SOLUTION

13–5.
The water-park ride consists of an 800-lb sled which slides
from rest down the incline and then into the pool. If the
frictional resistance on the incline is Fr = 30 lb, and in
the pool for a short distance Fr = 80 lb, determine how
fast the sled is traveling when s = 5 ft.
100 ft
SOLUTION
+ b a Fx = max;
800 sin 45° - 30 =
800
a
32.2
s
a = 21.561 ft>s
2
v21 = v20 + 2ac(s - s0)
v21 = 0 + 2(21.561)(100 22 - 0))
v1 = = 78.093 ft>s
+
;
a Fx = ma x;
- 80 =
800
a
32.2
a = -3.22 ft>s2
v22 = v21 + 2ac(s2 - s1)
v22 = (78.093)2 + 2(- 3.22)(5 - 0)
v2 = 77.9 ft>s
Ans.
100 ft
13–11.
The safe S has a weight of 200 lb and is supported by the
rope and pulley arrangement shown. If the end of the rope
is given to a boy B of weight 90 lb, determine his
acceleration if in the confusion he doesn’t let go of the rope.
Neglect the mass of the pulleys and rope.
S
SOLUTION
Equation of Motion: The tension T developed in the cord is the same throughout
the entire cord since the cord passes over the smooth pulleys.
From FBD(a),
+ c ©Fy = may;
T- 90 = - a
90
ba
32.2 B
(1)
From FBD(b),
+ c ©Fy = may ;
2T - 200 = - a
200
ba
32.2 S
(2)
Kinematic: Establish the position-coordinate equation, we have
2sS +sB = l
Taking time derivative twice yields
1+ T2
2aS + aB = 0
(3)
Solving Eqs.(1),(2), and (3) yields
aB = -2.30 ft>s2 = 2.30 ft>s2
aS = 1.15 ft>s2 T
T = 96.43 lb
c
Ans.
B
13–13.
The bullet of mass m is given a velocity due to gas pressure
caused by the burning of powder within the chamber of the
gun. Assuming this pressure creates a force of
F = F0 sin 1pt>t02 on the bullet, determine the velocity of
the bullet at any instant it is in the barrel. What is the
bullet’s maximum velocity? Also, determine the position of
the bullet in the barrel as a function of time.
F
F0
t0
SOLUTION
+ ©F = ma ;
:
x
x
a =
L0
F0 sin a
F0
pt
dv
= a b sin a b
m
dt
t0
v
dv =
v = a
t
L0
a
F0
pt
b sin a b dt
m
t0
v = -a
F0t0
pt t
b cos a b d
pm
t0 0
F0t0
pt
b a1 - cos a b b
pm
t0
vmax occurs when cos a
vmax =
t
ds =
L0
Ans.
pt
b = -1, or t = t0.
t0
2F0t0
pm
s
L0
pt
b = ma
t0
Ans.
a
F0t0
pt
b a 1 - cos a b b dt
pm
t0
s = a
t0
F0t0
pt t
b ct sin a b d
pm
p
t0 0
s = a
t0
F0t0
pt
b at sin a b b
p
pm
t0
Ans.
t
13–17.
The double inclined plane supports two blocks A and B,
each having a weight of 10 lb. If the coefficient of kinetic
friction between the blocks and the plane is mk = 0.1,
determine the acceleration of each block.
B
A
60
SOLUTION
Equation of Motion: Since blocks A and B are sliding along the plane, the friction
forces developed between the blocks and the plane are (Ff)A = mk NA = 0.1 NA
and (Ff)B = mk NB = 0.1NB. Here, aA = aB = a. Applying Eq. 13–7 to FBD(a),
we have
a + a Fy¿ = may¿;
Q + a Fx¿ = max¿;
NA - 10 cos 60° = a
10
b(0) NA = 5.00 lb
32.2
T + 0.1(5.00) - 10 sin 60° = - a
10
ba
32.2
(1)
From FBD(b),
10
b(0) NB = 8.660 lb
32.2
Q + a Fy¿ = may¿;
NB - 10 cos 30° = a
a + a Fx¿ = max¿;
T - 0.1(8.660) - 10 sin 30° = a
10
ba
32.2
(2)
Solving Eqs. (1) and (2) yields
a = 3.69 ft>s2
T = 7.013 lb
Ans.
