Boxes, cables, and girders (torque and equilibirum problem)

[SageOwl]

Usually I'm pretty good with torque and such but this one had me.
A box of mass 75 kg is put on a 10-m long steel beam of mass 150 kg and is connected to the wall and supported by a steel cable. The box is located 2.5 m from the wall and the cable makes 60o angle with the beam. What are the magnitude and the direction of the force exerted by the wall on the beam?

Homework Equations

Torque=Radius x Force, upwards/clockwise torque=downwards/counterclockwise torque, weight=mass x gravity

The Attempt at a Solution

I was able to find the tension in the cable, if that helps (14321.475N), but otherwise I'm clueless.

 

[Delphi51]

Kind of odd using a 10 m long beam to support a box only 2.5 m away from the wall. Or have I misunderstood?

The usual approach is to write that the sum of the moments about a point (in this case the point where the beam attaches to the wall) is zero. I can't go much further until you give it a try and show your work here.

 

[tomcue]

Hello,

It seems like you have made some progress in solving this problem by finding the tension in the cable. That is a good start! To find the magnitude and direction of the force exerted by the wall on the beam, we will need to use the concept of torque and equilibrium.

First, let's draw a diagram to visualize the situation. We have a box on a beam, connected to a wall by a cable at an angle of 60 degrees. The box is located 2.5 m from the wall. The beam is 10 m long and has a mass of 150 kg. The box has a mass of 75 kg.

Now, let's consider the forces acting on the beam. We have the weight of the beam acting downwards, the weight of the box acting downwards, and the tension in the cable acting upwards. We also have the force exerted by the wall on the beam, which we will call F.

To find the magnitude of F, we need to consider the torque acting on the beam. Torque is defined as the product of the force and the distance from the pivot point. In this case, the pivot point is where the beam meets the wall. Since the beam is in equilibrium, the sum of all torques acting on it must be zero.

We can start by considering the torque due to the weight of the beam. The weight of the beam acts downwards, at a distance of 5 m from the pivot point (half of the length of the beam). So the torque due to the weight of the beam is (150 kg x 9.8 m/s^2) x 5 m = 7350 Nm.

Next, we can consider the torque due to the weight of the box. The weight of the box acts downwards, at a distance of 7.5 m from the pivot point (2.5 m from the wall plus half of the length of the beam). So the torque due to the weight of the box is (75 kg x 9.8 m/s^2) x 7.5 m = 5512.5 Nm.

Now, let's consider the torque due to the tension in the cable. The tension in the cable acts upwards, at a distance of 2.5 m from the pivot point (where the cable connects to the beam). So the torque due to the tension in the cable is (14321.475 N) x