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andrezito [222]
3 years ago
12

A rod is pivoted about its center and oriented horizontally. A 5.0-N force directed upward is applied 4.0 m to the left of the p

ivot and another upward 5.0-N force is applied 1.5 m to the right of the pivot. What is magnitude of the total torque about the pivot?
Physics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

The total torque is 27.5 Nm

Explanation:

Given;

5.0-N force directed upward is applied 4.0 m to the left of the pivot,

5.0-N force directed upward is applied 1.5 m to the right of the pivot,

Taking the moment about the pivot, the total torque is given by;

τ = Fr

where;

F is the appllied force

r is the radius of the force arm

τ = (5 N x 4 m) + (5 N x 1.5 m)

τ = 27.5 Nm

Therefore, the total torque is 27.5 Nm

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Explanation:

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So, vertical component is given by :

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a bag of sugar is found to have the same weight on planet earth as an identical bag of dry sawdust on planet Jupiter explain why
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Cheers

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3 years ago
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