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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
7 months ago

Khimraj
3008 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..............

7 months ago
Rajdeep
230 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!

7 months 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