Forces and Equilibrium: Question 5

Worked solution

Step 1: Find the tension in the vertical part of the string

The parcel hangs in equilibrium from the vertical part of the string, so resolving vertically, the tension in that part must exactly balance the parcel’s weight: $T_{\text{vertical}} = 45\text{ N}$

Step 2: Apply the property of a light string over a smooth pulley

A pulley described as “smooth” is frictionless, and the string is “light” (has no mass) and inextensible. Together, these mean the string can change direction as it passes over the pulley without any change in tension. The tension is the same throughout the entire length of the string, on both sides of the pulley.

Step 3: State the tension in the horizontal part

Since the tension is the same throughout the string: $T_{\text{horizontal}} = T_{\text{vertical}} = 45\text{ N}$

The change in the string’s direction at the pulley does not affect the magnitude of the tension.

Why the other options are wrong

• B ($22.5\text{ N}$): this halves the tension, as if the load were shared between two separate strings rather than carried by one continuous string of constant tension.
• C ($90\text{ N}$): this doubles the tension, which would only be relevant in a pulley system with two strands directly supporting the same weight, not the case here.
• D ($0\text{ N}$): this assumes the horizontal part carries no force at all, ignoring that the same string (and hence the same tension) continues past the pulley.

Final answer

• The tension in the horizontal part of the string is $\boxed{45}\text{ N}$, option A.