30
13–22.
Determine the required mass of block A so that when it is
released from rest it moves the 5-kg block B a distance of
0.75 m up along the smooth inclined plane in t = 2 s.
Neglect the mass of the pulleys and cords.
E
C
D
SOLUTION
Kinematic: Applying equation s = s0 + v0 t +
(a+)
1
0.75 = 0 + 0 + aB A 22 B
2
B
1
a t2, we have
2 c
A
60
aB = 0.375 m>s2
Establishing the position - coordinate equation, we have
2sA + (sA - sB) = l
3sA - sB = l
Taking time derivative twice yields
3aA - aB = 0
(1)
From Eq.(1),
3aA - 0.375 = 0
aA = 0.125 m>s2
Equation of Motion: The tension T developed in the cord is the same throughout
the entire cord since the cord passes over the smooth pulleys. From FBD(b),
a+ ©Fy¿ = may¿ ;
T - 5(9.81) sin 60° = 5(0.375)
T = 44.35 N
From FBD(a),
+ c ©Fy = may ;
3(44.35) - 9.81mA = mA ( - 0.125)
mA = 13.7 kg
Ans.
13–25.
If the motor draws in the cable at a rate of v = (0.05t2) m>s,
where t is in seconds, determine the tension developed in the
cable when t = 5 s. The crate has a mass of 20 kg and the
coefficient of kinetic friction between the crate and the ground
is mk = 0.2.
s
v
M
SOLUTION
Kinematics: Since the motion of the crate is known, its acceleration a will be
determined first.
a =
dv
= 0.05(2t) = (0.1t) m>s2
dt
When t = 5 s,
a = 0.1(5) = 0.5 m>s2 :
Free-Body Diagram: The kinetic friction Ff = mkN = 0.2N must act to the left to
oppose the motion of the crate which is to the right, Fig. a.
Equations of Motion: Here, ay = 0. Thus,
+ c ©Fy = may;
N - 20(9.81) = 0
N = 196.2 N
Using the results of N and a,
+ ©F = ma ;
:
x
x
T - 0.2(196.2) = 20(0.5)
T = 49.2 N
Ans.
13–30.
F (N)
The force of the motor M on the cable is shown in the graph.
Determine the velocity of the 400-kg crate A when t = 2 s.
2500
F ⫽ 625 t 2
M
2
SOLUTION
Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a.
Equilibrium: For the crate to move, force 2F must overcome its weight. Thus, the
time required to move the crate is given by
+ c ©Fy = 0;
2(625t2) - 400(9.81) = 0
A
t = 1.772 s
Equations of Motion: F = A 625t2 B N. By referring to Fig. a,
+ c ©Fy = may;
2 A 625t2 B - 400(9.81) = 400a
a = (3.125t2 - 9.81) m>s2
Kinematics: The velocity of the crate can be obtained by integrating the kinematic
equation, dv = adt. For 1.772 s … t 6 2 s, v = 0 at t = 1.772 s will be used as the
lower integration limit. Thus,
(+ c )
L
dv =
v
L0
L
adt
t
dv =
L1.772 s
A 3.125t2 - 9.81 B dt
v = A 1.0417t3 - 9.81t B 2
t
1.772 s
= A 1.0417t - 9.81t + 11.587 B m>s
3
When t = 2 s,
v = 1.0417(23) - 9.81(2) + 11.587 = 0.301 m>s
Ans.
t (s)
13–33.
Each of the three plates has a mass of 10 kg. If the
coefficients of static and kinetic friction at each surface of
contact are ms = 0.3 and mk = 0.2, respectively, determine
the acceleration of each plate when the three horizontal
forces are applied.
