Question

A 3 kg box is hanged using 2 pieces of identical strings attached at the center of it's top face and fastened to a ceiling. The angle between the string is 60°. How do I solve for the tension in each string so that the box remains in equilibrium?

Answer 1

To solve for the tension in each string so that the box remains in equilibrium, we can analyze the forces acting on the box.

Let's label the tension in each string as T.

Since the box is in equilibrium, the net force acting on it must be zero in both the horizontal and vertical directions.

In the horizontal direction, the tensions in the strings cancel each other out, so there is no horizontal force.

In the vertical direction, we can resolve the forces into components:

1. The weight of the box acts vertically downward and can be calculated as W = mg, where m is the mass of the box (3 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

2. The vertical component of tension T can be calculated as T * cos(60°).

Since the box is in equilibrium, the vertical forces must balance:

T * cos(60°) + T * cos(60°) = W

Substituting the values:

2T * cos(60°) = 3 kg * 9.8 m/s^2

Simplifying the equation:

2T * 0.5 = 29.4

T * 1 = 29.4 / 1

T = 29.4 N

Therefore, the tension in each string should be approximately 29.4 N to keep the box in equilibrium.