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2 years ago

12

A rod 7.0 m long is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, and a downward fo

rce of 200 N acts at the right end. At what distance to the right of the pivot can a third force of 300 N acting upward be placed to produce rotational equilibrium?

1 answer:

kicyunya [14]2 years ago

6 0

If the rod is in rotational equilibrium, then the net torques acting on it is zero:

∑ τ = 0

Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:

• at the left end,

τ = + (50 N) (2.0 m) = 100 N•m

• at the right end,

τ = - (200 N) (5.0 m) = - 1000 N•m

• at a point a distance d to the right of the pivot point,

τ = + (300 N) d

Then

∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0

⇒ (300 N) d = 1100 N•m

⇒ d ≈ 3.7 m

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Explanation:

For this case we have the following system with the forces on the figure attached.

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Part b

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If we multiply all the equation by 2 we got:

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Now we can replace the values and we got:

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This might surely help you ☺️❤️

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