18 N
C
15 N
Plates B, C and D
100 - 15 - 18 - F = 0
F = 67 N
Fmax = 0.3(294.3) = 88.3 N 7 67 N
Plate B will not slip.
aB = 0
Ans.
Plates D and C
+ ©F = 0;
:
x
100 - 18 - F = 0
F = 82 N
Fmax = 0.3(196.2) = 58.86 N 6 82N
Slipping between B and C.
Assume no slipping between D and C,
+ ©F = ma ;
:
x
x
100 - 39.24 - 18 = 20 ax
ax = 2.138 m>s2 :
Check slipping between D and C.
+ ©F = m a ;
:
x
x
F - 18 = 10(2.138)
F = 39.38 N
Fmax = 0.3(98.1) = 29.43 N 6 39.38 N
Slipping between D and C.
Plate C:
+ ©F = m a ;
:
x
x
100 - 39.24 - 19.62 = 10 ac
ac = 4.11 m>s2 :
Ans.
Plate D:
+ ©F = m a ;
:
x
x
19.62 - 18 = 10 a D
aD = 0.162m>s2 :
B
A
SOLUTION
+ ©F = 0;
:
x
D
Ans.
100 N
13–35.
The conveyor belt is moving at 4 m>s. If the coefficient 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.
B
SOLUTION
+ ©F = m a ;
:
x
x
0.2198.12 = 10 a
a = 1.962 m>s2
+ 2v = v + a t
1:
0
c
4 = 0 + 1.962 t
t = 2.04 s
Ans.
13–37.
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 cylinder’s vertical position y so that
when F is applied the cylinder rises with a constant
acceleration aB. Neglect the mass of the cord, pulleys, hook
and chain.
d/2
d/2
F
y
SOLUTION
+ c ©Fy = may ;
2F cos u - mg = maB
2F £
F =
y
2y +
2
A d2 B 2
where cos u =
y
2y +
2
aB
A B
d 2
2
≥ - mg = maB
m(aB + g)24y2 + d2
4y
Ans.
B
13–41.
The 2-kg collar C is free to slide along the smooth shaft AB.
Determine the acceleration of collar C if (a) the shaft is
fixed from moving, (b) collar A, which is fixed to shaft AB,
moves downward at constant velocity along the vertical rod,
and (c) collar A is subjected to a downward acceleration of
2 m>s2. In all cases, the collar moves in the plane.
B
45
A
SOLUTION
(a) + b©Fx¿ = max¿ ;
219.812 sin 45° = 2aC
aC = 6.94 m>s2
(b) From part (a) a C>A = 6.94 m>s2
a C = a A + a C>A
Where a A = 0
= 6.94 m>s2
(c)
aC = aA + aC>A
= 2 + aC>A
b
√T
+ b©Fx¿ = max¿ ;
(1)
219.812 sin 45° = 212 cos 45°+ aC>A2
aC>A = 5.5225 m>s2 b
From Eq.(1)
aC = 2 + 5.52 25 = 3.905 + 5.905
;
b
T
T
aC = 23.9052 + 5.9052 = 7.08 m>s2
u = tan-1
5.905
= 56.5° ud
3.905
Ans.
Ans.
C
13–46.
Blocks A and B each have a mass m. Determine the largest
horizontal force P which can be applied to B so that A will
not move relative to B. All surfaces are smooth.
A
P
u B
C
SOLUTION
Require
aA = aB = a
Block A:
+ c ©Fy = 0;
+ ©F = ma ;
;
x
x
N cos u - mg = 0
N sin u = ma
a = g tan u
Block B:
+ ©F = ma ;
;
x
x
P - N sin u = ma
P - mg tan u = mg tan u
P = 2mg tan u
Ans.
13–47.
Blocks A and B each have a mass m. Determine the largest
horizontal force P which can be applied to B so that A will
not slip on B. The coefficient of static friction between A and
B is ms. Neglect any friction between B and C.
