Solution 50

Solution to Question 50

Step 1: Analyze Force Exerted by Rod on Block B

When the dumbbell is released, ball A tends to fall vertically, but the rigid rod constrains it. Since the rod is “light” and rigid, it transmits the component of A’s weight along the rod directly to B.

Decompose the weight of ball A ($mg$) into components relative to the rod:

Perpendicular to the rod: $mg \cos \theta$ (provides tangential acceleration).
Along the rod: $mg \sin \theta$ (pushes against the rod).

Note: Here $\theta$ is the angle with the horizontal. Therefore, the angle with the vertical is $90^\circ – \theta$.

Since the radial acceleration is initially zero (velocity is zero), the compressive force $P$ in the rod balances the component of gravity along it: $$ P = mg \sin \theta $$

Step 2: Force Analysis on Block B

The rod pushes block B with force $P$ at an angle $\theta$ below the horizontal. We resolve $P$ into horizontal and vertical components acting on B:

Horizontal driving force: $F_x = P \cos \theta = (mg \sin \theta) \cos \theta$
Vertical downward component: $F_y = P \sin \theta = (mg \sin \theta) \sin \theta = mg \sin^2 \theta$

The total Normal force $N_B$ on B from the ground is the sum of its own weight and the vertical push from the rod: $$ N_B = Mg + F_y = Mg + mg \sin^2 \theta $$

Step 3: Condition for Sliding

Block B will start sliding immediately if the horizontal driving force exceeds the maximum static friction: $$ F_x > f_{\text{max}} $$ $$ F_x > \mu N_B $$ Substituting the values derived above: $$ mg \sin \theta \cos \theta > \mu (Mg + mg \sin^2 \theta) $$

Rearranging to solve for the coefficient of friction $\mu$: $$ \mu < \frac{mg \sin \theta \cos \theta}{Mg + mg \sin^2 \theta} $$ Using the identity $2 \sin \theta \cos \theta = \sin 2\theta$, we can rewrite the numerator: $$ \mu < \frac{m \sin 2\theta}{2(M + m \sin^2 \theta)} $$