION
Time to accelerate to 60 m/s,
Time to decelerate to a stop,
Time to travel at 60 m/s,
Total time Ans.t = 10 + 7.5 + 6.25 = 23.8 s
t = 6.25 s
573 = 0 + 60t
( :
+ ) s = s0 + v0 t
900 - 300 - 225 = 375 m
s = 225 m
s = 0 + 60(7.5) - 1
2 (8)(7.52 )
( :
+ ) s = s0 + v0 t + 1
2 ac t2
t = 7.5 s
0 = 60 - 8t
( :
+ ) v = v0 + ac t
s = 300 m
s = 0 + 0 + 1
2 (6)(102 )
( :
+ ) s = s0 + v0 t + 1
2 ac t2
t = 10 s
06 = 0 + 6t
( :
+ ) v = v0 + ac t
A sports car can accelerate at and decelerate at
If the maximum speed it can attain is 60 m/s,
determine the shortest time it takes to travel 900 m starting
from rest and then stopping when s = 900 m.
8 m/s 2
.
6 m/s 2

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R1–23.
A 2-kg particle rests on a smooth horizontal plane and is acted
upon by forces and If
and when determine the equation
which describes the path.y = f1x2
t = 0,vy = 2 m/svx = 6 m/s,
y = 0,x = 0,Fy = 3 N.Fx = 0
SOLUTION
(1)
(2)
(3)
(4)
Eliminating t from Eq. (3) and (4) yields:
Ans.y = 0.0208x2 + 0.333x (Parabola)
x = 6t
L
x
0
dx = L
t
0
6dt
v x = dx
dt = 6
L
v x
6
dvx = L
t
0
0dt
ax = dvx
dt = 0
y = 0.75t2 + 2t
L
y
0
dy = L
t
0
(1.5t + 2) dt
vy = dy
dt = 1.5t + 2
L
v y
2
dvy = 1.5 L
t
0
dt
ay = dvy
dt = 1.5
:+ ©Fx = max; 0 = 2ax ax = 0
+ c ©Fy = may; 3 = 2ay ay = 1.5 m>s2

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*R1–24.
SOLUTION
Ans.Ns = 240 lb
+ c ©Fn = man ; N s - 120 = 120
32.2 c(31.08)2
30 d
r = 3 [1(dy
dx)2
]3>2
d2y
dx2
3 = (1 + 0)3>2
1
30
= 30 ft
d2y
dx2 = 1
30
dy
dx 2
x = 0
= 1
30x 2
x = 0
= 0
y = 1
60 x2 - 15
vB = 31.08 ft>s
0 + (120)(15) = 1
2 a120
32.2b v2
B + 0
TA + VA = TB + VB
A skier starts from rest at A (30 ft, 0) and descends the
smooth slope, which may be approximated by a parabola. If
she has a weight of 120 lb, determine the normal force she
exerts on the ground at the instant she arrives at point B. 30 ft
15 ft
y
B
A x
y = x2 – 151__
60

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R1–25.
SOLUTION
Block A:
(1)
Block B:
(2)
(3)
Block C:
(4)
Solving Eqs. (1)–(4) for a,
Ans.v = 2.68 ft>s
v2 = 0 + 2(1.79)(2 - 0)
(+ T)v2 = v2
0 + 2ac(s - s0)
a = 1.79 ft>s2
+ c ©Fy = may ; T2 - 6 = 6
32.2 a
+ c ©F y = may ; N B - 20 = 0
;+ ©Fx = max ; -T2 + T1 - 0.1NB = 20
32.2 a
+ T ©Fy = may ; -T1 + 10 = 10
32.2 a
The 20-lb block B rests on the surface of a table for which
the coefficient of kinetic friction is Determine the
speed of the 10-lb block A after it has moved downward 2 ft
from rest. Neglect the mass of the pulleys and cords.
mk = 0.1.
A
10 lb
C 6 lb
B
20 lb

