Two blocks of masses `m_1` = 4 kg and `m_2` = 2kg connected by a massless rod slide down on an inclined plane of inclination `37^o` as shown in figure. If coefficient of friction of `m_1` and `m_2` with inclined plane are `mu_1` = 0.75 and `mu_2` = 0.25 respectively, then tension (T) in the rod is (g = 10 `ms^-2`)

Options

A. `frac{16}{5}N`
B. `frac{16}{3}N`
C. 0N
D. 3N

Answer

To solve the problem, we need to find the tension (T) in the rod connecting the two blocks sliding down an inclined plane. Let's break it down step by step.

### Step 1: Identify the forces acting on each block
For block \( m_1 \) (4 kg):
- Weight: \( W_1 = m_1 \cdot g = 4 \cdot 10 = 40 \, \text{N} \)
- Component of weight parallel to the incline: \( W_{1\parallel} = W_1 \cdot \sin(37^\circ) = 40 \cdot \sin(37^\circ) \)
- Component of weight perpendicular to the incline: \( W_{1\perp} = W_1 \cdot \cos(37^\circ) = 40 \cdot \cos(37^\circ) \)
- Frictional force: \( F_{friction1} = \mu_1 \cdot W_{1\perp} = 0.75 \cdot W_{1\perp} \)

For block \( m_2 \) (2 kg):
- Weight: \( W_2 = m_2 \cdot g = 2 \cdot 10 = 20 \, \text{N} \)
- Component of weight parallel to the incline: \( W_{2\parallel} = W_2 \cdot \sin(37^\circ) = 20 \cdot \sin(37^\circ) \)
- Component of weight perpendicular to the incline: \( W_{2\perp} = W_2 \cdot \cos(37^\circ) = 20 \cdot \cos(37^\circ) \)
- Frictional force: \( F_{friction2} = \mu_2 \cdot W_{2\perp} = 0.25 \cdot W_{2\perp} \)

### Step 2: Calculate the net force acting on the system
The net force acting on the system (both blocks) can be expressed as:
\[ F_{net} = (W_{1\parallel} + W_{2\parallel}) - (F_{friction1} + F_{friction2}) \]
Substituting the expressions for the forces:
\[ F_{net} = (m_1 \cdot g \cdot \sin(37^\circ) + m_2 \cdot g \cdot \sin(37^\circ)) - (\mu_1 \cdot m_1 \cdot g \cdot \cos(37^\circ) + \mu_2 \cdot m_2 \cdot g \cdot \cos(37^\circ)) \]

### Step 3: Calculate the acceleration of the system
Using Newton's second law, the net force can also be expressed as:
\[ F_{net} = (m_1 + m_2) \cdot a \]
Setting the two expressions for \( F_{net} \) equal gives us:
\[ (m_1 + m_2) \cdot a = (m_1 \cdot g \cdot \sin(37^\circ) + m_2 \cdot g \cdot \sin(37^\circ)) - (\mu_1 \cdot m_1 \cdot g \cdot \cos(37^\circ) + \mu_2 \cdot m_2 \cdot g \cdot \cos(37^\circ)) \]

### Step 4: Solve for acceleration (a)
Substituting the values:
- \( g = 10 \, \text{m/s}^2 \)
- \( \sin(37^\circ) \approx 0.6 \)
- \( \cos(37^\circ) \approx 0.8 \)

Calculating:
\[ a = \frac{(4 \cdot 10 \cdot 0.6 + 2 \cdot 10 \cdot 0.6) - (0.75 \cdot 4 \cdot 10 \cdot 0.8 + 0.25 \cdot 2 \cdot 10 \cdot 0.8)}{4 + 2} \]
\[ = \frac{(24 + 12) - (24 + 4)}{6} = \frac{36 - 28}{6} = \frac{8}{6} \approx 1.33 \, \text{m/s}^2 \]

### Step 5: Calculate the tension (T) in the rod
For block \( m_1 \):
Using \( m_1 \):
\[ m_1 \cdot g \cdot \sin(37^\circ) - T - F_{friction1} = m_1 \cdot a \]
Substituting the known values:
\[ 4 \cdot 10 \cdot 0.6 - T - (0.75 \cdot 4 \cdot 10 \cdot 0.8) = 4 \cdot 1.33 \]
Solving for T gives:
\[ 24 - T - 24 = 5.32 \implies T = 0 \]

### Final Answer
The tension \( T \) in the rod is \( 0 \, \text{N} \).