Question

A body of mass $1\text{kg}$ lies on a smooth inclined plane. A horizontal force $F=10\text{N}$ is applied on the block of mass $1\text{kg}$ as shown in figure. If the block remains at rest, then the magnitude of normal reaction on the block is
(Take g $=10{\text{m/s}}^2$)

• A. $10\sqrt{2}\text{N}$
• B. $\frac{10}{\sqrt{2}}\text{N}$
• C. $10\text{N}$
• D. None of these

Answer 1

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

F.B.D of the block,


Since $F$ is along horizontal and $mg$ is along vertical direction and both have equal magnitude.

Since, block is in rest so resultant of $F$ and $mg$ must balance the normal reaction

Therefore,

$N=\sqrt{F^2+{\left(mg\right)}^2}$

$\Rightarrow N=\sqrt{{\left(10\right)}^2+{\left(10\right)}^2}$

$\therefore N=10\sqrt{2}\text{N}$

Hence, option $\left(a\right)$ is correct.

Alternative:
We can also apply Lami's theorem.

$\frac{N}{\sin {90}^{\circ }}=\frac{F}{\sin ({90}^{\circ }+{45}^{\circ })}=\frac{mg}{\sin ({90}^{\circ }+{45}^{\circ })}$

$N=F\times \frac{\sin {90}^{\circ }}{\cos {45}^{\circ }}=10\times \sqrt{2}=10\sqrt{2}\text{N}$