Consider three masses $m_1, m_2$ and $m_3\left(m_1>m_2>m_3\right)$ are at rest on a horizontal plane as…

Consider three masses $m_1, m_2$ and $m_3\left(m_1>m_2>m_3\right)$ are at rest on a horizontal plane as shown in the figure. Now the angle of inclination $(\theta)$ of the plane is gradually increased until the masses just begin to slide. (Assume the coefficient of static friction between the masses and the surface is constant). Then the correct statement of the following is

1. $m_3$ begins to slide at a higher inclination angle than $m_1$ and $m_2$.
2. $m_3$ begins to slide at a lower inclination angle than $m_1$ and $m_2$.
3. $m_1, m_2$ and $m_3$ begins to slide at the same inclination angle.
4. $m_2$ begins to slide at a higher inclination angle than $m_1$ and $m_3$.

Solution

An object started to slide down an inclined plane when angle of incline (also called angle of repose) is just equals to angle of friction. Let $\theta=$ angle at which a mass started to slide down an incline.
Then, force of friction = downward force . $\begin{aligned} \Rightarrow & \mu N & =m g \sin \theta \\ \Rightarrow & \mu m g \cos \theta & =m g \sin \theta\end{aligned}$ $\begin{aligned} \Rightarrow & & \tan \theta & =\mu \\ \Rightarrow & & \theta & =\tan ^{-1} \mu\end{aligned}$ Hence, angle does not depend on mass of body. So all three masses slides down at same instant at same inclination angle $(\theta)$.

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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