COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 2, Problem 118P

(a)

To determine

The magnitude of force Fc.

(a)

Expert Solution
Check Mark

Answer to Problem 118P

The magnitude of force Fc is 98 N.

Explanation of Solution

Draw the free body diagram.

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 2, Problem 118P

Write the equilibrium condition in x-direction.

    Fmcosθ=Fccosϕ

Here, θ is the angle made by Fm with horizontal.

Rewrite the expression for Fc.

  Fc=cosθcosϕFm        (I)

Write the equilibrium condition in y-direction.

    Fcsinϕ=Fmsinθ+W

Here, W is the weight of head.

Rewrite the expression for Fc.

  Fc=Fmsinθ+Wsinϕ        (II)

Rewrite the above equation by substituting equation (I).

    cosθcosϕFm=Fmsinθ+WsinϕsinϕcosθcosϕFm=Fmsinθ+Wtanϕ=Fmsinθ+Wtanϕ=Fmsinθ+WFmcosθ

Simplify and rewrite the equation for ϕ.

      tanϕ=tanθ+WFmcosθϕ=tan1(tanθ+WFmcosθ)        (III)

Conclusion:

Substitute 50 N for W, 35° for θ and 60 N for Fm in the above equation to find ϕ.

    ϕ=tan1(tan35°+50N(60N)(cos35°))=tan1(0.70+1.01)=tan1(1.71)=59.7°

Substitute 50 N for W, 35° for θ, 59.7° for ϕ and 60 N for Fm in the above equation to find ϕ.

    Fc=(60N)(sin35°)+50 Nsin59.7°=84.4N0.86=98N

Thus, the magnitude of force Fc is 98 N.

(a)

To determine

The direction of force Fc.

(a)

Expert Solution
Check Mark

Answer to Problem 118P

Fc is directed at 59.7° above the horizontal.

Explanation of Solution

Substitute 50 N for W, 35° for θ and 60 N for Fm in equation (III) to find ϕ.

    ϕ=tan1(tan35°+50N(60N)(cos35°))=tan1(0.70+1.01)=tan1(1.71)=59.7°

Conclusion:

Thus, Fc is directed at 59.7° above the horizontal.

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Chapter 2 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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