A board is resting on a support (fulcrum). A force of 294 N (F1) is applied down

Question

A board is resting on a support (fulcrum). A force of 294 N (F1) is applied downward a distance d1=1.76 m to the right of the support. You want to apply a force of 174 N downward on the left side of the board to keep the board in equilibrium (in this case think of equilibrium as balanced). A.) The torque from the 294-N force is 517 N*m in the clockwise direction (CW), so the torque from the 174-N force (F2) would need to be 517 N*m in the counterclockwise direction (CCW). How far away (distance d2) from the support should the left force be applied? B.) If the two forces are pushing down, what is the magnitude and direction of the force which the fulcrum must exert to keep the board in static equilibrium (stationary).

Explanation / Answer

a) 174*x = 294*1.76

x=2.97 m

B) F - 294 - 174 = 0

F = 294+174=468 N

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