A
C
SOLUTION
Require
aA = aB = a
Block A:
+ c ©Fy = 0;
+ ©F = ma ;
;
x
x
N cos u - ms N sin u - mg = 0
N sin u + ms N cos u = ma
N =
mg
cos u - ms sin u
a = ga
sin u + ms cos u
b
cos u - ms sin u
Block B:
+ ©F = ma ;
;
x
x
P - ms N cos u - N sin u = ma
P - mga
sin u + ms cos u
sin u + ms cos u
b = mga
b
cos u - ms sin u
cos u - ms sin u
P = 2mga
sin u + ms cos u
b
cos u - ms sin u
P
u B
Ans.
13–54.
The 2-kg block B and 15-kg cylinder A are connected to a
light cord that passes through a hole in the center of the
smooth table. If the block travels along a circular path of
radius r = 1.5 m, determine the speed of the block.
r
B
v
SOLUTION
Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The
tension in the cord is equal to the weight of cylinder A, i.e.,
T = 15(9.81) N = 147.15 N. Here, an must be directed towards the center of the
circular path (positive n axis).
Equations of Motion: Realizing that an =
©Fn = man;
147.15 = 2 a
v2
v2
and referring to Fig. (a),
=
r
1.5
2
v
b
1.5
v = 10.5 m>s
Ans.
A
13–57.
The block B, having a mass of 0.2 kg, is attached to the
vertex A of the right circular cone using a light cord. If the
block has a speed of 0.5 m>s around the cone, determine
the tension in the cord and the reaction which the cone
exerts on the block and the effect of friction.
z
200 mm
B
400 mm
SOLUTION
r
300
=
;
200
500
+Q©Fy = may;
(0.5)2
4
3
T - 0.2(9.81)a b = B 0.2 ¢
≤Ra b
5
0.120
5
T = 1.82 N
+a©Fx = max;
300 mm
r = 120 mm = 0.120 m
Ans.
(0.5)2
3
4
NB - 0.2(9.81)a b = - B 0.2 ¢
≤Ra b
5
0.120
5
NB = 0.844 N
Ans.
Also,
+ ©F = ma ;
:
n
n
+ c ©Fb = 0;
(0.5)2
3
4
b
T a b - NB a b = 0.2a
5
5
0.120
4
3
Ta b + NB a b - 0.2(9.81) = 0
5
5
T = 1.82 N
Ans.
NB = 0.844 N
Ans.
A
13–62.
The 10-lb suitcase slides down the curved ramp for which the
coefficient of kinetic friction is mk = 0.2. If at the instant it
reaches point A it has a speed of 5 ft>s, determine the normal
force on the suitcase and the rate of increase of its speed.
y
1 x2
y = ––
8
6 ft
A
SOLUTION
n =
x
1 2
x
8
dy
1
= - 1.5
= tan u = x 2
dx
4 x= -6
d2y
2
dx
=
u = - 56.31°
1
4
B1 + a
r =
2
3
dy 2 2
b R
dx
d2y
dx2
2
+Q©Fn = man ;
C 1 + (-1.5)2 D 2
3
=
212
4
= 23.436 ft
N - 10 cos 56.31° = a
(5)2
10
b¢
≤
32.2 23.436
N = 5.8783 = 5.88 lb
+R©Ft = mat;
-0.2(5.8783) + 10 sin 56.31° =
at = 23.0 ft s2
Ans.
10
a
32.2 t
Ans.
13–67.
The vehicle is designed to combine the feel of a motorcycle
with the comfort and safety of an automobile. If the vehicle
is traveling at a constant speed of 80 km> h along a circular
curved road of radius 100 m, determine the tilt angle u of
the vehicle so that only a normal force from the seat acts on
the driver. Neglect the size of the driver.
u
SOLUTION
Free-Body Diagram: The free-body diagram of the passenger is shown in Fig. (a).
Here, an must be directed towards the center of the circular path (positive n axis).
