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13–61.
At the instant the boy’s center of mass Gis
momentarily at rest. Determine his speed and the tension in
each of the two supporting cords of the swing when
The boy has a weight of 60 lb. Neglect his size and
the mass of the seat and cords.
u=90°.
u=60°,
SOLUTION
(1)
Ans.
From Eq. (1)
Ans.2T-60 sin 90° =60
32.2 a9.2892
10 b
T=38.0 lb
v=9.289 ft>s
Lv
0
v dn=L90°
60°
322 cos u du
v dn=a ds
however ds =10du
Q+©Fn=man;
2T-60 sin u=60
32.2 av2
10 b
+R©t=mat;
60 cos u=60
32.2 at
at=32.2 cos u
10 ft
u
13–62.
A girl having a mass of 25 kg sits at the edge of the merry-
go-round so her center of mass Gis at a distance of 1.5 m
from the axis of rotation. If the angular motion of the
platform is slowly increased so that the girl’s tangential
component of acceleration can be neglected, determine the
maximum speed which she can have before she begins to
slip off the merry-go-round.The coefficient of static friction
between the girl and the merry-go-round is ms=0.3.
SOLUTION
Ans.v=2.10 m>s
:
+©Fn=man; 0.3(245.25) =25(v2
1.5)
z
1.5 m
G
13–63.
The pendulum bob has a weight of 5 lb and is released from
rest in the position shown, .Determine the tension in
string just after the bob is released,,and also at the
instant the bob reaches point ,.Take .r=3ftu=45°D
u=0°BC
u=0°
B
SOLUTION
FBD(a), we have
Ans.
Applying Eq. 13–8 to FBD(b), we have
[1]
Kinematics:The speed of the bob at the instant when can be determined
using . However,,then .
Substitute and into Eq. [1] yields
Ans.T=10.6 lb
T-5 sin 45° =5
32.2 a136.61
3b
y2=136.61 ft2>s2
u=45°
y2=136.61 ft2>s2
Ly
0
ydy=3(32.2) L45°
0
cos udu
ydy=3atduds =3duydy=atds
u=45°
©F
n=man;T-5 sin u=5
32.2 ay2
3b
©F
t=mat;5cos u=5
32.2 atat=32.2 cos u
©F
n=man;T=5
32.2 a02
3b=0
C
r
B
308
*13–64.
SOLUTION
FBD(a), we have
Ans.
Applying Eq. 13–8 to FBD(b), we have
[1]
Kinematics:The speed of the bob at the arbitrary position can be detemined using
. However,,then .
Substitute into Eq. [1] yields
Ans.T=3mg sin u
T-mg sin u=ma2gr sin u
rb
y2=2gr sin u
y2=2gr sin u
Ly
0
ydy=gr Lu
0
cos udu
ydy=atrduds =rduydy=atds
u
©F
n=man;T-mg sin u=may2
rb
©F
t=mat;mg cos u=matat=gcos u
©F
n=man;T=ma02
rb=0
The pendulum bob Bhas a mass mand is released from rest
when .Determine the tension in string BC immediately
afterwards, and also at the instant the bob reaches the arbitrary
position .u
u=0°
C
r
B
13–65.
SOLUTION
Ans.
Ans.
Ans.
Ans. Fb=490 N
©Fb=m ab ;
Fb-490.5 =0
©Ft=m at;
Ft=0
©Fn=m an ;
Fn=50(16.3022
7)=283 N
v=6.30 m>s
T=906.2 N
+c©Fb=0;
T cos 30° -8019.812=0
;
+©Fn=m an ;
T sin 30° =80( v2
4+6 sin 30°)
Determine the constant speed of the passengers on the
amusement-park ride if it is observed that the supporting
cables are directed at from the vertical. Each chair
including its passenger has a mass of 80 kg. Also, what are
the components of force in the n, t, and bdirections which
the chair exerts on a 50-kg passenger during the motion?
u=30°
6 m
4 m
b
t
u
310
Ans:
v
=22.1 m>s
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 of
the vehicle so that only a normal force from the seat acts on
the driver. Neglect the size of the driver.
u
>u
SOLUTION
*13–68.
The 0.8-Mg car travels over the hill having the shape of a
parabola. If the driver maintains a constant speed of 9 ms,
determine both the resultant normal force and the
resultant frictional force that all the wheels of the car exert
on the road at the instant it reaches point A.Neglect the
size of the car.
>y
Ax
y20 (1 )
80 m
x2
6400
SOLUTION
d2y
dy
13–69.
The 0.8-Mg car travels over the hill having the shape of a
parabola. When the car is at point A,it is traveling at 9 ms
and increasing its speed at .Determine both the
resultant normal force and the resultant frictional force that
all the wheels of the car exert on the road at this instant.
Neglect the size of the car.
3m>s2>
SOLUTION
Ais given by
and the radius of curvature at point Ais
Equation of Motion: Applying Eq. 13–8 with and , we have
Ans.
