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A 30 kg block is to be moved up an inclined plane at an anglue 30° to the horizontal with a velocity of 5ms^(-1). If the frictional force retarding the motion is 150N find the horizontal force required to move the block up the plane. (g=10ms^(-2)) |
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Answer» Solution :The FORCE required to a body up an inclined plane is: `F = mg sin theta +` FRICTIONAL force. `=30(10) sin 30^(@) + 150 = 300 N` If P is the HORIZONTAL force, `F = P cos theta` `P = F/(cos theta) = 300/(cos theta) = (300 xx 2)/sqrt(3) = 200sqrt(3) = 346 N` |
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