A body is in limiting equilibrium on a rough plane inclined at $30^{o}$ with the horizontal. When the inclination is increased to $45^{o}$ , the body slides down with acceleration. Find this acceleration. (Take $g = 9.8 m s^{- 2}$ )

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Solution

Now
$\begin{matrix} m g sin (15 + 30) = m a g sin 45^{\circ} = a a = g sin 45^{\circ} a = g \times \frac{1}{\sqrt{2}} a = \frac{g}{\sqrt{2}} m / s^{2} \end{matrix}$

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A body is in limiting equilibrium on a rough plane inclined at 30o with the horizontal. When the inclination is increased to 45o, the body slides down with acceleration. Find this acceleration. (Take g=9.8ms−2)

Now mgsin(15+30)=magsin45∘=aa=gsin45∘a=g×1√2a=g√2m/s2

A body is in limiting equilibrium on a rough plane inclined at $30^{o}$ with the horizontal. When the inclination is increased to $45^{o}$ , the body slides down with acceleration. Find this acceleration. (Take $g = 9.8 m s^{- 2}$ )

Now
$\begin{matrix} m g sin (15 + 30) = m a g sin 45^{\circ} = a a = g sin 45^{\circ} a = g \times \frac{1}{\sqrt{2}} a = \frac{g}{\sqrt{2}} m / s^{2} \end{matrix}$