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 63P

(a)

To determine

To Calculate: The maximum and minimum values of applied force for which the block does not slip.

(a)

Expert Solution
Check Mark

Answer to Problem 63P

  Fmax=84NFmin=1.6N

Explanation of Solution

Given data:

Mass of the block, m=0.50kg

Mass of the wedge, m'=2.0kg

Coefficient of static friction, μs=0.80

Angle of wedge, θ=35°

Formula Used:

Newton’s second law of motion:

  F=ma

Where, m is the mass and a is the acceleration.

Calculation:

Free-body diagram:

  PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 5, Problem 63P

Maximum force, Fmax=mtamax

Minimum force, Fmin=mtamin

Where, mt is the total mass of the block and the wedge.

For the block,

  Fx=Nsinθfcosθ=max...(1)Fy=Ncosθ+fsinθmg=0...(2)fmax=μsNNcosθ+μsNsinθmg=0N=mgcosθ+μssinθ...(3)

Substitute (3) in (1) to find the minimum acceleration:

  mgcosθ+μssinθsinθμsmgcosθ+μssinθcosθ=maminamin=g(sinθμscosθ)cosθ+μssinθFmin=mtg(sinθμscosθ)cosθ+μssinθ

Substitute the values and solve:

  Fmin=|(2.0+0.5)(9.8)(sin35°0.80cos35°)cos35°+0.80sin35°|Fmin=1.6N

For maximum force, reverse the direction of f

  amax=g(sinθ+μscosθ)cosθμssinθFmax=mtg(sinθ+μscosθ)cosθμssinθ

Substitute the values and solve:

  Fmax=|(2.0+0.5)(9.8)(sin35°+0.80cos35°)cos35°0.80sin35°|Fmax=84N

Conclusion:

The maximum and minimum values of applied force for which the block does not slip are 84 N and 1.6 N respectively.

(b)

To determine

To Calculate: The maximum and minimum values of applied force for which the block does not slip.

(b)

Expert Solution
Check Mark

Answer to Problem 63P

  Fmax=37.5NFmin=5.8N

Explanation of Solution

Given data:

Mass of the block, m=0.50kg

Mass of the wedge, m'=2.0kg

Coefficient of static friction, μs=0.40

Angle of wedge, θ=35°

Formula Used:

From previous part:

  Fmin=mtg(sinθμscosθ)cosθ+μssinθ

  Fmax=mtg(sinθ+μscosθ)cosθμssinθ

Calculation:

Substitute the values and solve for minimum force:

  Fmin=mtg(sinθμscosθ)cosθ+μssinθ

  Fmin=|(2.0+0.5)(9.8)(sin35°0.40cos35°)cos35°+0.40sin35°|Fmin=5.8N

For maximum force, reverse the direction of f

  Fmax=mtg(sinθ+μscosθ)cosθμssinθ

Substitute the values and solve:

  Fmax=|(2.0+0.5)(9.8)(sin35°+0.40cos35°)cos35°0.40sin35°|Fmax=37.5N

Conclusion:

The maximum and minimum values of applied force for which the block does not slip are 37.5 N and 5.8 N respectively.

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Chapter 5 Solutions

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

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