PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 5, Problem 50P

(a)

To determine

The minimum acceleration so that box will not fall to the ground.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of box is 2.0kg .

Coefficient of friction between the box and the wall is 0.60 .

Formula used:

Draw free body diagram of the box.

  PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 5, Problem 50P , additional homework tip  1

Write expression for net force in x -direction.

  Fx=Fn

Here, Fn is the normal force on the box.

Substitute mamin for Fn in above expression.

  Fx=mamin

Here, m is mass of the box and amin is the minimum acceleration.

Write expression for net force in y -direction.

  Fy=0

Substitute fs,maxmg for Fy in above expression.

  fs,maxmg=0

Here, fs,max is the maximum friction force by the wall on the box.

Here, g is acceleration due to gravity.

Substitute μsFn for fs,max in above expression.

  μsFnmg=0

Here, μs is the coefficient of friction between box and wall.

Substitute mamin for Fn in above expression.

  μs(mamin)mg=0

Solve above expression for amin .

  amin=gμs.......(1)

Calculation:

Substitute 9.81m/s2 for g and 0.60 for μs in equation (1).

  amin=9.81m/ s 20.60=16.35m/s2

Conclusion:

Thus, the minimum acceleration is 16.35m/s2 .

(b)

To determine

The magnitude of frictional force.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of box is 2.0kg .

Coefficient of friction between the box and the wall is 0.60 .

Formula used:

Draw free body diagram of the box.

  PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 5, Problem 50P , additional homework tip  2

Write expression for net force in y -direction.

  Fy=0

Substitute fs,maxmg for Fy in above expression.

  fs,maxmg=0

Here, fs,max is the maximum friction force by the wall on the box.

Rearrange above expression for fs,max .

  fs,max=mg.......(1)

Calculation:

Substitute 2.0kg for m and 9.81m/s2 for g .

  fs,max=(2.0kg)(9.81m/ s 2)=19.62N

Conclusion:

Thus, the magnitude of frictional force is 19.62N .

(c)

To determine

The force of friction on box if the acceleration of box is twice the minimum acceleration.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of box is 2.0kg .

Coefficient of friction between the box and the wall is 0.60 .

Formula used:

Draw free body diagram of the box.

  PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 5, Problem 50P , additional homework tip  3

Write expression for net force in y -direction.

  Fy=0

Substitute fs,maxmg for Fy in above expression.

  fs,maxmg=0

Here, fs,max is the maximum friction force by the wall on the box.

Rearrange above expression for fs,max .

  fs,max=mg.......(1)

Calculation:

Substitute 2.0kg for m and 9.81m/s2 for g .

  fs,max=(2.0kg)(9.81m/ s 2)=19.62N

Conclusion:

The friction force is independent of minimum acceleration. Thus, the magnitude of frictional force is 19.62N .

(d)

To determine

To show that the box will not fall if the acceleration is greater or equal to gμs .

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

Mass of box is 2.0kg .

Coefficient of friction between the box and the wall is 0.60 .

Introduction:

Draw free body diagram of the box.

  PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 5, Problem 50P , additional homework tip  4

Write expression for net force in x -direction.

  Fx=Fn

Here, Fn is the normal force on the box.

Write expression for net force in y -direction.

  Fy=0

Substitute fs,maxmg for Fy in above expression.

  fs,maxmg=0

Here, fs,max is the maximum friction force by the wall on the box.

Here, g is acceleration due to gravity.

Substitute μsFn for fs,max in above expression.

  μsFnmg=0

Here, μs is the coefficient of friction between box and wall.

Substitute mamin for Fn in above expression.

  μs(mamin)mg=0

Solve above expression for amin .

  amin=gμs

The minimum acceleration required for the box to not fall in the ground is gμs .

If the acceleration is greater than gμs then also box will not fall.

Conclusion:

Thus, the box will not fall if the acceleration is greater or equal to gμs .

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PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

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