Question

In the figure shown, what minimum force $‘F^{\prime }$ should be applied perpendicular to the top face of the block for it to not move? Given the coefficient of static friction between the plane and the block $=0.5$. Take $g=10{\text{m/s}}^2$

• A. $10\sqrt{2}\text{N}$
• B. $20\sqrt{2}\text{N}$
• C. $\frac{5}{\sqrt{2}}\text{N}$
• D. $\frac{10}{\sqrt{2}}\text{N}$

Answer 1

The correct option is A $10\sqrt{2}\text{N}$


From the FBD of the block,
Perpendicular to the plane: $N=(F+10\sqrt{2})\text{N}$
Along the plane: $f=10\sqrt{2}$
Since friction force,
$f=\mu N$
$10\sqrt{2}=\mu (F+10\sqrt{2})$ $\Rightarrow 20\sqrt{2}=F+10\sqrt{2}$
$\Rightarrow F=10\sqrt{2}\text{N}$