Question

A force $F$ is applied to a block of mass $2\sqrt{3}\text{kg}$ as shown in the diagram. What should be the maximum value of force so that the block does not move?

• A. $10\text{N}$
• B. $20\text{N}$
• C. $30\text{N}$
• D. $40\text{N}$

Answer 1

The correct option is D $40\text{N}$

$F$ force will be having 2 components: Horizontal component will be $F\cos {60}^0$ and the veritical component will be $F\sin {60}^0$

Balancing vertical forces, we have
$\begin{array}{}N=F\sin {60}^{\circ }+mg& \text{(1)}\end{array}$
Balancing Horizontal forces
For the block not to move, we must have
$F\cos {60}^{\circ }=\mu N$
taking value of $N$ from equation $\text{1}$

i.e., $F\cos {60}^{\circ }=\mu (F\sin {60}^{\circ }+mg)$
$F\left(\frac{1}{2}\right)=\frac{1}{2\sqrt{3}}[F\left(\frac{\sqrt{3}}{2}\right)+(2\sqrt{3}\left)\right(10\left)\right]\text{N}$
or $F(\frac{1}{2}-\frac{1}{4})=10\text{N}$
or $F=4\times 10=40\text{N}$