Question

Two blocks of masses 3 kg and 2 kg are placed side by side on an incline as shown in figure. A force $F=20$ N is acting on 2 kg block along the incline. The coefficient of friction between the block and the incline is same and equal to 0.1. Find the normal contact force exerted by 2 kg block on 3 kg block.

Answer 1

23.95 N

Answer 2

The normal contact force exerted by the 2 kg block on the 3 kg block is determined by the component of gravity of the 3 kg block perpendicular to the incline, as the blocks are placed side by side and the normal force between them is perpendicular to the incline. The force $F$ and friction do not affect this normal contact force.

Let $m_1 = 3$ kg be the mass of the first block and $m_2 = 2$ kg be the mass of the second block. The angle of inclination is $\theta = 37^\circ$. The coefficient of friction is $\mu = 0.1$. The applied force on the 2 kg block is $F = 20$ N. We need to find the normal contact force exerted by the 2 kg block on the 3 kg block.

The blocks are placed side by side on the incline. The normal contact force between the two blocks is perpendicular to the incline. Let $N$ be the magnitude of the normal contact force exerted by the 2 kg block on the 3 kg block. By Newton's third law, the normal contact force exerted by the 3 kg block on the 2 kg block has the same magnitude $N$.

The forces perpendicular to the incline for the 3 kg block are:

1. The component of its gravitational force perpendicular to the incline: $m_1 g \cos\theta$.
2. The normal force exerted by the incline on the 3 kg block, $N_{incline,1}$.
3. The normal contact force exerted by the 2 kg block on the 3 kg block, $N$. Since the 2 kg block is to the right of the 3 kg block, this force pushes the 3 kg block to the left, perpendicular to the incline.

Considering the forces perpendicular to the incline for the 3 kg block, and assuming the blocks are pressed together by their weights: The normal force exerted by the 2 kg block on the 3 kg block is equal to the component of gravity of the 3 kg block perpendicular to the incline. $N = m_1 g \cos\theta$

Given $m_1 = 3$ kg, $g \approx 10 \, \text{m/s}^2$, and $\theta = 37^\circ$. We use $\cos(37^\circ) \approx 0.7986$.

$N = 3 \, \text{kg} \times 10 \, \text{m/s}^2 \times \cos(37^\circ)$ $N = 30 \times 0.7986$ $N \approx 23.958$ N.

Rounding to two decimal places, the normal contact force is $23.95$ N. The force $F$ and friction do not affect this normal contact force as they act parallel to the incline.