# What are the cord tension and force on the hinge for this angled beam?

• hangingwire
In summary, the task is to find the cord tension and force on the hinge of a 10kg, 8.0m long beam, given a picture of the equation. The attempt at a solution involves using the "Z-rule" for angles and taking into account the weight of the beam acting at an angle. The final answer for the cord tension is 49N. Regarding forces, the weight of the beam is purely vertical while the other forces may not be.
hangingwire

## Homework Statement

Find cord tension and force on the hinge on this beam given the following:

Picture of the equation (in attachments)
beam is 10 kg
beam is 8.0m long

## Homework Equations

[T=0
[F=0

[h2]The Attempt at a Solution[/h2]

Finding the Torque on the wire...
I attached my FBD I drew. Before I start, one question? Since the beam is on an angle, would there be a horizontal force on the x-axis resulting from the Mass of the beam? If so, It could be 60 degrees? Mg Sin60?

That is my attempt using the "Z-rule" in angles

Anyways:

[T= 0
Tcw = Tccw
mg(d) = Tsin( ) * (d)
(10)(9.8)(4) = T sin30 * (8m) ----> Sin 30* given that directly across is 90 degrees + 60 degrees given = 30 degrees left at where the beam and wire are attached.

392N = T sin30 * (8m)

392N divided by 8m = 49

49 = T sin30

49 divided by sin30 --> 98?

The answer is 49N. Why did I get double the answer? Is there something with my angles?

#### Attachments

• Beam1.jpg
7 KB · Views: 466
• Beam2.jpg
4.5 KB · Views: 479
The weight of the beam acts at angle to the "arm", so its torque must involve the sine of the angle. Intuitively that should be fairly clear: the "more vertical" the beam gets, the easier it is to hold it in place.

So instead of "Mg" just being "Mg". It is "Mgsin(30)"? Is that what you are trying to say? If I change that in my equation it works out?

If that is the case, waste no time to say... Would the horizontal axis on Mg also require a MgCos(30) to workout the force from the hinge in components?

It should be fairly obvious that 49 sin 30 = T sin 30 implies T = 49.

Regarding the forces, the answer is no. If you you use the horizontal and vertical axes, then the weight is purely vertical - it always is; the other forces, however, are not.

Figured it out! Thanks

Last edited:

## What is the purpose of beams and wires at an angle?

The purpose of beams and wires at an angle is to provide structural support and stability to a building or structure. They can also be used to distribute weight and maintain the integrity of the structure.

## How are beams and wires at an angle designed?

Beams and wires at an angle are designed using mathematical calculations and engineering principles. The size, shape, and material of the beams and wires are carefully chosen to ensure they can withstand the forces and weight they will be subjected to.

## What are the different types of beams and wires at an angle?

There are several types of beams and wires at an angle, including diagonal beams, trusses, and suspension wires. Each type has its own unique design and purpose depending on the specific needs of the structure.

## What factors should be considered when installing beams and wires at an angle?

When installing beams and wires at an angle, factors such as the weight of the structure, the angle of the beams and wires, and the type of material being used must be carefully considered. It is important to ensure the beams and wires can support the weight and forces they will be subjected to.

## How are beams and wires at an angle maintained?

Beams and wires at an angle should be regularly inspected and maintained to ensure their structural integrity. This may include repairing or replacing damaged beams or wires, checking for corrosion or wear, and making any necessary adjustments to maintain the structure's stability.

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