Consider the following massless board. The board is resting on a support (fulcru

Question

Consider the following massless board. The board is resting on a support (fulcrum). A force of 360. N (F1) is applied downward a distance d1=3.88 m to the right of the support. board. You want to apply a force of 221 N downward on the left side of the board to keep the board in equilibrium (in this case think of equilibrium as balanced). The torque from the 360.-N force is 1.40×103 N*m in the clockwise direction (CW), so the torque from the 221-N force (F2) would need to be 1.40×103 N*m in the counterclockwise direction (CCW).1. How far away (distance d2) from the support should the left force be applied? 2. 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

1)torque on clockwise direction = torque on anticlockwise direction

1.4 * 103  = 221 * d2

=>d2 = 1.4 * 103 /221 = 6.33 m

2) upward force = downward force

=> upward force = 221 + 360 = 581 N

ans - force of fulcrum = 581 N upwards

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