A light wire is attached to 2 points A and B which are at the same horizontal level with a distance 48m between them. A mass of 25N is tied on the midpoint of the wire and the mass... A light wire is attached to 2 points A and B which are at the same horizontal level with a distance 48m between them. A mass of 25N is tied on the midpoint of the wire and the mass moves 2m downwards from the straight line AB. What is the tension on the wire? Read more

Steps to Solve

1. Draw a free body diagram

Draw a free body diagram showing the forces acting on the mass. The forces are the weight of the mass acting downwards ($W$), and the tension in the wire on both sides ($T$). Because the mass is suspended at the midpoint of the wire, the tension in both halves of the wire will be equal.

2. Define the vertical angle

Define $\theta$ as the angle between the wire and the horizontal. We can calculate $\theta$ using trigonometry. We have a right triangle where the opposite side is the sag (10 cm = 0.1 m) and the adjacent side is half the distance between the two points (2 m / 2 = 1 m). $$ \tan(\theta) = \frac{0.1}{1} = 0.1 $$ $$ \theta = \arctan(0.1) \approx 5.71^\circ $$

3. Vertical Force Equilibrium

Since the mass is in equilibrium, the vertical forces must balance. The vertical component of the tension $T$ in each wire must equal half of the weight. The weight $W$ is $mg$, where $m = 100$ kg and $g = 9.81 m/s^2$. $$ 2T\sin(\theta) = W $$ $$ 2T\sin(\theta) = mg $$

4. Solve for Tension

Solve for the tension $T$: $$ T = \frac{mg}{2\sin(\theta)} $$ Substitute the values: $$ T = \frac{100 \times 9.81}{2\sin(5.71^\circ)} $$ $$ T \approx \frac{981}{2 \times 0.0995} $$ $$ T \approx 4929.6 \text{ N} $$