Question

A block of mass $10kg$ is held at rest against a rough vertical wall $[\mu =0.5]$ under the action a force $F$ as shown in figure. The minimum value of $F$ required for it is $(g=10m/s^2)$.

• A. $162.6N$
• B. $89.7N$
• C. $42.7N$
• D. $95.2N$

Answer 1

The correct option is A $162.6N$

For minimum value of $f$, forces acting on block is shown is shown in fig.

Normal, $N=F\sin \theta$

$f-$ frictional force

$f=\mu N=0.5\times F\sin {30}^{\circ }=\frac{0.5}{2}F$

Now, in vertical direction;

$F\cos {30}^{\circ }-f=mg$

$\frac{\sqrt{3}}{2}F-\frac{0.5}{2}F=10\times 10$

$\frac{\sqrt{3}-0.5}{2}F=100$

$F=\frac{100\times 2}{(\sqrt{3}-0.5)}$$=\frac{200}{(\sqrt{3}-0.5)}$$=\frac{200}{1.232}=162.3N$