A 1240 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1970 N crate hangs from the far end of the beamThe angle between the beam and the horizontal is 30 degrees upwards and the angle between the horizontal and the cable attached to the wall is 50 degrees upward.

a) Calculate the magnitude of the tension in the wire.


(b) Calculate the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.
Fx =
Fy =

To solve this problem, we can start by drawing a diagram to visualize the situation.

First, we have a uniform beam attached to a vertical wall. One end of the beam is attached to the wall, and the other end is supported by a cable. The crate hangs from the far end of the beam.

From the given information, we know that the weight of the crate (W) is 1970 N and the angle between the beam and the horizontal is 30 degrees upwards. The angle between the horizontal and the cable attached to the wall is 50 degrees upwards.

(a) To calculate the magnitude of the tension in the wire, we need to consider the forces acting on the beam. There are three forces to consider:

1. The weight of the beam itself. We'll denote this as WB.
2. The weight of the crate, which is acting at the far end of the beam.
3. The tension in the wire, denoted as T.

Now, let's break down the forces acting on the beam:

1. The weight of the beam (WB) acts vertically downward and is equal to the mass of the beam (m) multiplied by the acceleration due to gravity (g).
WB = m * g

2. The weight of the crate (W) acts vertically downward at the far end of the beam.

3. The tension in the wire (T) acts at an angle with respect to the horizontal.

To find the magnitude of the tension in the wire, we can consider the equilibrium of the forces in the horizontal and vertical directions.

In the vertical direction, the sum of the vertical forces must be zero:
W + WB - Tsin(30) = 0

In the horizontal direction, the sum of the horizontal forces must also be zero, since the beam is in equilibrium with no horizontal acceleration:
Tcos(30) - Fx = 0

Solving these equations will give us the magnitude of the tension in the wire.

(b) To calculate the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam, we need to consider the forces acting on the beam at the attachment point to the wall.

Since the beam is in equilibrium, the sum of the vertical forces at the attachment point must be zero:
Fy - WB sin(30) = 0

The horizontal force exerted by the wall is equal in magnitude but opposite in direction to the horizontal component of the tension in the wire:
Fx = Tcos(50)

Solving these equations will give us the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.