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        the force required just to move a body up an inclined plane is double the force required  just to prevent the body sliding down .if the coefficientof friction is 0.25, the angle of inclination of the plane
one year ago

Khimraj
3007 Points
							 force required just to move a body up an inclined planeF1 = mg(sin$\Theta$ + $\mu$cos$\Theta$)force required  just to prevent the body sliding downF2 = mg(sin$\Theta$ - $\mu$cos$\Theta$)F1 = 2F2 mg(sin$\Theta$ + $\mu$cos$\Theta$) = 2*mg(sin$\Theta$ - $\mu$cos$\Theta$)3$\mu$cos$\Theta$ = sin$\Theta$ tan$\Theta$ = 3$\mu$ = 3*0.25 = 0.75 $\Theta$ = tan-1(0.75)Hope it clears..............

one year ago
Rajdeep
231 Points
							HELLO THERE! First, just draw the Free body diagram of the body and then write the equations of motion: When the body is to be pushed up, the force mgsin$\dpi{100} \theta$ is acting downwards (parallel to the incline), and since the motion is upwards, the frictional force is also acting in the same direction. So, net Force required to push it up:$F_{1} = mgsin\theta + friction = mgsin\theta + \mu mgcos\theta$ When the body is to be stopped from sliding, a force of mgsin$\dpi{100} \theta$ will be acting downwards, but the frictional force this time will be acting in an opposite direction, so we get net force: $F_{2} = mgsin\theta - friction = mgsin\theta - \mu mgcos\theta$ Given that,$F_{1} = 2F_{2}$ On putting the values, we get:$\theta = tan^{-1}0.75$ THANKS!

one year ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions