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 beam.

(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 =

I assume that the beam is connected by a hinge at the wall (i.e. no bending moment is induced).

If the cable is vertical, there should be no horizontal force at the wall since ΣFx=0.

The load is applied at the "far end", is it the cable end or the wall end, the left end or the right end? Is the wall the right end?

Please understand that we do not see the figure you see, so a full description is naturally required.

To solve this problem, we can use the principles of equilibrium. In a state of equilibrium, the sum of the forces acting on an object is zero, and the sum of the torques (or moments) acting on an object is zero.

(a) To calculate the magnitude of 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 assume the length of the beam is L. The weight of the beam acts downward at its center, with a magnitude of 1240 N. The weight of the crate acts downward at the far end of the beam, with a magnitude of 1970 N.

To find the magnitude of the tension in the wire, we need to consider the vertical equilibrium of the beam. Since the beam is in equilibrium, the sum of the vertical forces acting on it must be zero.

The vertical forces acting on the beam are the weight of the beam (1240 N) and the weight of the crate (1970 N), both acting downward. The tension in the wire acts upward.

So, by summing up the forces in the vertical direction, we have:

Tension in wire - 1240 N - 1970 N = 0

From this equation, we can solve for the tension in the wire:

Tension in wire = 1240 N + 1970 N = 3210 N

Therefore, the magnitude of the tension in the wire is 3210 N.

(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 torque equation.

The torque about the left end of the beam is zero because the beam is in equilibrium. The torque is given by the product of the force and its perpendicular distance from the point of rotation (in this case, the left end of the beam).

Let's assume the distance from the left end of the beam to the point where the wire attaches is d.

The torque equation about the left end of the beam can be written as:

Torque = Force x Distance

The horizontal component of the force exerted by the wall doesn't produce any torque about the left end of the beam because the distance is zero.

So, the torque equation becomes:

0 = - Fy x d + F crate x L

Since we're interested in finding the horizontal and vertical components of the force exerted by the wall, we can rewrite the equation as:

Fy x d = F crate x L

Now, let's substitute the values we know:
- Fy is the vertical component of the force exerted by the wall, which is what we want to find.
- F crate is the weight of the crate, 1970 N.
- d is the distance from the left end of the beam to the point where the wire attaches.
- L is the length of the beam, which we assume to be known.

By rearranging the terms, we can solve for Fy:

Fy = (F crate x L) / d

Given the values from the problem, you can calculate the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.