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) at what angle of inclination θ 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 > θ 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.

Home » 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) at what angle of inclination θ 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 > θ 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.

CBSE | Class 11 | Laws of Motion | ncert exemplar | Physics

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) at what angle of inclination θ 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 > θ 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.

a) As the box starts to slide down the plane, $\mu=\tan \theta$
$\theta=\tan ^{-1}(\mu)$
b) If $a>\theta$ , the angle of inclination will be the angle of repose and the net force acting will be in the downward
$\begin{array}{l} \mathrm{F} 1=\mathrm{mg} \sin \mathrm{a}-\mathrm{f}=\mathrm{mg} \sin \mathrm{a}-\mu \mathrm{N} \\ =\mathrm{mg} \sin \mathrm{a}-\mu \mathrm{mg} \cos \mathrm{a} \\ \mathrm{F} 1=\mathrm{mg}(\sin \mathrm{a}-\mu \cos \mathrm{a}) \end{array}$
c) F2 is the force which is required to keep the box stationary and to move with uniform velocity
$\mathrm{F} 2-\mathrm{mg} \sin \mathrm{a}-\mathrm{f}=\mathrm{ma}$
Or
$\begin{array}{l} \text { F2 }-m g \sin a-\mu N=0 \\ F 2=m g \sin a-\mu N=0 \\ F 2=m g(\sin a-\mu \cos a) \end{array}$
d) $\mathrm{F} 3$ is the force that is required to move the box upward with a as acceleration
$F 3-m g \sin a-\mu m g \cos a=m a$
$\mathrm{F} 3=\mathrm{mg}(\sin \mathrm{a}+\mu \cos \mathrm{a})+\mathrm{ma}$