What is the equation for solving static equilibrium questions?

[XodoX]

I do not understand how to solve static equilibrium questions. Like those:

The figure shows a uniform, horizontal beam (length = 10 m, mass = 25 kg) that is
pivoted at the wall, with its far end supported by a cable that makes an angle of 51o with
the horizontal.(mass = 60 kg) stands 3.0 m from the pivot, what is the tension in the
cable?

Or this one

http://www.chegg.com/homework-help/questions-and-answers/a-uniform-120-lb-beam-is-supported-in-a-horizontal-positionby-a-pin-and-cable-as-shown-in--q701621

The second one with the pin is especially hard, imo.

I don't even know what the basic equation here is. The book dosen't tell you this. It's always \Sigma\tau=... something else. So I do have to create my own equation each time, basically. Is this correct? Depending on whatever the problem is. I don't get it. Can someone please explain this to me?

 

[tal444]

For these types of questions, essentially you make the torque provided by the cable equal to the torque caused by the weight of the beam. This must be true, because otherwise the beam would be rotating on its pivot. I can't really help you with anything until you show me an attempt at a solution.

 

[XodoX]

For these types of questions, essentially you make the torque provided by the cable equal to the torque caused by the weight of the beam. This must be true, because otherwise the beam would be rotating on its pivot. I can't really help you with anything until you show me an attempt at a solution.

I already have the solution... that's not my question. I'm wanting to know how to solve those kind of problems/concepts.

 

[tal444]

I essentially just told you in my last post. Think of torque as acting either clockwise or counterclockwise about the pivot. Take the perpendicular components and make them equal each other. Solve for missing variable. If you're looking for a basic equation, there isn't really one. All torque questions are based off the fact that the sum of torques must be equal to zero.

 

[XodoX]

I essentially just told you in my last post. Think of torque as acting either clockwise or counterclockwise about the pivot. Take the perpendicular components and make them equal each other. Solve for missing variable. If you're looking for a basic equation, there isn't really one. All torque questions are based off the fact that the sum of torques must be equal to zero.

What are the perpendicular components for the 1st example? Torque(0)=Tsin51*length of beam -weight of person*meters from the pivot - weight of beam*(I don't know). Don't know what the last component in this equation would be. But then I would solve it for T. I have no idea if this equation would get me the tension of the cable., but that's what I can come up with.

 

[tal444]

You've got it mostly right. The beam's center of gravity is in the middle, so how far away from the pivot is it acting on?

 

[XodoX]

Well, since it's in the middle, it's 5m.

 

[tal444]

Yes. With that information, you should be able to solve for T.

 

[XodoX]

But that is the tension of the cable. How would I do the second one? How do I know how to approach this ?

 

[tal444]

For the second question, the cable is acting at an angle right? This means that the cable has a horizontal force. Since the beam is not moving, therefore the pin must be supplying a horizontal force that either pushes out against the beam or pulls it toward the wall to stop it from moving sideways.

 

[XodoX]

For the second question, the cable is acting at an angle right? This means that the cable has a horizontal force. Since the beam is not moving, therefore the pin must be supplying a horizontal force that either pushes out against the beam or pulls it toward the wall to stop it from moving sideways.

I don't think I fully understand this, but would it look like this?

Basically the same I did above but just for this problem?

 

[tal444]

Find the horizontal component of the force that the cable is pulling on the beam. The force that the pin provides will be equal to that.