1h
km 1000 m
ba
ba
b
h
1 km
3600 s
= 22.22 m>s. Thus, the normal component of the passenger’s acceleration is given by
22.222
v2
= 4.938 m>s2. By referring to Fig. (a),
=
an =
r
100
Equations of Motion: The speed of the passenger is v = a80
+ c ©Fb = 0;
+ ©F = ma ;
;
n
n
N cos u - m(9.81) = 0
N =
9.81m
cos u
9.81m
sin u = m(4.938)
cos u
u = 26.7°
Ans.
13–71.
If the ball has a mass of 30 kg and a speed v = 4 m>s at the
instant it is at its lowest point, u = 0°, determine the tension
in the cord at this instant. Also, determine the angle u to
which the ball swings and momentarily stops. Neglect the
size of the ball.
u
4m
SOLUTION
+ c ©Fn = man;
T - 30(9.81) = 30a
(4)2
b
4
T = 414 N
+Q©Ft = mat;
Ans.
- 30(9.81) sin u = 30at
at = -9.81 sin u
at ds = v dv Since ds = 4 du, then
u
- 9.81
L0
0
sin u(4 du) =
L4
v dv
C 9.81(4)cos u D u0 = - (4)2
1
2
39.24(cos u - 1) = - 8
u = 37.2°
Ans.
13–73.
Determine the maximum speed at which the car with mass
m can pass over the top point A of the vertical curved road
and still maintain contact with the road. If the car maintains
this speed, what is the normal reaction the road exerts on
the car when it passes the lowest point B on the road?
r
A
r
B
r
SOLUTION
Free-Body Diagram: The free-body diagram of the car at the top and bottom of the
vertical curved road are shown in Figs. (a) and (b), respectively. Here, an must be
directed towards the center of curvature of the vertical curved road (positive n axis).
Equations of Motion: When the car is on top of the vertical curved road, it is
required that its tires are about to lose contact with the road surface. Thus, N = 0.
v2
v2
=
Realizing that an =
and referring to Fig. (a),
r
r
+ T ©Fn = man;
mg = m ¢
v2
≤
r
v = 2gr
Ans.
Using the result of v, the normal component of car acceleration is
gr
v2
an =
=
= g when it is at the lowest point on the road. By referring to Fig. (b),
r
r
+ c ©Fn = man;
N - mg = mg
N = 2mg
Ans.
r
13–74.
If the crest of the hill has a radius of curvature r = 200 ft,
determine the maximum constant speed at which the car
can travel over it without leaving the surface of the road.
Neglect the size of the car in the calculation. The car has a
weight of 3500 lb.
v
r
SOLUTION
T ©Fn = man;
3500 =
3500 v2
a
b
32.2 200
v = 80.2 ft>s
Ans.
200 ft
13–78.
A spring, having an unstretched length of 2 ft, has one end
attached to the 10-lb ball. Determine the angle u of the
spring if the ball has a speed of 6 ft> s tangent to the
horizontal circular path.
6 in.
A
u
SOLUTION
Free-Body Diagram: The free-body diagram of the bob is shown in Fig. (a). If we
denote the stretched length of the spring as l, then using the springforce formula,
Fsp = ks = 20(l - 2) lb. Here, an must be directed towards the center of the
horizontal circular path (positive n axis).
Equations of Motion: The radius of the horizontal circular path is r = 0.5 + l sin u.
62
v2
Since an =
, by referring to Fig. (a),
=
r
0.5 + l sin u
+ c ©Fb = 0;
+ ©F = ma ;
;
n
n
20(l - 2) cos u - 10 = 0
20(l - 2) sin u =
62
10
a
b
32.2 0.5 + l sin u
(1)
(2)
Solving Eqs. (1) and (2) yields
u = 31.26° = 31.3°
Ans.
l = 2.585 ft
Ans.
k
20 lb>ft
13–79.
u
The airplane, traveling at a constant speed of 50 m>s, is
executing a horizontal turn. If the plane is banked at
u = 15°, when the pilot experiences only a normal force on
the seat of the plane, determine the radius of curvature r of
the turn. Also, what is the normal force of the seat on the
pilot if he has a mass of 70 kg.
r
SOLUTION
+ c a Fb = mab;
NP sin 15° - 7019.812 = 0
NP = 2.65 kN
+
;
a Fn = man;
NP cos 15° = 70 a
r = 68.3 m
Ans.
