Question

A 25kg mass is suspended at the end of a horizontal, massless rope that extends from a wall on the left and from the end of a second massless rope connected to a wall on the right at an angle of 130* from the horizontal rope (up 50.0* from the horizontal). What are the tensions in the ropes?

Answer 1

To solve this problem, we can use the fact that the sum of the forces acting on an object must be zero in order for the object to be in equilibrium (not moving). In this case, the forces acting on the 25 kg mass are the tension in the horizontal rope, the tension in the angled rope, and the gravitational force acting downward.

We can write an equation representing the sum of the forces in the vertical direction, which is equal to zero because the mass is not moving vertically. This equation will have the form T1 * sin(130) - T2 * sin(50) - 25 * g = 0, where T1 and T2 are the tensions in the ropes, g is the gravitational acceleration (9.8 m/s^2), and the sine functions represent the vertical components of the forces.

Solving for T1 and T2, we find that T1 = (25 * g) / sin(130) = 178.8 N and T2 = (25 * g) / sin(50) = 107.5 N. These are the tensions in the ropes.