COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781711470832
Author: OpenStax
Publisher: XANEDU
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Chapter 9, Problem 12PE
To determine

(a)

The fraction of weight supported by the opposite shore.

Expert Solution
Check Mark

Answer to Problem 12PE

The fraction of weight supported by the opposite shore is 0.1667.

Explanation of Solution

Given:

The mass of the bridge is, m=2500kg.

Formula used:

The relation between the mass and weight is,

  w=mg

The expression for condition of equilibrium is,

  τ=0w(1.5m)Tsin40ο(9.0m)=0

The relation between the fraction of weight and tension is,

  wF=Tsin40οw

The free body diagram for the bridge is shown below.

  COLLEGE PHYSICS, Chapter 9, Problem 12PE

  Figure-(1)

Calculation:

The weight of the bridge is calculated as,

  w=mg=(2500kg)(9.8m/ s 2)=24525N

The tension is calculated as,

  w(1.5m)Tsin40°(9.0m)=0(24525N)(1.5m)Tsin40°(9.0m)=0Tsin40°(9.0m)=36787.5Nm

Solve further:

  T=36787.5N-msin 40ο( 9.0m)T=6359N

The fraction of weight is calculated as,

  wF=Tsin40°w=( 6359N)sin40°24525N=4087N24525N=0.1667

Conclusion:

The fraction of weight supported by the opposite shore is 0.1667.

To determine

(b)

The direction and magnitude of the force exerted on the bridge.

Expert Solution
Check Mark

Answer to Problem 12PE

The magnitude of the force is 21009N and the direction of force is 76.59° anti-clockwise.

Explanation of Solution

Given:

The mass of the bridge is m=2500kg

Formula used:

The expression for condition of equilibrium is,

  Fsinθ+Tsin40°=wFsinθ=wTsin40°

The expression for force along horizontal is,

  Fcosθ=Tcos40ο

Calculation:

The vertical component of the force is calculated as,

  Fsinθ=wTsin40°=24525N(6359)sin40°=20437.5N

The horizontal component of the force is calculated as,

  Fcosθ=Tcos40°=(6359N)cos40°=4871N

The magnitude of the force is given as,

  F= ( Fcosθ )2+ ( Fsinθ )2= ( 4871N )2+ ( 20437.5N )2=21009N

The direction of the force is given as,

  tanθ=FsinθFcosθ=20437.5N4871Ntanθ=4.195θ=76.59°

Conclusion:

The magnitude of the force is 21009N and the direction of force is 76.59° anti-clockwise.

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Chapter 9 Solutions

COLLEGE PHYSICS

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