1.

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 .

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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