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 $\alpha > \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 $a$ ?
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 $\alpha > \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 $a$ ?
Consider diagram shown in figure,
$(a)$ For box just starts to sliding down slope,
$f=m g \sin \theta$
$\mathrm{N}=m g \cos \theta$
$\tan \theta=\frac{f}{\mathrm{~N}}=\frac{m g \sin \theta}{m g \cos \theta}$
$\frac{f}{\mathrm{~N}}=\tan \theta$
$\frac{f}{\mathrm{~N}}=\mu$
$\mu=\tan \theta$
$\theta=\tan ^{-1}(\mu)$
$(b)$ When angle of inclination increased to $\alpha>\theta$. Resultant force be $F_{1}$
$\mathrm{F}_{1} =m g \sin \alpha-f$
f $=\mu \mathrm{N}$
$=\mu m g \cos \alpha$
$=m g \sin \alpha-\mu m g \cos \theta$
$=m g(\sin \alpha-\mu \cos \alpha)$
$(c)$ To keep box stationary or moving with uniform speed upward force ir required. Here friction force would be in downward direction.
$\mathrm{F}_{2} =m g \sin \alpha+f$
$=m g \sin \alpha+\mu \cos \alpha$
$=m g(\sin \alpha+\mu \cos \alpha)$
$(d)$ When box is to be moved with acceleration a upward along the plane net force be $F_{3}$ Friction would be in downward direction.
$\mathrm{F}_{3}=m \mathrm{~g}(\sin \alpha+\mu \cos \alpha)+m a$
Similar Questions
When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is such that it acts
When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is such that it acts
- [IIT 1990]
A block of mass $m$ (initially at rest) is sliding up (in vertical direction) against a rough vertical wall with the help of a force $F$ whose magnitude is constant but direction is changing. $\theta = {\theta _0}t$ where $t$ is time in sec. At $t$ = $0$ , the force is in vertical upward direction and then as time passes its direction is getting along normal, i.e., $\theta = \frac{\pi }{2}$ .The value of $F$ so that the block comes to rest when $\theta = \frac{\pi }{2}$ , is
A block of mass $m$ (initially at rest) is sliding up (in vertical direction) against a rough vertical wall with the help of a force $F$ whose magnitude is constant but direction is changing. $\theta = {\theta _0}t$ where $t$ is time in sec. At $t$ = $0$ , the force is in vertical upward direction and then as time passes its direction is getting along normal, i.e., $\theta = \frac{\pi }{2}$ .The value of $F$ so that the block comes to rest when $\theta = \frac{\pi }{2}$ , is
In the given figure the acceleration of $M$ is $(g = 10 \,ms^{-2})$
In the given figure the acceleration of $M$ is $(g = 10 \,ms^{-2})$
A block of mass $m$ is stationary on a rough plane of mass $M$ inclined at an angle $\theta$ to the horizontal, while the whole set up is accelerating upwards at an acceleration $\alpha$. If the coefficient of friction between the block and the plane is $\mu$, then the force that the plane exerts on the block is
A block of mass $m$ is stationary on a rough plane of mass $M$ inclined at an angle $\theta$ to the horizontal, while the whole set up is accelerating upwards at an acceleration $\alpha$. If the coefficient of friction between the block and the plane is $\mu$, then the force that the plane exerts on the block is
- [KVPY 2009]
The maximum static frictional force is
The maximum static frictional force is