Question

A weight of mass 7.5 kg hangs from a fixed point by a light inextensible string. It is pulled aside by a horizontal force, F, and then rests in equilibrium with the string inclined at 30 degrees to the vertical. Find the tension in the string and the value of F (take g=9.81 m/s^2).

a) Tension = 74.15 N, F=63.75 N
b) Tension = 63.75 N, F=74.15 N
c) Tension = 58.29 N, F=69.81 N
d) Tension = 69.81 N, F=58.29 N

Answer 1

To solve for the tension and horizontal force F, resolve the forces in vertical and horizontal directions for the hanging mass. Tension is found through gravity's vertical component, and horizontal force F through tension's horizontal component. The calculated tension and force don't match the given options, indicating potential errors in the question or answer choices.

Step-by-step explanation:

The question involves finding the tension in the string and the value of the horizontal force F when a weight is hanging in equilibrium at an angle. Given that the weight of 7.5 kg is being held at a 30-degree angle to the vertical, and using g = 9.81 m/s2, we start by resolving the forces in the vertical and horizontal directions.

First, we calculate the force of gravity acting on the mass (Fg = m * g):

Fg = 7.5 kg * 9.81 m/s2 = 73.575 N.

This is the vertical component of the tension T in the string. So, T * cos(30) = 73.575 N leads us to find T as follows:

T = 73.575 N / cos(30) = 85 N (approximately)

Now, let's find the horizontal force F. The horizontal component of the tension (T * sin(30)) should be equal to F, so:

F = T * sin(30) = 85 N * 0.5 = 42.5 N (approximately)

The tension in the string doesn't exactly match the options given, suggesting there might be a computation or conceptual error in the provided answer choices or the question details may be incomplete.