Ans.N
=
6729.67 N
=
6.73 kN
©Fn=man; 800(9.81) cos 26.57° -N=800a92
223.61
≤
Ff=1109.73 N =1.11 kN
©Ft=mat; 800(9.81) sin 26.57° -Ff=800(3)
r=223.61 mu=26.57°
r=
C
1+(dy>dx)2
D
3>2
|d2y>dx2|=
C
1+(-0.00625x)2
D
3>2
|-0.00625| 2x=80 m
=223.61 m
tan u=dy
dx 2x=80 m
=-0.00625(80)
u=-26.57°
d2y
dy
y
Ax
y20 (1 )
x2
6400
13–70.
SOLUTION
At ,
Ans.
Ans.
At
Ans.vB=21.0 ft s
f=45° +45° =90°
NC=7.91 lb
vC=19.933 ft>s=19.9 ft>s
f=45° +30° =75°
1
2v2-1
2(8)2=644(sin f-sin 45°)
Lv
g
vdv=Lf
45°
32.2 cos f(20 df)
vdv=atds
+a©Fn=man;N-5 sin f=5
32.2 ( v2
20)
at=32.2 cos f
+b ©Ft=mat;5cos f=5
32.2 at
The package has a weight of 5 lb and slides down the chute.
When it reaches the curved portion AB, it is traveling at
If the chute is smooth, determine the speed
of the package when it reaches the intermediate point
and when it reaches the horizontal plane
Also, find the normal force on the package at C.1u=45°2.
C1u=30°2
1u=0°2.8ft>s
20 ft
45°
45°
=30°
θ
B
A
8 ft/s
C
*13–72.
The 150-lb man lies against the cushion for which the
coefficient of static friction is If he rotates about
the zaxis with a constant speed determine the
smallest angle of the cushion at which he will begin to
slip off.
u
v=30 ft>s,
ms=0.5.
z
G
8 ft
13–73.
Determine the maximum speed at which the car with mass
mcan pass over the top point Aof 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 Bon the road? rr
rr
A
B
SOLUTION
13–75.
The box has a mass mand slides down the smooth chute
having the shape of a parabola. If it has an initial velocity of
at the origin, determine its velocity as a function of x.
Also, what is the normal force on the box, and the
tangential acceleration as a function of x?
v0
SOLUTION
(1)
Ans.
Ans.
From Eq. (1):
Ans.N=m
1+x2g-(v2
0+gx2)
(1 +x2)
v=2v2
0+gx2
1
2v2-1
2v2
0=gax2
2b
Lv
v0
vdv=Lx
0
gx dx
ds =
B
1+ady
dx b2
R
1
2
dx =
A
1+x2
B
1
2dx
vdv=atds =g
¢
x
21+x2
≤
ds
at=g
¢
x
21+x2
≤
+R©F
t=mat;mg
¢
x
21+x2
≤
=mat
+b©F
n=man;mg
¢
1
21+x2
≤
-N=m
¢
v2
(1 +x2)3
2
≤
r=
B
1+ady
dx b2
R
3
2
2d2y
dx22=
C
1+x2
D
3
2
|-1| =
A
1+x2
B
3
2
d2y
dx2=-1
dy
dx =-x
x=-
1
2x2
y
x
x
y=– 0.5 x
2
320
*13–76.
Prove that if the block is released from rest at point Bof a
smooth path of arbitrary shape, the speed it attains when it
reaches point Ais equal to the speed it attains when it falls
freely through a distance h; i.e., v=22gh.
SOLUTION
Q.E.D.v=22gh
v2
2=gh
Lv
0
vdv=Lh
0
gdy
vdv=atds =gsin uds
However
dy =ds sin u
+R©Ft=mat;mg sin u=matat=gsin u
A
B
h
321
Ans:
Fs=4.90 lb
322
13–78.
When crossing an intersection, a motorcyclist encounters
the slight bump or crown caused by the intersecting road.
If the crest of the bump has a radius of curvature r
=50 ft,
determine the maximum constant speed at which he can
travel without leaving the surface of the road. Neglect the
size of the motorcycle and rider in the calculation. The rider
and his motorcycle have a total weight of 450 lb.
Ans:
v
=40.1 ft>s
SOLUTION
+Tg
F
n=
ma
n
; 450 -0=
450
32.2
a
v
2
50
b
v
=40.1 ft>s
Ans.
r 50 ft
13–79.
The airplane, traveling at a constant speed of is
executing a horizontal turn. If the plane is banked at
when the pilot experiences only a normal force on
the seat of the plane, determine the radius of curvature of
the turn. Also, what is the normal force of the seat on the
pilot if he has a mass of 70 kg.
r
u=15°,
50 m>s,
SOLUTION
Ans.
Ans. r=68.3 m
;
+aFn=man;
NP cos 15° =70a502
rb
NP=2.65 kN
+caFb=mab;
NP sin 15° -7019.812=0
u
r
*13–80.
The 2-kg pendulum bob moves in the vertical plane with
a velocity of 8 m
>
s when
u=0°.
Determine the initial
tension in the cord and also at the instant the bob reaches
u=30°.
Neglect the size of the bob.
2 m
SOLUTION
