Minimizing Tension in Horizontal Girder w/ Suspended Objects

[Tonyt88]

Homework Statement

A heavy horizontal girder of length L has several objects suspended from it. It is supported by a frictionless pivot at its left end and a cable of negligible weight that is attached to an I-beam at a point a distance h directly above the girder's center. Where should the other end of the cable be attached to the girder so that the cable's tension is a minimum?

Homework Equations

F = Ma
Torque = I(alpha)

The Attempt at a Solution

In the y-axis I have:

mg = T sin(theta)

N = T cos(theta)

where do I go from here?

 

[OlderDan]

How have you defined theta, what is m in mg, and where are you calculating the torque? Where do h and L enter the problem?

 

[Tonyt88]

I have defined theta as the angle formed by the girder and the wire, m should be the scattered mass since it is composed of several objects spread about, I gander I should calculate torque from the pivot which is where h and L enter I assume? I guess a big question is how do I handle the fact that the mass is scattered when calculating the torque?

 

[OlderDan]

I have defined theta as the angle formed by the girder and the wire, m should be the scattered mass since it is composed of several objects spread about, I gander I should calculate torque from the pivot which is where h and L enter I assume? I guess a big question is how do I handle the fact that the mass is scattered when calculating the torque?

No matter how the mass is scattered, there is net torque do to all that mass tending to make the girder rotate downward. All of that downward rotational tendency has to be overcome by the torque from the cable. Surely the tension will depend on how much torque is needed from the cable, but does the point of connection that will minimize the tension depend on how much torque has to be overcome?

 

[Tonyt88]

Since you pose the question, it seems as if the answer to your question is no, but why then would there be no dependency on how much torque is to be overcome?

 

[Tonyt88]

Wait, would the torque just be:

(T)(L/2 + x) sin θ = R where R is some constant (that is the torque of the scattered mass)

And the angle would be:
sin θ = h/[x^2 + h^2]^2 = 1/[(x/h)^2 + 1]^1/2

Thus I would have to maximize:
(L/2 + x]/[(x/h)^2 + 1]^1/2

or is this incorrect?

 

[OlderDan]

Wait, would the torque just be:

(T)(L/2 + x) sin θ = R where R is some constant (that is the torque of the scattered mass)

And the angle would be:
sin θ = h/[x^2 + h^2]^2 = 1/[(x/h)^2 + 1]^1/2

Thus I would have to maximize:
(L/2 + x]/[(x/h)^2 + 1]^1/2

or is this incorrect?

That's the right idea. You have a ^2 there that should be a ^(1/2). But you do want to maximize the factor multiplying T to minimize T. You are correct that R is some constant determined by the mass distribution of stuff hanging on the girder. It makes no difference what R is; the position to attach the cable to minimize T is the same.