Question

Block A of mass 5 kg is kept between plank B of mass 10 kg and the inclined plane as shown below. Plank B is attached to an inextensible string which is parallel to inclined plane. Block A slides with constant velocity between plank and fixed wedge. Friction coefficient between all surfaces is same as $\mu$. Find value of 20$\mu$.

• A. 15
• B. 20
• C. 10
• D. 5

Answer 1

Answer: (A)

15

Answer 2

Block A slides with constant velocity, meaning the net force on it is zero. The forces acting on block A parallel to the inclined plane are its gravitational component ($m_A g \sin(\theta)$) downwards, and friction forces from plank B ($f_{BA}$) and the inclined plane ($f_{AW}$) upwards. Perpendicular to the incline, the forces are the gravitational component ($m_A g \cos(\theta)$) and the normal forces from plank B ($N_{AB}$) and the inclined plane ($N_{AW}$).

For equilibrium along the incline: $m_A g \sin(\theta) = f_{BA} + f_{AW}$ Since $f = \mu N$, we have: $m_A g \sin(\theta) = \mu N_{AB} + \mu N_{AW} = \mu (N_{AB} + N_{AW})$

For equilibrium perpendicular to the incline: $N_{AB} + N_{AW} = m_A g \cos(\theta)$

Substituting the second equation into the first: $m_A g \sin(\theta) = \mu (m_A g \cos(\theta))$

Dividing by $m_A g \cos(\theta)$: $\tan(\theta) = \mu$

Given $\theta = 37^\circ$, and $\tan(37^\circ) = \frac{3}{4} = 0.75$. So, $\mu = 0.75$.

The question asks for $20\mu$: $20\mu = 20 \times 0.75 = 15$.