A monkey A (mass = 5kg) is climbing up a rope tied to a rigid support. The monkey B (mass = 2kg) is holding on to the tail of monkey A. If the tail can tolerate a maximum tension of 30N, what force should monkey A apply on the rope in order to carry monkey B with it? `(g = 10 m//s^(2))`.

Answer

To solve the problem step by step, we will analyze the forces acting on both monkeys and use Newton's second law of motion.

### Step 1: Identify the Forces Acting on Monkey A
Monkey A has a mass of 5 kg. The forces acting on Monkey A are:
- The tension \( T \) in the rope acting upwards.
- The weight of Monkey A acting downwards, which is given by \( W_A = m_A \cdot g = 5 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 50 \, \text{N} \).

### Step 2: Write the Equation of Motion for Monkey A
Since Monkey A is climbing up with an acceleration \( a \), we can write the equation of motion as:
\[ T - W_A = m_A \cdot a \]
Substituting the values:
\[ T - 50 = 5a \quad \text{(1)} \]

### Step 3: Identify the Forces Acting on Monkey B
Monkey B has a mass of 2 kg. The forces acting on Monkey B are:
- The tension \( T \) in the tail acting upwards.
- The weight of Monkey B acting downwards, which is given by \( W_B = m_B \cdot g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \).

### Step 4: Write the Equation of Motion for Monkey B
The maximum tension that the tail can tolerate is 30 N. Therefore, we write the equation of motion for Monkey B as:
\[ T - W_B = m_B \cdot a \]
Substituting the values:
\[ T - 20 = 2a \quad \text{(2)} \]

### Step 5: Solve the Equations Simultaneously
From equation (1):
\[ T = 50 + 5a \quad \text{(3)} \]
From equation (2):
\[ T = 20 + 2a + 30 \quad \text{(4)} \]
Substituting \( T \) from equation (3) into equation (4):
\[ 50 + 5a = 20 + 2a + 30 \]
Simplifying:
\[ 50 + 5a = 50 + 2a \]
\[ 5a - 2a = 50 - 50 \]
\[ 3a = 0 \implies a = 0 \, \text{(not possible, we need to find a positive acceleration)} \]

### Step 6: Find the Maximum Tension
Since the maximum tension \( T \) that the tail can tolerate is 30 N, we can set \( T = 30 \) in equation (2):
\[ 30 - 20 = 2a \]
\[ 10 = 2a \implies a = 5 \, \text{m/s}^2 \]

### Step 7: Substitute Back to Find the Force Applied by Monkey A
Now substituting \( a = 5 \, \text{m/s}^2 \) back into equation (3):
\[ T = 50 + 5(5) = 50 + 25 = 75 \, \text{N} \]

### Conclusion
The force that Monkey A should apply on the rope to carry Monkey B with it is 75 N.

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