College Physics
College Physics
OER 2016 Edition
ISBN: 9781947172173
Author: OpenStax
Publisher: OpenStax College
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Chapter 9, Problem 13PE
To determine

(a)

The force on the cables.

Expert Solution
Check Mark

Answer to Problem 13PE

The force in the cables is 13213.5Nm.

Explanation of Solution

Given:

The mass of the bridge is M=2500kg.

The mass of the car is m=900kg.

The distance of weight of the bridge from point A is DA=1.5m.

The distance of weight of the car from point A is d=4.5m.

The distance of weight of the bridge from point B is DB=7.5m.

The length of the bridge is L=9.0m.

The angle of force from point B is β=40°.

Formula used:

The relation between the weight and the mass of the bridge is

  W=Mg  (1)

The relation between the weight and the mass of the car is

  w=mg  (2)

The relation between the torque and weight of bridge is

  τB=WDA  (3)

The relation between the torque sand weight of car is

  τC=wd  (4)

The relation between torque of cable is

  τ=LTsinθ  (5)

The relation between the torque of car, bridge and cable is

  τC+τB+τ=0  (6)

  College Physics, Chapter 9, Problem 13PE

Figure-(1)

Calculation:

Substituting the given values in equation (1), we get

  W=(2500kg)(9.8 m/s2)=24500N

Substituting the given values in equation (2), we get

  w=(900kg)(9.8 m/s2)=8820N

Substituting the given values in equation (3), we get

  τB=(24500N)(1.5m)=36750Nm

Substituting the given values in equation (4), we get

  τC=(8820N)(4.5m)=39690Nm

Substituting the given values in equation (5), we get

  τ=T(9.0m)sin40°=T(5.785m)

Substituting the given values in equation (6), we get

  39690Nm36750Nm+T(5.785m)=0T=76440Nm5.785m=13213.5N

Conclusion:

Thus, the force in the cables is 13213.5Nm.

To determine

(b)

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

Expert Solution
Check Mark

Answer to Problem 13PE

The magnitude of force exerted by the hinges on the bridge is 26811.7N and the direction is 68° above the horizontal axis towards the other side of the bridge.

Explanation of Solution

Formula used:

The relation between force in the horizontal direction and torque is

  Fx=Tcosθ  (7)

The relation between force in the vertical direction and torque is

  Fy=W+wTsinθ  (8)

The relation between the force and vertical and horizontal forces is

  F=Fx2+Fy2  (9)

The relation between the angle made by force and vertical and horizontal forces is

  β=tan1(FyFx)  (10)

Calculation:

Substituting the given values in equation (7), we get

  Fx=(13213.5N)cos40°=10122N

Substituting the given values in equation (8), we get

  Fy=24500N+8820N(13213.5N)sin40°=33320N8493.5N=24827.5N

Substituting the given values in equation (9), we get

  F= ( 10122N )2+ ( 24827.5N )226811.7N

Substituting the given values in equation (10), we get

  β=tan1( 24827.5N 10122N)68°

Conclusion:

Thus, the magnitude of force exerted by the hinges on the bridge is 26811.7N and the direction is 68° above the horizontal axis towards the other side of the bridge.

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Chapter 9 Solutions

College Physics

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