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, you can analyze the forces acting on the box and apply the principles of equilibrium. In this case, the forces acting on the box are the weight (mg) and the tensions in the strings.

Let's label the tensions in the strings as T₁ and T₂. Since the strings are identical, we assume they have the same tension.

To begin, resolve the weight of the box (mg) into vertical and horizontal components. The vertical component (mg_y) counteracts the tension in the strings, while the horizontal component (mg_x) doesn't affect the tension.

Now, let's analyze the forces acting on the box in the vertical direction:

1. Tension T₁ pulls upward at an angle of 30° from the vertical.
2. Tension T₂ pulls upward at an angle of 30° from the vertical.

Since the box is in equilibrium, the sum of the vertical forces must be zero. Therefore, we can write the following equation:

T₁sin(30°) + T₂sin(30°) - mg_y = 0

Next, let's analyze the forces acting on the box in the horizontal direction:

1. Tension T₁ pulls to the right at an angle of 30° from the horizontal.
2. Tension T₂ pulls to the left at an angle of 30° from the horizontal.

Since the box is in equilibrium, the sum of the horizontal forces must be zero. Therefore, we can write the following equation:

T₁cos(30°) - T₂cos(30°) - mg_x = 0

The horizontal component of the weight (mg_x) doesn't contribute to the tension in the strings since it cancels out.

Now, we can solve these two equations simultaneously to find the values of T₁ and T₂. Substituting mg_y = mg cos(30°) and mg_x = mg sin(30°), the equations become:

T₁sin(30°) + T₂sin(30°) = mg cos(30°) (Equation 1) T₁cos(30°) - T₂cos(30°) = 0 (Equation 2)

Dividing Equation 1 by Equation 2, we can eliminate cos(30°):

[T₁sin(30°) + T₂sin(30°)] / [T₁cos(30°) - T₂cos(30°)] = [mg cos(30°)] / 0

Simplifying, we get:

tan(30°) = mg / (T₁ - T₂)

Since T₁ = T₂ (the tensions are equal), we have:

tan(30°) = mg / 0

As there is no solution for T₁ and T₂ that would satisfy the given conditions, the box cannot remain in equilibrium with the described configuration of the strings.