Question

Two blocks ($P$ and $Q$) with respectively masses $2\text{ kg}$ and $1.5\text{ kg}$ are joined by a massless thread. These blocks are mounted on a frictionless pully which is fixed on the edge of a cube $(S)$, as shown in the figure below. Block $P$ is positioned on the top surface which has no friction and block $Q$ is in contact with side-surface, having coefficient friction $\mu$. The cube $(S)$ moves towards the right with acceleration of $\frac{g}{2}$, where $g$ is gravitational acceleration. During this movement the block $P$ and $Q$ remain stationary. The value of $\mu$ is ______. (take $g = 10\text{ m/s}^2$)

• A. 0.33
• B. 0.67
• C. 1
• D. 0.5

Hint:

Use a non-inertial frame of reference (the cube) and apply pseudo-forces. Ensure that the required static friction to keep block Q from sliding down is less than or equal to $\mu N$.

Answer 1

The Correct Option is B

We can also solve this using the principles of Newtonian mechanics in an inertial (ground) frame. In this frame, both blocks $P$ and $Q$ are accelerating to the right with an acceleration $a = \frac{g}{2}$ along with the cube $S$.

For block $P$ on the top surface:
The horizontal net force is provided by the tension $T$ in the string. According to Newton's second law:
$$T = m_P \times a = 2 \times \frac{g}{2} = g$$

For block $Q$ against the vertical side surface:
Horizontally, it has an acceleration $a = \frac{g}{2}$. This acceleration is caused by the normal force $N$ exerted by the wall of the cube:
$$N = m_Q \times a = 1.5 \times \frac{g}{2} = 0.75g$$
Vertically, block $Q$ does not accelerate (it is stationary relative to the cube). Therefore, the forces must balance:
$$T + f = m_Q g$$
Substitute the values: $g + f = 1.5g$, which gives the friction force $f = 0.5g$ acting upwards.

We know that the static friction force $f$ must satisfy the condition $f \leq \mu N$. Substituting our derived values:
$$0.5g \leq \mu(0.75g)$$
Dividing both sides by $g$ and rearranging for $\mu$:
$$\mu \geq \frac{0.5}{0.75} = \frac{2}{3}$$
$$\mu \geq 0.666...$$
Rounding to two decimal places, we get $\mu = 0.67$.