Question

A block of mass $10\text{kg}$ is kept on a rough inclined plane of inclination ${45}^{\circ }$. A force of $3\text{N}$ is applied on the block down the incline. The coefficient of static friction between the plane and the block is $0.6$. What should be the minimum value of force $F$, such that the block does not move downward? (Take $g=10{\text{ms}}^{-2}$)

• A. $32\text{N}$
• B. $25\text{N}$
• C. $23\text{N}$
• D. $18\text{N}$

Answer 1

The correct option is A $32\text{N}$

Free body diagram, for the given figure is as follows,


Let the $x$-axis be along the downward incline direction.
Since the block is just about to move downwards, limiting friction ($f_l$) acts up the incline. Since minimum value of $F$ is asked, it is up the incline and also parallel to the incline.
For the block to be in equilibrium i.e., so that it does not move downward, then $\sum f_x=0$
$\therefore 3+Mg\sin \theta -F-f_l=0$
or $3+Mg\sin \theta =F+f_l$
As, frictional force, $f_l=\mu R$
$\therefore 3+Mg\sin \theta =F+\mu R$ ... (i)
Similarly, $\sum f_y=0$
$-Mg\cos \theta +R=0$
or $Mg\cos \theta =R$ ... (ii)
Substituting the value of ${}^{\prime }{R}^{\prime }$ from Eq. (ii) to Eq. (i), we get
$3+Mg\sin \theta =F+\mu \left(Mg\cos \theta \right)$ ... (iii)
Here, $M=10\text{kg},\theta ={45}^{\circ },g=10{\text{m/s}}^2$
and $\mu =0.6$
Substituting these values in Eq. (iii), we get
$3+(10\times 10\sin {45}^{\circ })-(0.6\times 10\times 10\cos {45}^{\circ })=F$
$\Rightarrow F=3+\frac{100}{\sqrt{2}}-\frac{60}{\sqrt{2}}$
$=3+\frac{40}{\sqrt{2}}$
$=3+20\sqrt{2}=31.8\text{N}$ or $F\simeq 32\text{N}$