Question

A block of mass 15 kg is resting on a rough inclined plane as shown in Fig. The block is tied by a horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contact is

• A. 1/2
• B. 2/3
• C. 3/4
• D. 1/4

Answer 1

The correct option is A 1/2

In horizontal direction,

$\sum F_x=0$ given

$mg\sin \theta =f+T\cos \theta$ (f = frictional force)

$\Rightarrow f=\mu N=mg\sin \theta -T\cos \theta$.........(1)

and $\sum F_y=0$ given: -

$mg\cos \theta +T\sin \theta =N$......(2)

Divide (1) by (2) to get, Put values :-

$\mu =\frac{mg\sin \theta -T\cos \theta }{mg\cos \theta +T\sin \theta }=\frac{150\sin \left({45}^0\right)-50\cos \left({45}^0\right)}{150\cos \left({45}^0\right)+50\sin \left({45}^0\right)}=0.5$