Two weights `w_(1)` and `w_(2)` are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up at an acceleration g , the tension in the string will be

Options

A. `(4w_(1)w_(2))/(w_(1)+w_(2))`
B. `(2w_(1)w_(2))/(w_(1)+w_(2))`
C. `(w_(1)w_(2))/(w_(1)+w_(2))`
D. `(w_(1)w_(2))/(2(w_(1)+w_(2))`

Answer

To find the tension in the string when two weights \( W_1 \) and \( W_2 \) are suspended from a pulley that is being pulled upward with an acceleration equal to \( g \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Weights
1. The weight \( W_1 \) exerts a downward force equal to \( W_1 \) (which is \( m_1 g \), where \( m_1 \) is the mass corresponding to weight \( W_1 \)).
2. The weight \( W_2 \) exerts a downward force equal to \( W_2 \) (which is \( m_2 g \), where \( m_2 \) is the mass corresponding to weight \( W_2 \)).
3. The tension \( T \) in the string acts upward on both weights.

### Step 2: Apply Newton's Second Law
For weight \( W_2 \) (assuming \( W_2 > W_1 \)):
- The net force acting on \( W_2 \) is given by:
\[ W_2 - T = m_2 a \]
where \( a \) is the acceleration of the system.

For weight \( W_1 \):
- The net force acting on \( W_1 \) is given by:
\[ T - W_1 = m_1 a \]

### Step 3: Substitute the Masses
Since \( W_1 = m_1 g \) and \( W_2 = m_2 g \), we can rewrite the equations:
1. For \( W_2 \):
\[ W_2 - T = \frac{W_2}{g} a \]
Rearranging gives:
\[ T = W_2 - \frac{W_2}{g} a \]

2. For \( W_1 \):
\[ T - W_1 = \frac{W_1}{g} a \]
Rearranging gives:
\[ T = W_1 + \frac{W_1}{g} a \]

### Step 4: Set the Two Expressions for Tension Equal
Now we have two expressions for \( T \):
\[ W_2 - \frac{W_2}{g} a = W_1 + \frac{W_1}{g} a \]

### Step 5: Solve for Acceleration \( a \)
Combine the terms:
\[ W_2 - W_1 = \left(\frac{W_2 + W_1}{g}\right) a \]
Thus,
\[ a = \frac{g(W_2 - W_1)}{W_2 + W_1} \]

### Step 6: Substitute \( a \) Back into One of the Tension Equations
Using the expression for \( T \):
\[ T = W_1 + \frac{W_1}{g} \left(\frac{g(W_2 - W_1)}{W_2 + W_1}\right) \]
This simplifies to:
\[ T = W_1 + \frac{W_1(W_2 - W_1)}{W_2 + W_1} \]

### Step 7: Final Expression for Tension
After simplifying, we find:
\[ T = \frac{2W_1W_2}{W_1 + W_2} \]

### Conclusion
Thus, the tension in the string when the pulley is pulled up with an acceleration equal to \( g \) is:
\[ T = \frac{2W_1W_2}{W_1 + W_2} \]