502
b
r
Ans.
13–83.
The ball has a mass m and is attached to the cord of length l.
The cord is tied at the top to a swivel and the ball is given a
velocity v0. Show that the angle u which the cord makes with
the vertical as the ball travels around the circular path
must satisfy the equation tan u sin u = v20>gl. Neglect air
resistance and the size of the ball.
O
u
l
SOLUTION
+ ©F = ma ;
:
n
n
+ c ©Fb = 0;
Since r = l sin u
T sin u = m a
v20
b
r
T cos u - mg = 0
T =
a
v0
mv20
l sin2 u
mv20
cos u
b a 2 b = mg
l
sin u
tan u sin u =
v20
gl
Q.E.D.
13–87.
The path of motion of a 5-lb particle in the horizontal plane
is described in terms of polar coordinates as r = (2t + 1) ft
and u = (0.5t2 - t) rad, where t is in seconds. Determine
the magnitude of the resultant force acting on the particle
when t = 2 s.
SOLUTION
#
$
r = 2 ft>s
r = 0
#
= 0 rad
u = t - 1|t = 2 s = 1 rad>s
r = 2t + 1|t = 2 s = 5 ft
u = 0.5t2 - t|t = 2 s
#
$
ar = r - ru2 = 0 - 5(1)2 = -5 ft>s2
$
##
au = ru + 2ru = 5(1) + 2(2)(1) = 9 ft>s2
©Fr = mar;
Fr =
5
( -5) = -0.7764 lb
32.2
©Fu = mau;
Fu =
5
(9) = 1.398 lb
32.2
F = 2F2r + F2u = 2(-0.7764)2 + (1.398)2 = 1.60 lb
$
u = 1 rad>s2
Ans.
13–89.
Rod OA rotates
# counterclockwise with a constant angular
velocity of u = 5 rad>s. The double collar B is pinconnected together such that one collar slides over the
rotating rod and the other slides over the horizontal curved
rod, of which the shape is described by the equation
r = 1.512 - cos u2 ft. If both collars weigh 0.75 lb,
determine the normal force which the curved rod exerts on
one collar at the instant u = 120°. Neglect friction.
A
B
r
SOLUTION
·
#
$
Kinematic: Here, u = 5 rad>s and u = 0. Taking the required time derivatives at
u = 120°, we have
r = 1.5 (2 – cos ) ft.
r = 1.5(2 - cos u)|u = 120° = 3.75 ft
#
#
r = 1.5 sin uu|u = 120° = 6.495 ft>s
$
#
$
r = 1.5(sin uu + cos uu 2)|u = 120° = - 18.75 ft>s2
Applying Eqs. 12–29, we have
#
$
ar = r - ru2 = - 18.75 - 3.75(52) = - 112.5 ft>s2
$
# #
au = r u + 2r u = 3.75(0) + 2(6.495)(5) = 64.952 ft>s2
Equation of Motion: The angle c must be obtained first.
tan c =
1.5(2 - cos u)
r
2
=
= 2.8867
dr>du
1.5 sin u
u = 120°
c = 70.89°
Applying Eq. 13–9, we have
a Fr = mar ;
-N cos 19.11° =
0.75
(-112.5)
32.2
N = 2.773 lb = 2.77 lb
a Fu = mau ;
FOA + 2.773 sin 19.11° =
FOA = 0.605 lb
Ans.
0.75
(64.952)
32.2
O
= 5 rad/s
13–94.
