Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 14, Problem 98P

a.

To determine

The value of coefficient of static friction between box and floor.

a.

Expert Solution
Check Mark

Explanation of Solution

Given:

Period of oscillation is 0.80s .

Box started to slip when amplitude reaches 10cm .

Formula used:

For equilibrium:

Write expression for force on x-direction.

  fx=0

Substitute fsm2amax for fx in above expression.

  fsm2amax=0  ......(1)

Here, fs is the maximum friction on box and amax is the maximum acceleration.

Write expression for force on y-direction.

  fy=0

Substitute Fnm2g for fy in above expression.

  Fnm2g=0  ......(2)

Here, Fn is the normal force on the box and g is the acceleration due to gravity.

Write expression for maximum acceleration related to angular frequency and amplitude.

  amax=Aω2  ......(3)

Here, A is the amplitude and ω is the angular frequency.

Substitute μsFn for fs in equation (1) and solve for Fn .

  μsFnm2amax=0Fn=m2a maxμs

Substitute m2amaxμs for Fn in equation (2) and solve for amax .

  m2a maxμsm2g=0m2a maxμs=m2ga maxμs=gamax=μsg

Substitute μsg for amax and in equation (3) and solve for μs .

  μsg=Aω2μs=Aω2g

Substitute 2πT for ω in above expression.

  μs=4π2AT2g  ......(4)

Calculation:

Substitute 10cm for A , 0.80s for T and 9.81m/s2 for g in equation (4).

  μs=4π2( 10cm) ( 0.80s )2( 9.81m/ s 2 )( 1m 100cm)=0.63

Conclusion:

Thus, the value of coefficient of static friction between box and floor is 0.63 .

(b)

To determine

The maximum amplitude of vibration for given value of μs .

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The value of μs is 0.40 .

Formula used:

For equilibrium:

Write expression for force on x-direction.

  fx=0

Substitute fsm2amax for fx in above expression.

  fsm2amax=0  ......(1)

Here, fs is the maximum friction on box and amax is the maximum acceleration.

Write expression for force on y-direction.

  fy=0

Substitute Fnm2g for fy in above expression.

  Fnm2g=0  ......(2)

Here, Fn is the normal force on the box and g is the acceleration due to gravity.

Write expression for maximum acceleration related to angular frequency and amplitude.

  amax=Aω2  ......(3)

Here, A is the amplitude and ω is the angular frequency.

Substitute μsFn for fs in equation (1) and solve for Fn .

  μsFnm2amax=0Fn=m2a maxμs

Substitute m2amaxμs for Fn in equation (2) and solve for amax .

  m2a maxμsm2g=0m2a maxμs=m2ga maxμs=gamax=μsg

Substitute μsg for amax and in equation (3) and solve for A .

  μsg=Aω2A=μsgω2

Substitute 2πT for ω in above expression.

  A=μsgT24π2   ...... (4)

Calculation:

Substitute 0.40 for μs , 9.81m/s2 for g and 0.80s for T

  A=( 0.40)( 9.81m/ s 2 ) ( 0.80s )24π2=0.63m

Conclusion:

Thus, the maximum value of amplitude is 0.63m .

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Chapter 14 Solutions

Physics for Scientists and Engineers

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