A 180-kg uniform horizontal beam is attached to a vertical wall at one end and is supported by a vertical

cable at the other end making an angle of 30 degree above the x axis. What is the magnitude of the tension in the cable?

To find the magnitude of the tension in the cable, we can use the principles of equilibrium. Let's break down the forces acting on the beam:

1. Weight: The weight acts vertically downward and can be calculated as W = m * g, where m is the mass of the beam and g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, the weight of the beam is W = 180 kg * 9.8 m/s^2 = 1764 N.

2. Tension: The tension in the cable acts at an angle of 30 degrees above the x-axis. Let's denote the tension as T.

3. Horizontal Component of Tension: T_h = T * cos(30°)
Vertical Component of Tension: T_v = T * sin(30°)

Now, let's consider equilibrium in the horizontal direction:
The horizontal forces acting on the beam are the horizontal component of the tension (T_h) and there is no other horizontal force acting. Therefore, T_h = 0.

Next, let's consider equilibrium in the vertical direction:
The vertical forces acting on the beam are the vertical component of the tension (T_v) and the weight (W). Therefore, the upward force (T_v) must balance the downward force (W). This gives us the following equation:
T_v - W = 0

Substituting the values, we get:
T * sin(30°) - 1764 N = 0

Now we can solve for T, the tension in the cable:
T = 1764 N / sin(30°)
T ≈ 3528 N

Therefore, the magnitude of the tension in the cable is approximately 3528 N.