The collar has a mass of 2 kg and travels along the smooth
horizontal rod defined by the equiangular spiral
r = 1eu2 m, where u is in radians. Determine the tangential
force F and the normal force N acting on the collar when
u = 90°, if the force F maintains a constant angular motion
#
u = 2 rad>s.
F
SOLUTION
r
r = eu
u
#
#
r = euu
#
$
$
r = eu(u)2 + euu
At u = 90°
#
u = 2 rad>s
$
u = 0
r = 4.8105
#
r = 9.6210
$
r = 19.242
#
$
ar = r - r(u)2 = 19.242 - 4.8105(2)2 = 0
$
##
au = ru + 2ru = 0 + 2(9.6210)(2) = 38.4838 m>s2
tan c =
r
A B
dr
du
= eu>eu = 1
c = 45°
+ c a Fr = mar;
-NC cos 45° + F cos 45° = 2(0)
+
;
a Fu = mau;
F sin 45° + NC sin 45° = 2(38.4838)
NC = 54.4 N
Ans.
F = 54.4 N
Ans.
r eu
13–99.
For a short time, the 250-kg roller coaster car is traveling
along the spiral track such that its position measured from
the top of the track has components r = 8 m,
u = 10.1t + 0.52 rad, and z = 1-0.2t2 m, where t is in
seconds. Determine the magnitudes of the components of
force which the track exerts on the car in the r, u, and z
directions at the instant t = 2 s. Neglect the size of the car.
r=8m
SOLUTION
#
$
Kinematic: Here, r = 8 m, r = r = 0. Taking the required time derivatives at
t = 2 s, we have
#
$
u = 0.100 rad>s
u = 0
u = 0.1t + 0.5|t = 2s = 0.700 rad
z = - 0.2t|t = 2 s = - 0.400 m
#
z = - 0.200 m>s
$
z = 0
Applying Eqs. 12–29, we have
$
$
ar = r - ru2 = 0 - 8(0.1002) = - 0.0800 m>s2
$
##
au = r u + 2 ru = 8(0) + 2(0)(0.200) = 0
$
az = z = 0
Equation of Motion:
©Fr = mar ;
Fr = 250(-0.0800) = - 20.0 N
Ans.
©Fu = mau ;
Fu = 250(0) = 0
Ans.
©Fz = maz ;
Fz - 250(9.81) = 250(0)
Fz = 2452.5 N = 2.45 kN
Ans.
13–103.
A ride in an amusement park consists of a cart which is
supported by small wheels. Initially the cart is traveling in a
circular path# of radius r0 = 16 ft such that the angular rate of
rotation is u0 = 0.2 rad>s. If the attached cable OC is drawn
#
inward at a constant speed of r = - 0.5 ft>s, determine the
tension it exerts on the cart at the instant r = 4 ft. The cart
and its passengers have a total weight of 400 lb. Neglect
the effects of friction. Hint: First show that the equation
of
..
..
motion in# the u direction yields # au = ru + 2ru =
11>r2 d1r2u2>dt = 0. When integrated, r2u = c, where the
constant c is determined from the problem data.
r
u
O
SOLUTION
+Q©Fr = mar ;
-T = ¢
+a©Fu = mau ;
0 = ¢
#
400
$
≤ ¢ r - ru 2 ≤
32.2
$
400
##
≤ ¢ ru + 2ru ≤
32.2
#
1 d
From Eq. (2), ¢ ≤ ¢ r2u ≤ = 0
r dt
(1)
(2)
#
r 2u = c
#
Since u0 = 0.2 rad>s when r0 = 16 ft, c = 51.2.
Hence, when r = 4 ft,
#
51.2
u = ¢ 2 ≤ = 3.2 rad>s
(4)
$
Since r = - 0.5 ft>s, r = 0, Eq. (1) becomes
-T =
400
a 0 - (4)(3.2)2 b
32.2
T = 509 lb
C
Ans.
·
u0
13–107.
