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  • A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is μ. Let the mass of the box be m.

A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is \mu. Let the mass of the box be m.

a) at what angle of inclination \theta of the plane to the horizontal will the box just start to slide down the plane?

b) what is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a > \theta

c) what is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?

d) what is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a.

Answers (1)

(a) By angle of repose\mu = \tan \theta , as the box start to slide down the plane.

Therefore,\theta = \tan^{-1} \mu

(b) the angle of inclination of the plane with horizontal will slide down if angle \alpha >\theta, as \theta is the angle of repose.

Thus, the net downward force will be,

F_{1} = mg \sin\alpha - f

    = mg \sin\alpha -\mu N

  = mg \sin\alpha -\mu mg \cos \alpha

    = mg [ \sin \alpha - \mu \cos \alpha]

(c) To move the box upward with uniform velocity or to keep it stationary,

F_{2} - mg \sin \alpha - f = ma

Or F_{2} - mg \sin \alpha - \mu N =0          ….. (since a=0)

F_{2} = mg \sin \alpha - \mu N =0

Therefore, F_{2} = mg (\sin\alpha - \mu \cos \alpha)

(d) Let F3 be the force applied to move the box upward with acceleration a.

F_{3} - mg \sin \alpha - \mu mg \cos \alpha = ma

Thus, F_{3} = mg( \sin \alpha - \mu \cos \alpha) + ma

Posted by

infoexpert24

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