Question

In a pulley system, two blocks are connected by a string over a frictionless pulley. If tensions \(T_1\) and \(T_2\) are given in two segments of the string, what is their relation?

• A. \( T_1 = T_2 \)
• B. \( T_1>T_2 \)
• C. \( T_1<T_2 \)
• D. Depends on masses only

Hint:

Tension only changes across a pulley if the pulley has mass and friction (which requires torque to rotate). In most introductory physics problems, unless stated otherwise, assume the tension is uniform.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In basic classical mechanics problems involving pulley systems, unless stated otherwise, the string connecting the blocks is considered ideal (massless and inextensible) and the pulley is considered frictionless and massless.
Step 2: Key Formula or Approach:
For an ideal string passing over a frictionless, massless pulley, the rotational equation is $\tau = (T_1 - T_2)R = I\alpha$. Since $I = 0$ for a massless pulley, we find $T_1 = T_2$.
Step 3: Detailed Explanation:
Let's consider the rotational dynamics of the pulley. If the pulley had a moment of inertia $I$ and angular acceleration $\alpha$, the net torque would be $(T_1 - T_2)R = I\alpha$.
However, since the pulley is ideal (frictionless and massless $\implies I = 0$), the equation becomes: \[ (T_1 - T_2)R = 0 \implies T_1 = T_2 \] Thus, the tension in both segments of the string is exactly the same, irrespective of the masses of the blocks hanging from it.
Step 4: Final Answer:
The relation between the tensions is $T_1 = T_2$.