The forked rod is used to move the smooth
2-lb particle around the horizontal path in the shape of a
limaçon, r = (2 + cos u) ft. If u = (0.5t2) rad, where t is in
seconds, determine the force which the rod exerts on the
particle at the instant t = 1 s. The fork and path contact the
particle on only one side.
2 ft
u
·
u
SOLUTION
r = 2 + cos u
#
r = -sin uu
#
$
$
r = -cos uu2 - sin uu
u = 0.5t2
#
u = t
$
u = 1 rad>s2
$
At t = 1 s, u = 0.5 rad, u = 1 rad>s, and u = 1 rad>s2
r = 2 + cos 0.5 = 2.8776 ft
#
r = -sin 0.5(1) = -0.4974 ft>s2
$
r = -cos 0.5(1)2 - sin 0.5(1) = -1.357 ft>s2
#
$
ar = r - ru2 = -1.375 - 2.8776(1)2 = -4.2346 ft>s2
$
##
au = ru + 2ru = 2.8776(1) + 2( -0.4794)(1) = 1.9187 ft>s2
tan c =
r
r
2 + cos u 2
=
= -6.002
dr>du
-sin u u = 0.5 rad
c = -80.54°
2
(-4.2346)
32.2
+Q©Fr = mar;
-N cos 9.46° =
+a©Fu = mau;
F - 0.2666 sin 9.46° =
F = 0.163 lb
N = 0.2666 lb
2
(1.9187)
32.2
Ans.
3 ft
13–109.
The smooth particle has a mass of 80 g. It is attached to an
elastic cord extending from O to P and due to the slotted
arm guide moves along the horizontal circular path
r = 10.8 sin u2 m. If the cord has a stiffness k = 30 N>m and
an unstretched length of 0.25 m, determine the force of the
guide on the particle
when u = 60°. The guide has a constant
#
angular velocity u = 5 rad>s.
P
r
·
u
0.4 m
SOLUTION
u
r = 0.8 sin u
#
#
r = 0.8 cos u u
O
#
$
$
r = -0.8 sin u (u)2 + 0.8 cos uu
#
$
u = 5,
u = 0
r = 0.6928
At u = 60°,
#
r = 2
$
r = -17.321
#
$
ar = r - r(u)2 = -17.321 - 0.6928(5)2 = -34.641
$
##
au = ru + 2 ru = 0 + 2(2)(5) = 20
Fs = ks;
Fs = 30(0.6928 - 0.25) = 13.284 N
Q+ ©Fr = m ar;
-13.284 + NP cos 30° = 0.08(- 34.641)
a+ ©Fu = mau;
F - NP sin 30° = 0.08(20)
F = 7.67 N
NP = 12.1 N
Ans.
5 rad/s
13–110.
The smooth particle has a mass of 80 g. It is attached to an
elastic cord extending from O to P and due to the slotted arm
guide moves along the horizontal circular path r = (0.8 sin u)
m. If the cord has a stiffness k = 30 N>m and an unstretched
length of 0.25$ m, determine
# the force of the guide on the
particle when u = 2 rad>s2, u = 5 rad>s, and u = 60°.
P
r
·
u
0.4 m
SOLUTION
r = 0.8 sin u
u
#
#
r = 0.8 cos u u
O
$
#
$
r = -0.8 sin u (u)2 + 0.8 cos uu
$
#
u = 2
u = 5,
At u = 60°,
r = 0.6928
#
r = 2
$
r = -16.521
#
$
ar = r - r(u)2 = -16.521 - 0.6928(5)2 = -33.841
$
##
au = r u + 2 ru = 0.6925(2) + 2(2)(5) = 21.386
Fs = ks;
Q+ ©Fr = m ar;
+a©Fu = mau;
Fs = 30(0.6928 - 0.25) = 13.284 N
- 13.284 + NP cos 30° = 0.08(-33.841)
F - NP sin 30° = 0.08(21.386)
F = 7.82 N
NP = 12.2 N
Ans.
5 rad/s