Question 58: Laws of Motion

Question

Two blocks ( $P$ and $Q$ ) with respectively masses 2 kg and 1.5 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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

Since the cube is accelerating at $\mathrm{a}=\frac{\mathrm{g}}{2}$ to the right, when analysing the blocks from the cube's non-inertial frame, we must apply a pseudo-force ( $\mathrm{F}_{\mathrm{p}}=\mathrm{m} 	imes \mathrm{a}$ ) on both blocks in the opposite direction to the acceleration of block S.
For block P (Mass $\mathrm{m}_{\mathrm{P}}=2 \mathrm{~kg}$ )
Along horizontal the tension (T) in the thread pulls it to the right. The pseudo-force ( $\mathrm{m}_{\mathrm{P}} \cdot \mathrm{a}$ ) pulls it to the left. Along vertical direction gravity ( $\mathrm{m}_{\mathrm{P}} \mathrm{g}$ ) acts downward and normal force ( $\mathrm{N}_{\mathrm{P}}$ ) from the top surface acts upward. Since the block is stationary relative to the cube :
$$ \mathrm{T}=\mathrm{m}_{\mathrm{P}} \cdot \mathrm{a}=\mathrm{m}_{\mathrm{P}} \cdot \frac{\mathrm{~g}}{2} $$
$$\Rightarrow $$ $$ \mathrm{T}=2 	imes \frac{10}{2}=10 \mathrm{~N} $$
For block Q (Mass $\mathrm{m}_{\mathrm{Q}}=1.5 \mathrm{~kg}$ )
Along horizontal direction the cube pushes against the block with a normal force $\left(\mathrm{N}_{\mathrm{Q}}ight)$ to the right. The pseudoforce $\left(\mathrm{m}_{\mathrm{Q}} \cdot \mathrm{a}ight)$ acts to the left.
$$ \mathrm{N}_{\mathrm{Q}}=\mathrm{m}_{\mathrm{Q}} \cdot \mathrm{a}=1.5 	imes \frac{\mathrm{g}}{2} $$
Along vertical direction, gravity ( $\mathrm{m}_{\mathrm{Q}} \mathrm{g}$ ) acts downward.
$$ \mathrm{F}_{\mathrm{gQ}}=1.5 	imes 10=15 \mathrm{~N} $$
Tension (T) in the thread acts upward, $\mathrm{T}=10 \mathrm{~N}$ [as the string is massless and inextensible]
Friction (f) acts upward to prevent the block from falling.
For equilibrium in the vertical direction:
$$ \mathrm{T}+\mathrm{f}=\mathrm{m}_{\mathrm{Q}} \mathrm{~g} $$
$$\Rightarrow $$ $$ 10+\mathrm{f}=15 \Rightarrow \mathrm{f}=5 \mathrm{~N} $$
The friction force is given by $\mathrm{f}=\mu \mathrm{N}_{\mathrm{Q}}$.
$$ 5=\mu 	imes\left(1.5 	imes \frac{10}{2}ight) $$
$$\Rightarrow $$ $5=\mu 	imes(1.5 	imes 5)$
$$\Rightarrow $$ $5=\mu 	imes 7.5$
$$\Rightarrow $$ $$ \mu=\frac{5}{7.5}=\frac{2}{3} \approx 0.67 $$
Therefore, the value of coefficient of friction between the block Q and S is $\mu=0.67$.
Thus, the correct option is (B).

Answer

0.33

Answer

0.67

Answer

1

Answer

0.5

Laws of Motion

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