Question

Two blocks of mass "2kg" & "3kg" respectively are placed on rough inclined having inclination angle $\theta$ = 37° as shown in the diagram (N is the normal reaction between the two blocks):-

• A. If μ₁ = μ2 = 0.5 ; a₁ = a2 ≠ 0, N = 0
• B. If μ₁ = μ₂ = 0.5 ; a₁ = a2 ≠ 0, N≠0
• C. If μ₁ = 0.6, μ2 = 0.1 ; a₁ = a2 = 3.6 m/s², N≠0
• D. If μ₁ = 0.1, μ2 = 0.6 ; a₁ > a2, N = 0

Answer 1

Answer: (A), (C), (D)

If μ₁ = μ2 = 0.5 ; a₁ = a2 ≠ 0, N = 0, If μ₁ = 0.6, μ2 = 0.1 ; a₁ = a2 = 3.6 m/s², N≠0, If μ₁ = 0.1, μ2 = 0.6 ; a₁ > a2, N = 0

Answer 2

To analyze the motion of the two blocks, we first calculate the acceleration each block would have if it were moving independently down the incline. The acceleration of a block of mass $m$ with coefficient of kinetic friction $\mu$ on an incline of angle $\theta$ is given by:

$a_{ind} = g \sin\theta - \mu g \cos\theta$

Given: $m_1 = 2 \text{ kg}$, $m_2 = 3 \text{ kg}$ $\theta = 37^\circ$ We use $g = 10 \text{ m/s}^2$. $\sin(37^\circ) \approx 0.6$ $\cos(37^\circ) \approx 0.8$

So, $g \sin\theta = 10 \times 0.6 = 6 \text{ m/s}^2$. And $g \cos\theta = 10 \times 0.8 = 8 \text{ m/s}^2$.

Thus, $a_{ind} = 6 - 8\mu$.

Now, let's analyze each option:

Option 1: If $\mu_1 = \mu_2 = 0.5$; $a_1 = a_2 \neq 0$, $N = 0$

Calculate independent accelerations: $a_1^{ind} = 6 - 8(0.5) = 6 - 4 = 2 \text{ m/s}^2$ $a_2^{ind} = 6 - 8(0.5) = 6 - 4 = 2 \text{ m/s}^2$

Since $a_1^{ind} = a_2^{ind}$, both blocks tend to move with the same acceleration. Therefore, they will move together with $a_1 = a_2 = 2 \text{ m/s}^2$. As there is no relative motion tendency, there will be no normal force between them, so $N=0$. The conditions $a_1 = a_2 \neq 0$ and $N = 0$ are met. This option is correct.

Option 2: If $\mu_1 = \mu_2 = 0.5$; $a_1 = a_2 \neq 0$, $N \neq 0$

From the analysis of Option 1, when $\mu_1 = \mu_2 = 0.5$, we found that $N=0$. This option states $N \neq 0$, which contradicts our finding. This option is incorrect.

Option 3: If $\mu_1 = 0.6, \mu_2 = 0.1$; $a_1 = a_2 = 3.6 \text{ m/s}^2$, $N \neq 0$

Calculate independent accelerations: $a_1^{ind} = 6 - 8(0.6) = 6 - 4.8 = 1.2 \text{ m/s}^2$ $a_2^{ind} = 6 - 8(0.1) = 6 - 0.8 = 5.2 \text{ m/s}^2$

Since $a_2^{ind} > a_1^{ind}$, block 2 (the one behind) tends to accelerate faster than block 1 (the one in front). This means block 2 will push block 1, so they will move together with a common acceleration $a$, and $N \neq 0$. To find the common acceleration $a$, treat the two blocks as a single system with total mass $M = m_1 + m_2 = 2 + 3 = 5 \text{ kg}$. The net force down the incline on the combined system is: $F_{net} = (m_1+m_2)g \sin\theta - (\mu_1 m_1 g \cos\theta + \mu_2 m_2 g \cos\theta)$ $F_{net} = (2+3)(6) - (0.6 \times 2 \times 8 + 0.1 \times 3 \times 8)$ $F_{net} = 5 \times 6 - (9.6 + 2.4)$ $F_{net} = 30 - 12 = 18 \text{ N}$ The common acceleration is $a = \frac{F_{net}}{M} = \frac{18 \text{ N}}{5 \text{ kg}} = 3.6 \text{ m/s}^2$. This matches the given acceleration $a_1 = a_2 = 3.6 \text{ m/s}^2$. Since $a_2^{ind} > a_1^{ind}$, there must be a normal force $N \neq 0$ for them to move together. (We can calculate $N$ for verification. For $m_1$: $m_1 a = m_1 g \sin\theta + N - \mu_1 m_1 g \cos\theta$. $2 \times 3.6 = 2 \times 6 + N - 0.6 \times 2 \times 8$ $7.2 = 12 + N - 9.6$ $7.2 = 2.4 + N \implies N = 4.8 \text{ N}$, which is non-zero.) The conditions $a_1 = a_2 = 3.6 \text{ m/s}^2$ and $N \neq 0$ are met. This option is correct.

Option 4: If $\mu_1 = 0.1, \mu_2 = 0.6$; $a_1 > a_2$, $N = 0$

Calculate independent accelerations: $a_1^{ind} = 6 - 8(0.1) = 6 - 0.8 = 5.2 \text{ m/s}^2$ $a_2^{ind} = 6 - 8(0.6) = 6 - 4.8 = 1.2 \text{ m/s}^2$

Since $a_1^{ind} > a_2^{ind}$, block 1 (the one in front) tends to accelerate faster than block 2 (the one behind). This means block 1 will move away from block 2, and they will separate. In this case, there will be no normal force between them, so $N=0$. The accelerations will be their independent accelerations: $a_1 = 5.2 \text{ m/s}^2$ and $a_2 = 1.2 \text{ m/s}^2$. The conditions $a_1 > a_2$ ($5.2 > 1.2$) and $N = 0$ are met. This option is correct.

Therefore, options 1, 3, and 4 are correct.