Question

Consider three masses \(M_1\), \(M_2\) and \(M_3\), where \(M_1 \gt M_2 \gt M_3\), 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 is

• A. \(M_3\) begins to slide at a higher inclination angle than \(M_1\) and \(M_2\)
• B. \(M_3\) begins to slide at a lower inclination angle than \(M_1\) and \(M_2\)
• C. \(M_1\), \(M_2\) and \(M_3\) begin to slide at the same inclination angle
• D. \(M_2\) begins to slide at a higher inclination angle than \(M_1\) and \(M_3\)

Hint:

For a body on a rough inclined plane, the angle of repose is given by \[ \tan\theta=\mu_s. \] It depends only on the coefficient of static friction and not on the mass of the body.

Answer 1

The Correct Option is C

Step 1: Identify the forces acting on a block on an inclined plane.
For any block of mass \(m\) placed on an inclined plane, the component of weight acting down the plane is \[ mg\sin\theta \] The normal reaction on the block is \[ N=mg\cos\theta \] The maximum static friction is \[ f_{\max}=\mu_s N \] Therefore, \[ f_{\max}=\mu_s mg\cos\theta \]

Step 2: Apply the condition for just sliding.
The block begins to slide when the component of weight down the plane becomes equal to the maximum static friction.
So, \[ mg\sin\theta=\mu_s mg\cos\theta \]

Step 3: Simplify the equation.
Cancelling \(mg\) from both sides, we get \[ \sin\theta=\mu_s\cos\theta \] Dividing by \(\cos\theta\), \[ \tan\theta=\mu_s \] Hence, \[ \theta=\tan^{-1}(\mu_s) \]

Step 4: Understand the dependence on mass.
The angle at which sliding begins is \[ \theta=\tan^{-1}(\mu_s) \] This expression does not contain mass \(m\).
Therefore, the angle of sliding is independent of the mass of the object.
Since the coefficient of static friction \(\mu_s\) is the same for \(M_1\), \(M_2\), and \(M_3\), all three masses will begin to slide at the same inclination angle.

Step 5: Final conclusion.
Therefore, \[ \boxed{M_1,\ M_2\ \text{and}\ M_3\ \text{begin to slide at the same inclination angle}} \]