Question

In the figure shown, find the tension $‘T^{\prime }$ in the string to prevent the body from sliding down the plane. Assume the string to be horizontal.

• A. $2.42\text{N}$
• B. $3.5\text{N}$
• C. $4.2\text{N}$
• D. $5.6\text{N}$

Answer 1

The correct option is A $2.42\text{N}$

As the body tends to slide down the plane, friction will act upwards.
Perpendicular to the plane:
$N=mg\cos {30}^{\circ }+T\sin {30}^{\circ }--\left(1\right)$
Along the plane:
$mg\sin {30}^{\circ }=T\cos {30}^{\circ }+\mu N$
$\Rightarrow \frac{mg}{2}=\frac{T\sqrt{3}}{2}+\mu [\frac{mg\sqrt{3}}{2}+\frac{T}{2}]$
(substituting $N$ from eqn. (1))
$\frac{mg}{2}=\frac{T}{2}[\sqrt{3}+\frac{1}{2}]+\frac{mg\sqrt{3}}{4}$
(substituting $\mu =\frac{1}{2}$)
$\Rightarrow T=\frac{mg[2-\sqrt{3}]}{(2\sqrt{3}+1)}$
$=\frac{40(2-\sqrt{3})}{(2\sqrt{3}+1)}$
$=2.42\text{N}$