Physics Solution – Question 7

Solution to Question 7

1. Free Body Diagram and Force Analysis

Let $a$ be the acceleration of the prism. Since the disc accelerates up the slant face, the prism must move to the right due to the reaction forces. We analyze the forces in the frame of the prism (which is non-inertial).

Forces on Disc (mass $m$, charge $q$):
- Gravitational force: $mg$ (Down)
- Electric force: $qE$ (Right)
- Normal force: $N$ (Perpendicular to incline, Outward)
- Frictional force: $f = \mu N$ (Down the incline, opposing the upward slip)
- Pseudo force: $ma$ (Left, since prism accelerates Right)

Figure 3: Free Body Diagram of the Disc on the Incline.

2. Equation for Normal Force ($N$)

The disc is in equilibrium perpendicular to the incline. We resolve forces along the normal direction.

Component of $mg$ (Down) into the incline: $mg \cos\theta$.
Component of $qE$ (Right) into the incline: $qE \sin\theta$. (Right is $90^\circ+\theta$ from Normal).
Component of $ma$ (Left) out of the incline: $ma \sin\theta$.

Balancing inward and outward forces:

$$ N + ma \sin\theta = mg \cos\theta + qE \sin\theta $$ $$ N = mg \cos\theta + qE \sin\theta – ma \sin\theta \quad \dots(1) $$

3. Equation of Motion for the Prism

Now consider the Prism (Mass $M$). It accelerates to the right with acceleration $a$. The forces exerted by the disc on the prism are equal and opposite to those on the disc:

Normal Reaction $N$ pushes the prism Down-Right.
Friction $f = \mu N$ pushes the prism Up-Right (since friction on disc is Down-Left).

Horizontal components driving the prism:

Horizontal component of $N$: $N \sin\theta$.
Horizontal component of $f$: $f \cos\theta = \mu N \cos\theta$.

Newton’s Second Law for the prism:

$$ Ma = N \sin\theta + \mu N \cos\theta $$ $$ Ma = N (\sin\theta + \mu \cos\theta) \quad \dots(2) $$

4. Solving for Acceleration ($a$)

Substitute the expression for $N$ from equation (1) into equation (2):

$$ Ma = (mg \cos\theta + qE \sin\theta – ma \sin\theta)(\sin\theta + \mu \cos\theta) $$

Group terms containing $a$ on the left side:

$$ Ma = (mg \cos\theta + qE \sin\theta)(\sin\theta + \mu \cos\theta) – ma \sin\theta(\sin\theta + \mu \cos\theta) $$ $$ Ma + ma \sin\theta(\sin\theta + \mu \cos\theta) = (mg \cos\theta + qE \sin\theta)(\sin\theta + \mu \cos\theta) $$ $$ a \left[ M + m \sin\theta(\sin\theta + \mu \cos\theta) \right] = (mg \cos\theta + qE \sin\theta)(\sin\theta + \mu \cos\theta) $$

Thus, the acceleration is:

$$ a = \frac{(qE \sin\theta + mg \cos\theta)(\sin\theta + \mu \cos\theta)}{M + m \sin\theta (\sin\theta + \mu \cos\theta)} $$

Final Answer

$$ a = \frac{(qE \sin\theta + mg \cos\theta)(\sin\theta + \mu \cos\theta)}{M + m \sin\theta (\sin\theta + \mu \cos\theta)} $$