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A body is sliding down an inclined plane have coefficient of friction 0.5. If he normal reaction is twice that of resultant downward force along the incline. Find the angle between the inclined plane and the horizontal. |
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Answer» Solution :`MU = 0.5` `N=2mg(sin THETA-mu cos theta)` `mg cos theta = 2mg (sin theta - mu cos theta)` `cos theta = 2 cos theta(tan theta - mu)` `(1)/(2)=tan theta-(1)/(2)` `rArr tan theta = 1 rArr = 45^(@)` |
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