EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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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:

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, 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

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 55PCh. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - Prob. 58PCh. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - Prob. 61PCh. 5 - Prob. 62PCh. 5 - Prob. 63PCh. 5 - Prob. 65PCh. 5 - Prob. 67PCh. 5 - Prob. 68PCh. 5 - Prob. 69PCh. 5 - Prob. 70PCh. 5 - Prob. 71PCh. 5 - Prob. 72PCh. 5 - Prob. 73PCh. 5 - Prob. 74PCh. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 80PCh. 5 - Prob. 82PCh. 5 - Prob. 83PCh. 5 - Prob. 84PCh. 5 - Prob. 85PCh. 5 - Prob. 86PCh. 5 - Prob. 87PCh. 5 - Prob. 88PCh. 5 - Prob. 89PCh. 5 - Prob. 90PCh. 5 - Prob. 91PCh. 5 - Prob. 92PCh. 5 - Prob. 93PCh. 5 - Prob. 94PCh. 5 - Prob. 95PCh. 5 - Prob. 96PCh. 5 - Prob. 97PCh. 5 - Prob. 101PCh. 5 - Prob. 102PCh. 5 - Prob. 103PCh. 5 - Prob. 104PCh. 5 - Prob. 105PCh. 5 - Prob. 106PCh. 5 - Prob. 107PCh. 5 - Prob. 108PCh. 5 - Prob. 109PCh. 5 - Prob. 110PCh. 5 - Prob. 111PCh. 5 - Prob. 112PCh. 5 - Prob. 113PCh. 5 - Prob. 114PCh. 5 - Prob. 115PCh. 5 - Prob. 116PCh. 5 - Prob. 117PCh. 5 - Prob. 118PCh. 5 - Prob. 119PCh. 5 - Prob. 120PCh. 5 - Prob. 121PCh. 5 - Prob. 122PCh. 5 - Prob. 123PCh. 5 - Prob. 124PCh. 5 - Prob. 125PCh. 5 - Prob. 126PCh. 5 - Prob. 127PCh. 5 - Prob. 128PCh. 5 - Prob. 129PCh. 5 - Prob. 130PCh. 5 - Prob. 131PCh. 5 - Prob. 132PCh. 5 - Prob. 133PCh. 5 - Prob. 134PCh. 5 - Prob. 135PCh. 5 - Prob. 136PCh. 5 - Prob. 137PCh. 5 - Prob. 138PCh. 5 - Prob. 139P
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