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Margaret
25 days ago
14

A uniform bridge span weighs 50.0 x 10^3 N and is 40.0 m long. An automobile weighing 15.0 x 10^3 N is parked with its center of

gravity located 12.0 m from the right pier. What upward support force does the left pier provide?
Physics
1 answer:
Sav [3.1K]25 days ago
3 0

Answer:

(Left Force absolute value) Fl=29500N

Explanation:

When considering the bridge in a static state, it is essential to acknowledge its equilibrium. At this point, torques can be utilized to solve for the answer, as illustrated below:

-Fl*(40m)-Wa*(12m)-Wb(20m)+Fr*(0m)=0

If assessing the torque around the right pier, the distance Fl measures 40 m, the weight of the car (Wa) is located 12 m from it, and the bridge's center of mass (Wb) is 20 m from that point, while the Right Force pier sits at 0m, where it effectively has no contribution to rotation.

The signs signify the direction of torque, whether positive or negative, based on the chosen reference point (in this scenario, the torque point is set at the right pier).

Ultimately, substituting in the values provides the Left Force result:

-Wa*(12m)-Wp*(20m)-Fl*(40m)+Fr*(0m)=0-Wa*(12m)-Wp*(20m)=Fl*(40m)\frac{-Wa*(12m)-Wp*(20m)}{40m}=Fl \frac{-15000N*12m-50000N*20m}{40m}=Fl-29500N=Fl

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Answer:

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2 months ago
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Answer:

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