A block of mass 15 kg is resting on a rough inclined plane as shown in figure. 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 ($g = 10\ ms^{- 2}$) [image:5_image132.emf]

Question

Dr. Abubacker · Accepted Answer

(a) Figure shows free body diagram of the block
[image:5_image286.emf]
For equilibrium, along the place
$\mu R + T\cos{45{^\circ} = mg\sin 45{^\circ}}$
$\mu R + \frac{T}{\sqrt{2}} = \frac{mg}{\sqrt{2}}\ \ \ \ \ \ \ \ $…(i)
For equilibrium, in direction perpendicular to inclined plane,
$R = T\sin 45{^\circ} = mg\cos{45{^\circ}}$
$= \frac{T}{\sqrt{2}} + \frac{mg}{\sqrt{2}}$
Put in Eq. (i), $\frac{\mu}{\sqrt{2}}(T + mg) = - \frac{1}{\sqrt{2}}(mg - T)$
$\mu$(50+15$	imes 10$) = (15$	imes 10 - 50$)
$\mu = \frac{100}{200} = \frac{1}{2}$

Q. A block of mass 15 kg is resting on a rough inclined plane as shown in figure. 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 ($g = 10\ ms^{- 2}$)

Explanation

Reviewed by Dr. Abubacker

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