978-0073380308 Chapter 3 Solution Manual Part 24

636 Solutions Manual

Problem 3.142

Two particles

A

and

B

, with masses

mA

and

mB

, respectively, are a distance

r0

apart. Particle

B

is

fixed in space, and

A

is initially at rest. Using Eq. (1.5) on p. 3 and assuming that the diameters of the

masses are negligible, determine the time it takes for the two particles to come into contact if

mAD1kg

,

mBD2kg

, and

r0D1

m. Assume that the two masses are infinitely far from any other mass. Hint:

Rr0

0pr0r=.r0r/dr D1

2⇡r3=2

0.

Solution

We model

A

as a particle subject only to the gravitational attraction due to its interaction with

B

.

The motion of Awill only occur along the line connecting Aand B. Particle Bdoes not move.

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Problem 3.143

The system shown is initially at rest when the bent bar starts to rotate about

the vertical axis

AB

with constant angular acceleration

˛0D3rad=s2

. The

coefficient of static friction between the collar of mass

mD2kg

and the bent

bar is

sD0:6

, the angle of the bend in the bar is

✓D30ı

, and the collar is

initially at dD70 cm from the spin axis AB.

Assuming the motion starts at

tD0

, determine the time at which the

collar starts to slip relative to the bent bar.

Solution

The figure at the right shows a side and a top view of the collar’s FBD. Until the

collar slides, it moves along a circle with center

O

and radius

dcos ✓

. We use

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Problem 3.144

The system shown is initially at rest when the bent bar starts to rotate about

the vertical axis

AB

with constant angular acceleration

˛0D3rad=s2

. The

coefficient of static friction between the collar of mass

mD2kg

and the bent

bar is

sD0:6

, the angle of the bend in the bar is

✓D30ı

, and the collar is

initially at dD70 cm from the spin axis AB.

Determine the number of rotations undergone by the bent bar when the

collar starts to slip relative to it.

Solution

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Problem 3.145

The centers of two spheres

A

and

B

, with weights

WAD3lb

and

WBD7lb

,

respectively, are a distance

r0D5ft

apart when they are released from rest.

Using Eq. (1.5) on p. 3, determine the speed with which they collide if the

diameters of spheres

A

and

B

are

dAD2:5 in:

and

dBD4in:

, respectively.

Assume that the two masses are only inﬂuenced by their mutual gravitational

attraction.

Solution

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Problem 3.146

A roller coaster goes over the top

A

of the track shown with a speed

vD135 km=h

. If the radius of

curvature at

A

is

⇢D60

m, determine the minimum force that a restraint must apply to a person with a

mass of 85 kg to keep the person on his or her seat.

Solution

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Problem 3.147

A

50;000 lb

aircraft is ﬂying along a rectilinear path at a constant altitude with a

speed

vD720 mph

when the pilot initiates a turn by banking the plane

20ı

to the

right. Assuming that the initial rate of change of speed is negligible, determine

the components of the acceleration of the aircraft right at the beginning of the

turn if the pilot does not adjust the attitude of the aircraft so that the magnitude of

the lift remains the same as when the plane is ﬂying straight and the aerodynamic

drag remains in the horizontal plane. Also determine the radius of curvature at

the beginning of the turn.

3

Solution

3

First we determine the components of the acceleration of the aircraft using a convenient

component system. Then we will determine the radius of curvature of the trajectory at the

start of the maneuver. Referring to the figure at the right, we consider the FBD of the aircraft

at the start of the maneuver. We sketched a view from the rear and a view from the top in

which we have indicated a Cartesian component system with base vectors

O{

,

O|

,

O

k

, such that

O|

is oriented as the velocity of the plane at the instant shown, and

O

k

is oriented opposite

to gravity. The coordinate system corresponding to these base vectors is fixed in space and

has its origin coinciding with the aircraft at the instant shown. We model the aircraft as a

particle subject to its own weight

mg

, the thrust

T

, the drag force

D

, and the lift force

FL

.

The lift force is shown forming an angle

✓

with the vertical direction, where

✓D20ı

is the

bank angle.

Balance Principles. Applying Newton’s second law, we have

XFxWFLsin ✓Dmax;(1)

XFyWTDDmay;(2)

XF´WFLcos ✓mg Dma´;(3)

where ax,ay, and a´are the desired acceleration components.

Force Laws.

As indicated in the problem statement, at the instant shown, the magnitudes of the lift and

drag forces are the same as right before the start of the turn. Since the aircraft was traveling horizontally at

constant velocity, we must have

FLDmg and DDT: (4)

Kinematic Equations.

No kinematic relations are needed since the components of acceleration are the

primary unknowns.

Computation. Substituting Eq. (4) into Eqs. (1)–(3),after simplification we have

axDgsin ✓; ayD0; a´Dg.cos ✓1/: (5)

Recalling that gD32:2 ft=s2and ✓D20ı, we have

axD11:01 ft=s2;a

yD0; a´D1:942 ft=s2:(6)

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July 17, 2012