A 1210 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1820 N crate hangs from the far end of the beam. The 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 find the magnitude of the tension in the wire, you can use the concept of equilibrium. In equilibrium, the sum of the forces acting on an object is zero.

(a) To begin, let's draw a free-body diagram of the beam.

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| Tension (T)
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______| Wall
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Here, T represents the tension in the wire, and the vertical line represents the beam. Since the beam is uniform, its weight acts at its center of mass, which is at the midpoint. The weight of the beam can be calculated as:

Weight of the beam = Mass of the beam * g
= (1210 N) * 9.8 m/s^2
= 11858 N

In equilibrium, the sum of the vertical forces must be zero. Considering upward as positive, we have:

T - 11858 N * sin(30) = 0

Now, you can solve for the tension in the wire:

T = 11858 N * sin(30)
T = 5929 N

So, the magnitude of the tension in the wire is 5929 N.

(b) To find the horizontal and vertical components of the force that the wall exerts on the left end of the beam, we need to consider the sum of the forces in both directions and calculate the angle between the wall and the beam.

Let's call the horizontal component Fx and the vertical component Fy. Since the beam is in equilibrium, the sum of the horizontal forces is zero:

Fx - T * cos(50) = 0

Now, you can solve for Fx using the previously calculated tension in the wire:

Fx = T * cos(50)
Fx = 5929 N * cos(50)
Fx ≈ 3838 N

Therefore, the magnitude of the horizontal component of the force exerted by the wall on the left end of the beam (Fx) is approximately 3838 N.

To find the vertical component Fy, we consider the sum of the vertical forces:

Fy + T * sin(50) - (Weight of the crate) = 0

Now, you can solve for Fy:

Fy = (Weight of the crate) - T * sin(50)
Fy = 1820 N - 5929 N * sin(50)
Fy ≈ 68.6 N

Thus, the magnitude of the vertical component of the force exerted by the wall on the left end of the beam (Fy) is approximately 68.6 N.