# Minimizing Tension in Horizontal Girder w/ Suspended Objects

• Tonyt88
In summary, the problem involves finding the position to attach a cable to a horizontal girder with a pivot at one end and a cable attached to an I-beam at a point above the girder's center, so that the tension in the cable is minimized. Torque is calculated from the pivot and the angle formed by the girder and the cable. The angle is determined by the distance h and the unknown position x where the cable is attached. The tension in the cable can be minimized by maximizing the factor multiplying T in the torque equation. The value of R, determined by the mass distribution, does not affect the position to attach the cable.
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?

Last edited:
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?

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?

Tonyt88 said:
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?

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?

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?

Tonyt88 said:
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.

## 1. How does the weight of the suspended objects affect the tension in the horizontal girder?

The weight of the suspended objects directly affects the tension in the horizontal girder. The more weight that is added, the greater the tension will be in the girder. This is because the girder must support the weight of the objects and distribute it evenly along its length.

## 2. What factors contribute to the tension in a horizontal girder?

There are several factors that contribute to the tension in a horizontal girder. These include the weight of the suspended objects, the length and material of the girder, and any external forces such as wind or seismic activity. The design and construction of the girder also play a role in determining its tension.

## 3. How can the tension in a horizontal girder be minimized?

There are a few methods for minimizing tension in a horizontal girder with suspended objects. One approach is to evenly distribute the weight of the objects along the length of the girder, using multiple attachment points if necessary. Another method is to use a stronger and stiffer material for the girder, which can better withstand the tension. Proper design and construction techniques are also crucial in minimizing tension.

## 4. Can the suspended objects be rearranged to reduce tension in the horizontal girder?

In some cases, rearranging the suspended objects can help to reduce tension in the horizontal girder. For example, if the objects are currently clustered at one end of the girder, spreading them out evenly can distribute the weight more evenly and reduce tension. However, this may not always be possible or practical depending on the type and placement of the objects.

## 5. How does the angle of the suspended objects affect the tension in the horizontal girder?

The angle of the suspended objects can have a significant impact on the tension in the horizontal girder. Objects that are suspended at an angle have a greater downward force on the girder, leading to higher tension. It is important to consider the angle of the suspended objects when designing and constructing the girder to minimize tension.

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