# Solving Beam Hinge Torque for Horizontal Force

• NAkid
In summary, the horizontal force at pivot is equal and opposite to the horizontal force of tension. So, the hinge exerts a force on the beam of (L/2)mgcos(14o), or -0.59N.
NAkid

## Homework Statement

A 35.0kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of theta=14.0o with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam? (Use the `to the right' as + for the horizontal direction.)

## The Attempt at a Solution

i know that the horizontal force at pivot must be equal and opposite to horizontal force of tension. so for torques i have
LTsin(theta) = (L/2)mgcos(theta)
Ls cancel
so Tsin(theta) = (1/2)mgcos(theta)

but I don't know what I'm solving for exactly.. when I draw the free body diagram I have that F(horiz)=Fcos(theta), which seems to contradict my equations... help please!

#### Attachments

• prob19a_beamhinge1.gif
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NAkid said:
LTsin(theta) = (L/2)mgcos(theta)

About which point are you taking the moments?

but I don't know what I'm solving for exactly.. when I draw the free body diagram I have that F(horiz)=Fcos(theta),...

What is F?

First label the forces properly. Resolve into horizontal and vertical components. Without a neat diagram, you will keep making mistakes. Then take the torques about a suitable point.

Draw a complete diagram!

Hi NAkid!

Shooting Star is right.

You must draw in the force F at the hinge, and an angle (phi, perhaps?) for it.

And the tension force and the weight force.

It's no good keeping half the diagram in your head - "Without a neat diagram, you will keep making mistakes", and even miss some things completely.
NAkid said:
i know that the horizontal force at pivot must be equal and opposite to horizontal force of tension.

Yes, and that's a linear force equation, not a torque, isn't it?

And you haven't written it down, anyway!
so for torques i have
LTsin(theta) = (L/2)mgcos(theta)
Ls cancel
so Tsin(theta) = (1/2)mgcos(theta)

I'm not sure what you've done here.

After you've written the linear force equation, do the torque about the hinge, and that should immediatly give you the value of T.

And then …

## 1. What is beam hinge torque?

Beam hinge torque refers to the twisting force experienced by a beam at its point of connection to a hinge. This torque is caused by a horizontal force acting on the beam and can be calculated using mathematical equations.

## 2. Why is it important to solve for beam hinge torque?

It is important to solve for beam hinge torque in order to ensure the structural stability and safety of the beam and its supporting structure. Calculating the torque allows engineers to determine the appropriate size and strength of the hinge and other supporting elements.

## 3. What factors affect the calculation of beam hinge torque?

The factors that affect the calculation of beam hinge torque include the magnitude and direction of the horizontal force, the length of the beam, and the properties of the materials used for the beam and hinge.

## 4. What equations are used to solve for beam hinge torque?

The equations used to solve for beam hinge torque depend on the type of beam and the complexity of the supporting structure. However, the general equation for calculating torque is T = F x d, where T is the torque, F is the horizontal force, and d is the distance from the force to the point of rotation.

## 5. Are there any software programs available for solving beam hinge torque?

Yes, there are various software programs available for solving beam hinge torque, such as CAD and structural analysis software. These programs use advanced mathematical algorithms to accurately calculate the torque and provide detailed results for engineers to analyze and use in their designs.

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