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A body of mass 10 kg is placed on an inclined surface of angle `30^(@)` . If coefficient of limiting friction is `1 //sqrt3` , find the inclined plane. Force is being exerted parallel to the inclined plane . |
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Answer» Here , ` m = 10 kg , theta = 30^(@) , mu = (1)/sqrt3 ` As is clear from force required just to push the body up the inclined plane is ` F = mg sin theta + f ` ` = mg sin theta + mu R ` ` = mg sin theta + mu mg cos theta ` `= mg (sin theta + mu cos theta) ` ` = 10 xx 9.8 (sin 30^(@) + (1)/sqrt(3 ) cos 30^(@)) ` ` F = 98 (0.5 + (1)/(sqrt3) xx sqrt(3)/(2)) = 98 N ` . |
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