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, let's break it down step-by-step:

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

We can start by resolving the forces acting on the beam. The weight of the crate (W) acts downwards and can be split into two components: one component parallel to the beam (Wx) and another component perpendicular to the beam (Wy).

Given that the angle between the beam and the horizontal is 30 degrees, we can calculate Wx and Wy:

Wx = W * cos(30°)
= 1970 N * cos(30°)
≈ 1703.97 N

Wy = W * sin(30°)
= 1970 N * sin(30°)
≈ 985 N

Since the beam is in equilibrium, the sum of the vertical forces must be zero. So, the total force acting upwards must equal the total force acting downwards:

T + Wy = 1240 N

Solving for T, we have:

T = 1240 N - Wy
= 1240 N - 985 N
≈ 255 N

Therefore, the magnitude of the tension in the wire is approximately 255 N.

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

To find the horizontal component (Fx) and vertical component (Fy) of the force that the wall exerts on the left end of the beam, we need to consider the forces acting on the beam.

The force exerted by the wall on the beam (Tension) can be resolved into two components: one perpendicular to the beam (Tension Y) and another parallel to the beam (Tension X).

Given that the angle between the cable attached to the wall and the horizontal is 50 degrees upward, we can calculate Tension X and Tension Y:

Tension X = T * sin(50°)
= 255 N * sin(50°)
≈ 196.85 N

Tension Y = T * cos(50°)
= 255 N * cos(50°)
≈ 163.23 N

Hence, the magnitude of the horizontal component (Fx) is approximately 196.85 N and the magnitude of the vertical component (Fy) is approximately 163.23 N.

To solve this problem, we can split it into two parts. First, let's calculate the tension in the wire.

(a) Magnitude of the tension in the wire:

To find the tension in the wire, we can consider the forces acting on the beam. There are three forces acting on the beam: the weight of the beam itself, the weight of the crate, and the tension in the wire.

Let's break down the forces acting on the beam:

1. The weight of the beam: This force acts vertically downward at the center of the beam and has a magnitude of 1240 N. We can consider this force to be acting at the midpoint of the beam.

2. The weight of the crate: This force acts vertically downward at the far end of the beam and has a magnitude of W = 1970 N.

3. The tension in the wire: This force acts horizontally at the far end of the beam in the direction towards the wall. We need to find its magnitude.

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

Since the beam is in equilibrium, the sum of the vertical forces acting on the beam should be zero.

1. The weight of the beam (1240 N) acts downward.
2. The weight of the crate (1970 N) acts downward.

Therefore, the vertical component of the tension in the wire should be equal to the sum of these two forces:

Vertical Component of Tension = Weight of Beam + Weight of Crate
= 1240 N + 1970 N
= 3210 N

Now, let's analyze the forces acting on the beam in the horizontal direction. Again, the beam is in equilibrium, so the sum of the horizontal forces acting on the beam should be zero.

1. The horizontal component of the tension in the wire acts towards the wall.

Therefore, the horizontal component of the tension in the wire should be equal to the horizontal force exerted by the wall:

Horizontal Component of Tension = Horizontal Force Exerted by Wall

To find the horizontal component of the tension in the wire, we can use trigonometry.

Using the given angles, the horizontal component of the tension in the wire can be calculated as:

Horizontal Component of Tension = Tension in Wire * cos(angle between the horizontal and the cable attached to the wall)
= Tension in Wire * cos(50 degrees)

Since the angle between the beam and the horizontal is 30 degrees, we have:

Tension in Wire * sin(30 degrees) = Vertical Component of Tension
Tension in Wire * sin(30 degrees) = 3210 N

Solving this equation for Tension in Wire, we get:

Tension in Wire = 3210 N / sin(30 degrees)
= 6420 N

Therefore, the magnitude of the tension in the wire is 6420 N.

(b) Magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam:

To find the components of the force that the wall exerts on the left end of the beam, we can use trigonometry.

The horizontal component of the force exerted by the wall can be calculated using the following equation:

Horizontal Component of Force = Tension in Wire * sin(angle between the horizontal and the cable attached to the wall)
= 6420 N * sin(50 degrees)

The vertical component of the force exerted by the wall can be calculated using the following equation:

Vertical Component of Force = Tension in Wire * cos(angle between the horizontal and the cable attached to the wall)
= 6420 N * cos(50 degrees)

Therefore, the magnitude of the horizontal component of the force that the wall exerts on the left end of the beam is 6420 N * sin(50 degrees).

And the magnitude of the vertical component of the force that the wall exerts on the left end of the beam is 6420 N * cos(50 degrees).