Question

A block of mass $15kg$ 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 $50N$. The coefficient of friction between the surfaces of contact is $\left(g=10m/s^2\right)$

• A. 1/2
• B. 2/3
• C. 3/4
• D. 1/4

Answer 1

Answer: (A)

$T=50N$
Component of force (in y direction)
$N=\frac{T}{\sqrt{2}}+\frac{150}{\sqrt{2}}$
$\Rightarrow N=200/\sqrt{2}$
$f=\mu \frac{200}{\sqrt{2}}$
Component of force in $x$ direction
$\frac{150}{\sqrt{2}}=\frac{T}{\sqrt{2}}+f_r$
$\frac{150}{\sqrt{2}}-\frac{50}{\sqrt{2}}=\frac{\mu \times 200}{\sqrt{2}}$
$\Rightarrow \mu =\frac{1}{2}$