Question

A block of mass 15 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 50N. Calculate the coefficient of friction between the block and inclined plane.

Answer 1

As $N=\frac{50}{\sqrt{2}}+\frac{15}{\sqrt{2}}\Rightarrow \frac{50}{\sqrt{2}}+\frac{150}{\sqrt{2}}=\frac{200}{\sqrt{2}}$

thus, from the forula of friction ,

$F_{max}=\mu N=\mu \times \frac{200}{\sqrt{2}}$

Under equillibrium along incline

$\frac{150}{\sqrt{2}}=\frac{50}{\sqrt{2}}+\mu \frac{200}{\sqrt{2}}$

$100=\mu 200\Rightarrow \mu =\frac{1}{2}=0.5$