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Grade: 12
        
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

Answers : (2)

Khimraj
3008 Points
							
 force required just to move a body up an inclined plane
F1 = mg(sin\Theta + \mucos\Theta)
force required  just to prevent the body sliding down
F2 = mg(sin\Theta - \mucos\Theta)
F1 = 2F2
 mg(sin\Theta + \mucos\Theta) = 2*mg(sin\Theta - \mucos\Theta)
3\mucos\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\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\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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