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<div class=\"question_summary user_post\">A 3.0-m-long board of weight 200 N has

ID: 2022926 • Letter: #

Question

<div class="question_summary user_post">A 3.0-m-long board of weight 200 N has a mass weighing 500 N sitting on it, as in the diagram. A fulcrum is 30 cm from the right end, and the mass's center of mass is 1.00 m from the fulcrum. A force F is exerted upward, 30 cm from the left end of the board. The entire system is static: it is not moving. Assume that the board is uniform: its CM is at its center.Phys Find the value of the force F 1 (Don't forget to include the torque due to the weight of the board.)</div>

Explanation / Answer

Length of the board L = 3 m

Weight of the board W = 200 N

Weight of the mass W ' = 500 N

Apply moment of forces law with respect to fulcrum, you get

(500 N x 1 m) +(200 N x (1.5 m-0.3 m) = F x 2.4 m

500 Nm +200N(1.2m) = (2.4m) F

500 Nm + 240 Nm = (2.4m) F

(2.4 m) F = 740 Nm

            F = 308.33 N

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