Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
11th Edition
ISBN: 9780077687342
Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 12, Problem 12.122RP

In the braking test of a sports car, its velocity is reduced from 70 mi/h to zero in a distance of 170 ft with slipping impending. Knowing that the coefficient of kinetic friction is 80 percent of the coefficient of static friction, determine (a) the coefficient of static friction, (b) the stopping distance for the same initial velocity if the car skids. Ignore air resistance and rolling resistance.

Expert Solution
Check Mark
To determine

(a)

The co-efficient of static friction.

Answer to Problem 12.122RP

μs=0.9627

Explanation of Solution

Given information:

At a braking test,

The velocity is reduced from 70mi/h to zero

The stopping distance is equal to 170ft

Co-efficient of kinematic friction is 80% of static friction

For a uniformly accelerated motion,

v2=v02+2a(xx0)

In the above equation,

v - End velocity

v0 - Start velocity

a - Acceleration

x - End position

x0 - Start position

The static friction force is defined as,

Ffriction=μsR

In the above equation,

μs - Co-efficient of static friction

R - Reaction force

Calculation:

Convert,

70mi/h=102.667ft/s

For a uniformly accelerated motion,

v2=v02+2a(xx0)0=(102.667ft/s)2+2a(170ft)a=31ft/s2

For the force balance in upwards direction,

F=RW=0

Therefore,

R=W=mg

Apply Newton’s second law of motion,

F=maμsR=ma

Then,

μs=mamg=ag

Therefore,

μs=ag=(31ft/s2)32.2ft/s2μs=0.9627

Conclusion:

The co-efficient of static friction is equal to μs=0.9627

Expert Solution
Check Mark
To determine

(b)

The stopping distance if a car skids.

Answer to Problem 12.122RP

Stopping distance if car skids is 212.51ft.

Explanation of Solution

Given information:

At a braking test,

The velocity is reduced from 70mi/h to zero.

The stopping distance is equal to 170ft without sliding.

Co-efficient of kinematic friction is 80% of static friction.

For a uniformly accelerated motion,

v2=v02+2a(xx0)

In the above equation,

v - End velocity

v0 - Start velocity

a - Acceleration

x - End position

x0 - Start position

The kinematic friction force is defined as,

Ffriction=μkR

In the above equation,

μs - Co-efficient of kinematic friction

R - Reaction force

Calculation:

Convert,

70mi/h=102.667ft/s

According to the given information,

μk=0.8μs=0.8(0.9627)=0.77

Apply Newton’s second law of motion,

F=maμkR=ma

Therefore,

a=μkRm=μkg

Substitute,

a=μkg=(0.77)(32.2ft/s2)=24.8ft/s2

Now,

v2=v02+2a(xx0)

Rearrange to find the stopping distance,

xx0=v2v022a

Substitute,

xx0=0(102.667ft/s)22(24.8ft/s2)=212.51ft

Conclusion:

For skidding, the stopping distance is equal to 212.51ft.

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Chapter 12 Solutions

Vector Mechanics for Engineers: Dynamics

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