Solving Tension of a Wire Problem

[dzidziaud]

Homework Statement

This is actually a sound wave problem, but I think I'll be fine when I actually get to that part; my issue is that it is a cumulative problem that involves torque, which I haven't had practice with since the fall. It's embarrassing how little I remember how to do from just a few months ago. Can you look at what I'm doing and make sure I'm heading in the right direction?

Problem:
A thin wire of mass m and length l is suspended between two beams of mass M and length L that are attached to the ground with hinges as shown in the figure. The system is symmetric such that the wire is horizontal and the two beams each make an angle θ with the ground.
The problem then asks for the frequency and various other things that I'll know how to do if I get the tension right.
I couldn't figure out how to copy or attach the figure, but this system basically forms a trapezoid with the top (the wire) being longer than the bottom, so the wire is holding up the two beams.

The Attempt at a Solution

The mass of each beam is being supported in part by both the ground and the wire, so I tried to find each of these components to get the mass being supported by just the wire to get the tension.
cosθ = mass_horizontal / mass_total
And the horizontal component will give me the tension, right? And the length of the beam L needs to be relevant somehow, so I figure that the tension needs to come from the torque, so I multiply by L. And since there are two beams, I also multiplied by 2 to give me:
tension = 2MgLcosθ
Is that at all correct?

 

[SteamKing]

No figure attached.

 

[dzidziaud]

I know; I described it instead. It is a trapezoid with a longer top than bottom. I guess I should also say that θ is each outside (acute) angle that the beams make with the ground.

 

[PhanthomJay]

You need to restudy your equilibrium equations, as you presently are just taking a stab in the dark. Try looking at one beam only and draw a free body diagram of it if you remember how. The tension force acts horizontally at the top, the beam weight acts down at its mass center, and there are horizontal and vertical reaction forces at the hinge. Use the 3 equilibrium equations to solve for the unknowns. Remember Torque is force times perpendicular distance to pivot, or use cross product rule.

 

[dzidziaud]

You need to restudy your equilibrium equations, as you presently are just taking a stab in the dark. Try looking at one beam only and draw a free body diagram of it if you remember how. The tension force acts horizontally at the top, the beam weight acts down at its mass center, and there are horizontal and vertical reaction forces at the hinge. Use the 3 equilibrium equations to solve for the unknowns. Remember Torque is force times perpendicular distance to pivot, or use cross product rule.

Ah. The clockwise torque should equal the counterclockwise torque. Would counterclockwise be MgL? And clockwise would be... the tension? That seems much too simple. :(

 

[PhanthomJay]

Ah. The clockwise torque should equal the counterclockwise torque. Would counterclockwise be MgL? And clockwise would be... the tension? That seems much too simple. :(

Sorry to leave you hanging dz*dz*
Torque is force times perpendicular distance from the line of action of the force to the pivot. The weight force, mg, acts through the center of gravity of the beam, downward, through its midpoint. Draw a straight vertical line through that force extending downward to the base level. That line is the 'line of action' of the weight force. Then you can get the perpendicular distance from that line to the pivot using simple trig...prove to yourself that the perpendicular distance is (L/2)(cosΘ), and thus the torque is (mg)(L/2)(cosΘ), ccw.
Now using the same principles, determine the cw torque of the tension force about the pivot. Set cw torque = ccw torque, and solve for T.