A man of 80 kg fastns a rope to a mass of 120kg, and passes the rope over a smooth pulley. In order to raise it up he climbs the riope with acceleration `(3g)/2` relative to the rope. Calculate the tension in the rope in N. (Take `g=10m//s^(2)`)

Explanation

To solve the problem of finding the tension in the rope when a man of 80 kg climbs a rope attached to a mass of 120 kg with an acceleration of \( \frac{3g}{2} \) relative to the rope, we can follow these steps:

### Step 1: Understand the System
We have a man of mass \( m_1 = 80 \, \text{kg} \) climbing a rope attached to a mass \( m_2 = 120 \, \text{kg} \) over a smooth pulley. The acceleration of the man relative to the rope is given as \( a_{rel} = \frac{3g}{2} \).

### Step 2: Define the Acceleration
Let \( g = 10 \, \text{m/s}^2 \). The weight of the man is:
\[ W_1 = m_1 \cdot g = 80 \cdot 10 = 800 \, \text{N} \]
The weight of the block is:
\[ W_2 = m_2 \cdot g = 120 \cdot 10 = 1200 \, \text{N} \]

### Step 3: Determine the Acceleration of the Man
The acceleration of the man with respect to the ground \( a_p \) can be expressed as:
\[ a_p = a_{rel} - a \]
where \( a \) is the acceleration of the block. Since the man is climbing up, the block will move down with acceleration \( a \).

### Step 4: Write the Equation for the Man
Using Newton's second law for the man:
\[ T - W_1 = m_1 \cdot a_p \]
Substituting \( a_p \):
\[ T - 800 = 80 \left(\frac{3g}{2} - a\right) \]
Substituting \( g = 10 \):
\[ T - 800 = 80 \left(15 - a\right) \]
\[ T - 800 = 1200 - 80a \quad \text{(Equation 1)} \]

### Step 5: Write the Equation for the Block
Using Newton's second law for the block:
\[ T - W_2 = -m_2 \cdot a \]
This gives:
\[ T - 1200 = -120a \quad \text{(Equation 2)} \]

### Step 6: Solve the Equations
Now we have two equations:
1. \( T - 800 = 1200 - 80a \)
2. \( T - 1200 = -120a \)

From Equation 2, we can express \( T \):
\[ T = 1200 - 120a \]

Substituting \( T \) from Equation 2 into Equation 1:
\[ 1200 - 120a - 800 = 1200 - 80a \]
Simplifying:
\[ 400 - 120a = 1200 - 80a \]
Rearranging gives:
\[ 400 = 1200 + 40a \]
\[ 40a = 400 - 1200 \]
\[ 40a = -800 \]
\[ a = -20 \, \text{m/s}^2 \]

### Step 7: Substitute Back to Find Tension
Now substitute \( a \) back into the equation for \( T \):
\[ T = 1200 - 120(-20) \]
\[ T = 1200 + 2400 \]
\[ T = 3600 \, \text{N} \]

### Final Answer
The tension in the rope is:
\[ \boxed{3600 \, \text{N}} \]