A body is in limiting equilibrium on a rough inclined plane at angle $30^{\circ}$ with horizontal. Calculate the acceleration with which the body will slide down when the inclination of the plane is changed to $60^{\circ}$ . (Take $g = 10 m / s^{2}$ )

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Solution

Since body is in limiting equilibrium,
$m g sin 30 = μ s m g cos 30$
$μ s = tan 30 ⟹ \frac{1}{\sqrt{3}}$
Acceleration of body= $\frac{F_{n e t}}{m} = \frac{(m g sin θ - μ s m g cos θ)}{m}$
$= g (sin θ - μ s cos θ)$
$= g (sin 60 - \frac{1}{\sqrt{3}} cos 60) ⟹ g (\frac{\sqrt{3}}{2} - \frac{1}{2 \sqrt{3}})$
$g = \frac{3 - 1}{2 \sqrt{3}} ⟹ \frac{g}{\sqrt{3}} ⟹ \frac{10}{\sqrt{3}}$

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A body is in limiting equilibrium on a rough inclined plane at angle 30∘ with horizontal. Calculate the acceleration with which the body will slide down when the inclination of the plane is changed to 60∘. (Take g=10m/s2)

Since body is in limiting equilibrium,mgsin30=μsmgcos30μs=tan30⟹1√3Acceleration of body=Fnetm=(mgsinθ−μsmgcosθ)m=g(sinθ−μscosθ)=g(sin60−1√3cos60)⟹g(√32−12√3)g=3−12√3⟹g√3⟹10√3

A body is in limiting equilibrium on a rough inclined plane at angle $30^{\circ}$ with horizontal. Calculate the acceleration with which the body will slide down when the inclination of the plane is changed to $60^{\circ}$ . (Take $g = 10 m / s^{2}$ )