Question:
A block of mass \(m\) is lying on the edge having inclination angle \(\alpha = \tan^{-1}\!\!\left(\dfrac{1}{5}\right)\). Wedge is moving with a constant acceleration \(a = 2\) ms\(^{-2}\). The minimum value of coefficient of friction \(\mu\), so that \(m\) remains stationary with respect to wedge is
A block of mass \(m\) is lying on the edge having inclination angle \(\alpha = \tan^{-1}\!\!\left(\dfrac{1}{5}\right)\). Wedge is moving with a constant acceleration \(a = 2\) ms\(^{-2}\). The minimum value of coefficient of friction \(\mu\), so that \(m\) remains stationary with respect to wedge is
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In an accelerating frame, treat the pseudo-force as a real force. Resolve along and perpendicular to the incline to find the friction requirement.
Updated On: Apr 20, 2026
- \(\dfrac{2}{a}\)
- \(\dfrac{5}{12}\)
- \(\dfrac{1}{5}\)
- \(\dfrac{2}{5}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
\begin{itemize} \item In non-inertial frame of wedge: \begin{itemize} \item Pseudo-force \(ma\) acts horizontally backward on block \end{itemize} \end{itemize}
Step 2: Detailed Explanation:
\begin{itemize} \item Normal reaction: \[ N = mg\cos\alpha - ma\sin\alpha \] \item Friction: \[ f = \mu N = ma\cos\alpha + mg\sin\alpha \] \item Therefore: \[ \mu = \frac{a\cos\alpha + g\sin\alpha}{g\cos\alpha - a\sin\alpha} \] \item Simplifying: \[ \mu = \frac{a + g\tan\alpha}{g - a\tan\alpha} \] \item Substituting values: \[ \mu = \frac{2 + 10\times\frac{1}{5}}{10 - 2\times\frac{1}{5}} = \frac{4}{\frac{48}{5}} = \frac{5}{12} \] \end{itemize}
Step 3: Final Answer:
Minimum coefficient of friction \(\mu = \dfrac{5}{12}\).
\begin{itemize} \item In non-inertial frame of wedge: \begin{itemize} \item Pseudo-force \(ma\) acts horizontally backward on block \end{itemize} \end{itemize}
Step 2: Detailed Explanation:
\begin{itemize} \item Normal reaction: \[ N = mg\cos\alpha - ma\sin\alpha \] \item Friction: \[ f = \mu N = ma\cos\alpha + mg\sin\alpha \] \item Therefore: \[ \mu = \frac{a\cos\alpha + g\sin\alpha}{g\cos\alpha - a\sin\alpha} \] \item Simplifying: \[ \mu = \frac{a + g\tan\alpha}{g - a\tan\alpha} \] \item Substituting values: \[ \mu = \frac{2 + 10\times\frac{1}{5}}{10 - 2\times\frac{1}{5}} = \frac{4}{\frac{48}{5}} = \frac{5}{12} \] \end{itemize}
Step 3: Final Answer:
Minimum coefficient of friction \(\mu = \dfrac{5}{12}\).
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