P PROBLEM 11.

138

A robot arm moves so that P travels in a circle about Point B, which

0.8 m is not moving. Knowing that P starts from rest, and its speed
2
increases at a constant rate of 10 mm/s , determine (a) the magnitude
of the acceleration when t = 4 s, (b) the time for the magnitude of
2
the acceleration to be 80 mm/s .

O B

SOLUTION

Tangential acceleration: at = 10 mm/s2

Speed: v = at t

v2 at2 t 2
Normal acceleration: an = =
r r
where r = 0.8 m = 800 mm

(a) When t = 4 s v = (10)(4) = 40 mm/s

(40)2
an = = 2 mm/s2
800

Acceleration: a = at2 + an2 = (10)2 + (2)2

a = 10.20 mm/s2 ◀

(b) Time when a = 80 mm/s2

a 2 = an2 + at2
é (10)2 t 2 ù 2
(80) = ê
2 ú + 10 2 t 4 = 403200 s 4
ëê 800 ûú
t = 25.2 s ◀

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 400 m PROBLEM 11.142
A
At a given instant in an airplane race, airplane A is flying
450 km/h horizontally in a straight line, and its speed is being increased
B
at the rate of 8 m/s2. Airplane B is flying at the same altitude
300 m
as airplane A and, as it rounds a pylon, is following a circular
path of 300-m radius. Knowing that at the given instant the
30°
speed of B is being decreased at the rate of 3 m/s2, determine,
540 km/h for the positions shown, (a) the velocity of B relative to A,
(b) the acceleration of B relative to A.

SOLUTION

First note v A = 450 km/h v B = 540 km/h = 150 m/s

(a ) We have v B = v A + v B /A

The graphical representation of this equation is then as shown.

We have v B2 /A = 450 2 + 540 2 - 2(450)(540) cos 60

vB/A = 501.10 km/h

540 501.10
and =
sin a sin 60
a = 68.9 

v B / A = 501 km/h 68.9 ◀

(b) First note a A = 8 m/s2 (a B )t = 3 m/s2 60

vB2 (150 m/s)2

Now (aB )n = =
rB 300 m

(aB )n = 75 m/s2 30

a B = ( a B )t + ( a B ) n
Then = 3(- cos60i + sin 60 j) + 75(- cos30i - sin 30 j)
= -(66.452 m/s2 )i - (34.902 m/s2 ) j

Finally aB = a A + aB / A
a B / A = -(66.452 i - 34.902 j) - (8i )
= -(74.452 m/s2 )i - (34.902 m/s2 ) j

a B / A = 82.2 m/s2 25.1 ◀

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PROBLEM 11.158
A satellite will travel indefinitely in a circular orbit around the earth if the normal component of its
acceleration is equal to g(R/r )2, where g = 9.81 m/s2, R = radius of the earth = 6370 km, and r = distance
from the center of the earth to the satellite. Assuming that the orbit of the moon is a circle of radius
384 ´103 km, determine the speed of the moon relative to the earth.

SOLUTION

gR 2 v2 v 2
Normal acceleration: an = and an = r =
r2 r

gR 2
v2 = ran =
2
Solve for v :
r

Data: g = 9.81 m/s2, R = 6370 km = 6.370 ´106 m

r = 384 ´103 km = 384 ´106 m

2
(9.81)(6.370 ´106 )
v2 = = 1.0366 ´106 m 2 /s2
384 ´106
v = 1.018 m/s v = 3670 km/h ◀

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PROBLEM 12.9
If an automobile’s braking distance from 108 km/h is 75 m on level pavement, determine the
automobile’s braking distance from 108 km/h when it is (a) going up a 5° incline, (b) going down a
3-percent incline. Assume the braking force is independent of grade.

SOLUTION
Assume uniformly decelerated motion in all cases.

For braking on the level surface,

v0 = 108 km/h = 30 m/s, v f = 0
x f - x0 = 75 m
v 2f = v02 + 2 a( x f - x0 ) y
v 2f - v02 FBD KD
a= x
2( x f - x0 ) mg
0 - (30)2 FB
= ma
(2)(75)
= -6.00 m/s2
Braking force. N
Fb = ma, Fb = 6 m

y
FBD KD
(a) Going up a 5° incline. mg
x

 SF = ma ma
FB
-Fb - mg sin 5 = ma
5°
Fb + mg sin 5
a =-
m N
6m + m(9.81) sin 5
a =-
m
= -(6 + 9.81sin(5))
= -6.855 m/s2
v 2f - v02
x f - x0 =
2a
0 - (30)2
= x f - x0 = 65.6 m ◀
(2)(- 6.855)

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 y
KD
PROBLEM 12.9 (CONTINUED) FBD
mg x

FB
(b) Going down a 3 percent incline. ma
β
3
tan b = b = 1.71835 N
100
-Fb + mg sin b = ma
-6 m + m(9.81) sin 1.71835
a=
m
a = -5.7058

0 - (30)2
x f = x0 = x f - x0 = 78.9 m ◀
(2)(-5.7058)

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 PROBLEM 12.34

The 15-kg block B is supported by the 25-kg block A and is

attached to a cord to which a 225-N horizontal force is applied
as shown. Neglecting friction, determine (a) the acceleration of
block A, (b) the acceleration of block B relative to A.

SOLUTION

(a) First we note a=

B a A + a B/A , where a B/A is directed along the inclined surface of A.

B: =ΣFx mB ax : P − W=
B sin 25° mB a A cos 25° + mB aB/A
B:

or 225 − 15
= g sin 25° 15(a A cos 25° + aB/A )

or 15 − g=
sin 25° a A cos 25° + aB/A (1)

ΣFy = mB a y : N AB − WB cos 25° = −mB a A sin 25°

or N AB = 15( g cos 25° − a A sin 25°)

A: ΣFx′ mA a A : P − P cos 25° + N AB

= = sin 25° mA a A A:

or N AB = [25a A − 225(1 − cos 25°)]/ sin 25°

Equating the two expressions for N AB

25a A − 225(1 − cos 25°)

15( g cos 25° − a A sin 25°) =
sin 25°
3(9.81) cos 25° sin 25° + 45(1 − cos 25°)
or aA =
5 + 3sin 2 25°
2
= 2.7979 m/s

a A = 2.80 m/s 2 W

(b) From Eq. (1)

aB/A = 15 − (9.81)sin 25° − 2.7979cos 25°

or a B/A = 8.32 m/s 2 25° W

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2188
 PROBLEM 12.36

A 450-g tetherball A is moving along a horizontal circular path at a

constant speed of 4 m/s. Determine (a)) the angle  that the cord
forms with pole BC, (b)) the tension in the cord.

SOLUTION

v A2
First we note a A  an 


where   l AB sin 

(a) Fy  0: TAB cos  WA  0

mA g
or TAB 
cos 

v A2
Fx  mA a A : TAB sin   mA


Substituting for TAB and 

mA g v A2
sin   mA sin 2   1  cos2 
cos  l AB sin 
(4 m/s) 2
1  cos 2   cos 
1.8 m  9.81 m/s 2

or cos 2   0.906105cos   1  0

Solving cos   0.64479

or   49.9 

mA g 0.450 kg  9.81 m/s 2

(b) From above TAB  
cos  0.64479
or TAB  6.85 N 

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2191