A 40 kg boy climbs a rope that passes over an ideal pulley. The other end of the rope is attached to a 60 kg weight placed on the ground. What is the maximum upward acceleration the boy can have without lifting the weight ? If he climbs the rope with upward acceleration 2 g, with what acceleration the weight will rise up ?

Answer

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

### Step 1: Identify the forces acting on the boy and the weight
- The weight of the boy (W_boy) = mass × gravity = 40 kg × g = 40g.
- The weight of the block (W_block) = mass × gravity = 60 kg × g = 60g.
- The tension in the rope (T) acts upwards on the boy and downwards on the block.

### Step 2: Determine the maximum upward acceleration of the boy without lifting the weight
For the weight not to lift off the ground, the tension in the rope must not exceed the weight of the block. Therefore, we set up the equation:

\[ T \leq 60g \]

For the boy climbing with an upward acceleration (a), the net force acting on him can be expressed as:

\[ T - 40g = 40a \]

From this, we can express the tension (T):

\[ T = 40g + 40a \]

### Step 3: Set the maximum tension equal to the weight of the block
To find the maximum upward acceleration (a_max) of the boy without lifting the weight, we substitute T from the previous equation into the maximum tension condition:

\[ 40g + 40a_{max} \leq 60g \]

### Step 4: Solve for a_max
Rearranging the equation gives:

\[ 40a_{max} \leq 60g - 40g \]
\[ 40a_{max} \leq 20g \]
\[ a_{max} \leq \frac{20g}{40} \]
\[ a_{max} \leq \frac{g}{2} \]

Thus, the maximum upward acceleration of the boy without lifting the weight is:

\[ a_{max} = \frac{g}{2} \]

### Step 5: Analyze the scenario when the boy climbs with an upward acceleration of 2g
If the boy climbs with an upward acceleration of 2g, we substitute this value into the tension equation:

\[ T = 40g + 40(2g) = 40g + 80g = 120g \]

### Step 6: Determine the acceleration of the weight
Now, we analyze the forces acting on the weight:

The net force acting on the weight is given by:

\[ T - 60g = 60a \]

Substituting the value of T:

\[ 120g - 60g = 60a \]
\[ 60g = 60a \]
\[ a = g \]

Thus, when the boy climbs with an upward acceleration of 2g, the weight will rise with an acceleration of:

\[ a = g \]

### Summary of Results
1. The maximum upward acceleration the boy can have without lifting the weight is \( \frac{g}{2} \).
2. If the boy climbs the rope with an upward acceleration of 2g, the weight will rise with an acceleration of g.

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