Question

A block of mass $15\text{kg}$ is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of $50\text{N}$. The coefficient of friction between the surfaces of contact is $(g=10{\text{m/s}}^2)$

• A. $3/4$
• B. $1/4$
• C. $1/2$
• D. $2/3$

Answer 1

The correct option is C $1/2$

FBD of system
In horizontal direction, $\Sigma F_x=0$$mg\sin \theta =f+T\cos \theta$
$f=\mu N=mg\sin \theta -T\cos \theta$ ......$\left(1\right)$

In vertical direction, $\Sigma F_y=0$
$N=mg\cos \theta +T\sin \theta$ ......$\left(2\right)$

Divide Eq. $\left(1\right)$ by Eq. $\left(2\right)$ to get,

$\frac{\mu N}{N}=\frac{mg\sin \theta -T\cos \theta }{mg\cos \theta +T\sin \theta }$

$\mu =\frac{150\sin {45}^{\circ }-50\cos {45}^{\circ }}{150\cos {45}^{\circ }+50\sin {45}^{\circ }}$

$\mu =\frac{1}{2}$

Hence, option $\left(A\right)$ is correct