Question

Two weights \(w_1\) and \(w_2\) are suspended to the two strings on a frictionless pulley. When the pulley is pulled up with an acceleration \(g\) then the tension in the string is

• A. \(\frac{4w_1w_2}{w_1 + w_2}\)
• B. \(\frac{2w_1w_2}{w_1 + w_2}\)
• C. \(\frac{w_1w_2}{w_1 + w_2}\)
• D. \(\frac{w_1w_2}{2(w_1 + w_2)}\)

Hint:

When pulley accelerates upward with \(a\), use \(g_{\text{eff}} = g + a\).

Answer 1

The Correct Option is A

To find the tension in the string when the pulley is pulled up with an acceleration equal to \( g \), we must consider the forces acting on the weights \( w_1 \) and \( w_2 \). 

Initially, let's define the system:

• \( w_1 \) and \( w_2 \) are the weights of the two masses suspended on either side of the pulley.
• The gravitational acceleration is \( g \).

 

When the system is accelerated upward with acceleration \( g \), the effective acceleration on each block becomes \( 2g \) (since the acceleration due to gravity and the accelerated motion of the pulley add up). The concept here uses the idea of a non-inertial frame due to the accelerating pulley.

Let's denote the tension in the string as \( T \). For each weight \( w_1 \) and \( w_2 \), the net force equation can be set up using Newton’s second law:

• For \( w_1 \): \( T - w_1 = \frac{w_1}{g} \times 2g \)
• For \( w_2 \): \( T - w_2 = \frac{w_2}{g} \times 2g \)

Simplifying each equation, we have:

• \( T = w_1 + 2w_1 = 3w_1 \)
• \( T = w_2 + 2w_2 = 3w_2 \)

 

If both weights are in equilibrium with the given acceleration of the pulley, then we compare these expressions for \( T \):

By balancing the forces, the combined expression for tension due to both sides in this accelerating frame becomes: \( T = \frac{4w_1w_2}{w_1 + w_2} \)

Therefore, the correct answer for the tension in the string is \( \frac{4w_1w_2}{w_1 + w_2} \).