Question

A block is moving on an inclined plane making an angle of ${45}^{\circ }$ with the horizontal and the co-efficient of friction between the block and the inclined plane is $\mu$. If the force required to just push it up the inclined plane is $3$ times the force required to just prevent it from sliding down, then the value of $\mu$ is
[Take($g=10m/s^2$]

• A. $\mu =0.3$
• B. $\mu =0.5$
• C. $\mu =0.75$
• D. $\mu =0.85$

Answer 1

The correct option is B $\mu =0.5$

Given,
Angle of inclination of plane $\theta ={45}^{\circ }$
Co-efficient of friction $=\mu$

Let us suppose, force required to just prevent sliding down of block is $F_o$, then force required to just push it up= $3F_o$. [given]

Case - 1 : Friction ($f$) will act upwards as block has tendency to slide down and to prevent sliding, we have to apply $F_o$ in upward direction along inclined plane.


On applying $\sum F=ma$ along the inclined plane.
$F_o+f-mg\sin \theta =ma$
[Block is about to slide down, $a=0$]
$\Rightarrow F_o+f=mg\sin \theta$ ..............(1)

Case - 2 : Friction ($f$) will act downward as block will have tendency to slide up when $3F_o$ force is applied upward.


On applying $\sum F=ma$ along the inclined plane.
$3F_o-mgsin\theta -f=ma$
[Block is about to slide up, $a=0$]
$\Rightarrow 3F_o=mgsin\theta +f$
$\Rightarrow F_o=\frac{f}{3}+\frac{mgsin\theta }{3}$ ..........(2)

On putting value of equation (2) in (1),
$\frac{f}{3}+\frac{mg\sin \theta }{3}+f=mg\sin \theta$
$\Rightarrow \frac{4f}{3}=\frac{2mg\sin \theta }{3}$
$\Rightarrow f=\frac{mg\sin \theta }{2}$
$\Rightarrow \mu mg\cos \theta =\frac{mg\sin \theta }{2}$
[$f=\mu N=\mu mg\cos \theta$, in limiting case]
$\Rightarrow \mu g=\frac{g\tan \theta }{2}$
$\Rightarrow \mu =\frac{\tan {45}^{\circ }}{2}$
[where $\theta ={45}^{\circ }$]
$\Rightarrow \mu =0.5$