Question

Two 25.0 N weights are suspended at opposite end of a rope that passes over a light, frictionless pulley. The pulley is attached to a chain that goes to the ceiling.
A. What is the tension in the rope?
b. What is the tension in the chain?

Answer 1

Answer on question #61316, Physics / Mechanics | Relativity

Two 25.0 N weights are suspended at opposite end of a rope that passes over a light, frictionless pulley. The pulley is attached to a chain that goes to the ceiling.

a. What is the tension in the rope?

b. What is the tension in the chain?

Solution:

The pulley has negligible mass. Let $T_r$ be the tension in the rope and let $T_c$ be the tension in the chain. $P = 25.0 \, \text{N}$

a) The diagram for the weight is given here

Write Newton's second law in the vector form:

$$\vec{F} = \vec{T_r} + \vec{P}$$

The equation will look like in the projection on the axis of $+Y$:

$$ma = T_r - P \quad (where\ a = 0)$$ $$-T_r = -P \quad (multiplied\ by\ -1)$$ $$T_r = P$$ $$T_r = 25.0 \, \text{N}$$

As the rope will be stationary (same weight on the other side), therefore tension in the rope will be $25.0 \, \text{N}$.

b) The diagram for the pulley is given here

Write Newton's second law in the vector form:

$$\vec {F} = \overrightarrow {T _ {c}} + 2 \vec {P}$$

The equation will look like in the projection on the axis of $+Y$ :

$$\begin{array}{l} m a = T _ {c} - 2 P \text{ (where } a = 0) \\ - T _ {c} = - 2 P \\ T _ {c} = 2 P \\ T _ {c} = 2 \times 25.0 \, N = 50.0 \, N \\ \end{array}$$

Therefore, tension in chain= 50.0 N.

Answer: a) 25.0 N; b) 50.0 N

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