Solve Static Equilibrium: Beam, 520N Force, Cable Tension

AI Thread Summary
A uniform vertical beam with a mass of 40kg is subjected to a horizontal force of 520N at its top and is supported by a cable at a height of 3m, forming a 28-degree angle with the ground. The discussion revolves around drawing a free body diagram, calculating cable tension, and determining ground reaction forces. Participants note confusion regarding the torque calculations, particularly how the applied force affects the equilibrium equations. There is also uncertainty about whether friction should be considered, as the problem does not specify it. The conversation emphasizes the importance of accurately accounting for all forces and moments to solve the static equilibrium problem effectively.
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Homework Statement


A uniform vertical beam of mass 40kg is acted on by a horizontal force of 520N at its top and is held, in the vertical position, by a cable as shown.
a) Draw a free body diagram for the beam, clearly labelling all the forces acting on it
b) Calculate the tension in the cable.
c) Determine the reaction forces acting on the beam by the ground.

Homework Equations



ΣFx = 0

ΣFy = 0

Στ = 0

The Attempt at a Solution


a) I can't seem to add images to this. Whenever I try it only gives me the option of adding an image url which I don't have. Maybe someone can tell me how to do that!

For the image provided the uniform beam is standing up right (height of 5m). Force applied (520 N) acts on the top end of the beam to the left. A cable attaches the beam from 3m height down to the ground on the right. The angle between the ground and the cable is 28degrees.

In my free-body diagram I have the following forces acting on the beam.
Fa = 520N
Force tension from the cable
mg
Fn

Am I missing a force here? Should there be a force of friction between the beam and the ground? The question does not explicitly state anything about that.

b)
ΣFx = Ftx - Fa = 0

ΣFy = Fn-mg-Fty=0

Στ = Fa * 2 = 0
where the axis is located where the cable attaches to the beam at 3m. Ft does not create a torque force as it acts at the axis, and mg and Fn also do not create a torque force as the act directly away from the axis

I get here and I realize something is missing...based on the torque equation Fa would = 0, but Fa is given as 520N. Here's where I'm confused.

Please help!

Thanks
 
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To add an image, you can try putting your image in a Word document, say, and then "cut" the image from the document and "paste" the image into your post. Or, when entering a post, you should see an "upload" button that you can use to upload a picture that is stored as a file on your computer.

You can see that your torque equation leads to a contradiction. So, maybe you were on the right track when you were wondering if friction between the beam and the ground should be included. Note that the question asks for the reaction forces on the beam by the ground.
 
Use the upload button then the insert thumbnail or full size.
 
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