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R1–26.
SOLUTION
Block A:
Block B:
Kinematics:
Solving,
Ans.vA = 26.8 ft>s T
vA = 6 + 10.40(2)
(+ T) vA = (vA)0 + aAt
aB = -20.81 ft>s2
aA = 10.40 ft>s2
T = 3.38 lb
2aA = -aB
2sA + sB = l
;+ ©F x = max ; -T + 0.2(4) = 4
32.2 aB
+ T ©Fy = may ; 10 - 2T = 10
32.2 aA
At a given instant the 10-lb block A is moving downward
with a speed of 6 ft/s. Determine its speed 2 s later. Block
B has a weight of 4 lb, and the coefficient of kinetic friction
between it and the horizontal plane is Neglect the
mass of the pulleys and cord.
mk = 0.2.
A
B

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R1–27.
Two smooth billiard balls A and B have an equal mass of
If A strikes B with a velocity of
as shown, determine their final velocities just after collision.
Ball B is originally at rest and the coefficient of restitution
is e = 0.75.
1vA21 = 2 m/sm = 200 g.
SOLUTION
(1)
(2)
Solving Eqs. (1) and (2)
For A:
For B:
Hence Ans.
Ans.
Ans.(uA)2 = tan- 1 a0.1915
1.285 b = 8.47° e
(vA)2 = 2(-0.1915)2 + (1.285)2 = 1.30 m>s
(vB)2 = (vB)x2 = 1.34 m>s ;
(vB)y2 = 0
(+ c) m B (vB)y1 = mB (vB)y2
(vA)y2 = 1.285 m>s
(+ T) m A (vA)y1 = mA (vA)y2
(vB)x2 = -1.3405 m>s
(vA)x2 = -0.1915 m>s
57.0 = (vA)x2 - (vB)x2
1.532
( :+ ) e = (vrel)2
(vrel)1
-2(1.532) + 0 = 0.2(vA)x2 + 0.2(vB)x2
( :+ ) mA(vA)x1 + mB (vB)x1 = mA(vA)x2 + mB (vB)x2
(vA)y1 = -2 sin 40° = -1.285 m>s
(vA)x1 = -2 cos 40° = -1.532 m>s
y
x
40°
B
A
(vA) 1

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*R1–28.
SOLUTION
Ans.
Ans.U1-2 = 14.4 ft # kip
U1 - 2 = (744.4 cos 15°)(20) = 14 380.7 ft # lb
NC = 1307.3 lb
T = 744.4 lb = 744 lb
:+ ©Fx = 0; T cos 15° - 0.55N C = 0
+ c ©Fy = 0; N C - 1500 + T sin 15° = 0
A crate has a weight of 1500 lb. If it is pulled along the
ground at a constant speed for a distance of
20 ft, and the towing cable makes an angle of 15° with the
horizontal, determine the tension in the cable and the work
done by the towing force. The coefficient of kinetic friction
between the crate and the ground is mk = 0.55.

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Loading page 30...

R1–30.
Determine the tension developed in the two cords and the
acceleration of each block. Neglect the mass of the pulleys
and cords. Hint: Since the system consists of two cords, relate
the motion of block A to C, and of block B to C. Then, by
elimination, relate the motion of A to B.
SOLUTION
Block A:
(1)
Block B:
(2)
Pulley C:
(3)
Kinematics:
Taking the two time derivatives:
Also,
So that
Since
(4)
Solving Eqs.(1)–(4),
Ans.
Ans.
Ans.
Ans.TB = 45.3 N
TA = 90.6 N
aB = 1.51 m>s2
aA = 0.755 m>s2
a B = 2 aA
a C¿ = -aC
2aC¿ = aB
sC¿ + (sC¿ - sB) = l¿
aA = -aC
sA + sC = l
+ c ©Fy = 0; TA - 2TB = 0
+ c ©Fy = may ; TB - 4(9.81) = 4aB
+ T ©Fy = may ; 10(9.81) - TA = 10aA
C
B 4 kg
A10 kg
3 m

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Solution Manual for Engineering Mechanics Dynamics, 13th Edition

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