15.8) A uniform horizontal lever of length 4.8 m pivoted at its mid-point is in

Question

15.8) A uniform horizontal lever of length 4.8 m pivoted at its mid-point is in equilibrium. The following forces are acting on the lever: an unknown upward force (force with upward vertical component) that makes an angle of 40 deg with the horizontal-left acting at the right end of the lever, a 5 N vertically upward force acting at the left end of thelever, and an and a 3 N vertically downward force acting at a point on the lever 0.4 m away from the right end. Calculate the magnitude of the unknown force.

The correct answer is 11.668 N, but I need to know how to get there (formulas used, each step taken). Thank you!

Explanation / Answer

The second condition of equilibrium is that the sum of the torques (moments) must be zero, or the object is accelerating rotationally, so

T = 0

Then we have the torque for each force

Torque = r x F

- For the unknow force:

F1*2.4*sin40

- For the force at the left end

5*2.4

- For the force 0.4 m away from the right end

3*2

So 0 = F*2.4*sin40 + 2.4*5 + 3*2

F = 11.6679 N

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