# Finding weight of a beam using torque & equilibrium?

• chmergatroyd
In summary, you need to add 2.5 to the weight force of a beam that is suspended by a cord at X which is 2m from one end of the beam and 3m from the other end.f

## Homework Statement

A uniform beam at equilibrium is suspended by a cord at X which is 2m from one end of the beam and and 3m from the other end. There is a mass of 28N attached to the 2m end and a mass of 10N on the 3m end. Determine the weight force of the beam

## Homework Equations

Tclockwise=Tanticlockwise. T=Fd.[/B]

## The Attempt at a Solution

(28×2)+2.5F=10×3
56+2.5F=30
2.5F=-26
F=-10.4N

EMMAYAH
Welcome to PF!

Be sure you are clear on which point you are choosing as the origin for calculating the torques. The torques for each force must be determined using the same origin.

## Homework Statement

A uniform beam at equilibrium is suspended by a cord at X which is 2m from one end of the beam and and 3m from the other end. There is a mass of 28N attached to the 2m end and a mass of 10N on the 3m end. Determine the weight force of the beam

## Homework Equations

Tclockwise=Tanticlockwise. T=Fd.[/B]

## The Attempt at a Solution

(28×2)+2.5F=10×3
56+2.5F=30
2.5F=-26
F=-10.4N

T
Thanks I just confirmed it and it turns out to be correct.

## Homework Statement

A uniform beam at equilibrium is suspended by a cord at X which is 2m from one end of the beam and and 3m from the other end. There is a mass of 28N attached to the 2m end and a mass of 10N on the 3m end. Determine the weight force of the beam

## Homework Equations

Tclockwise=Tanticlockwise. T=Fd.[/B]

(28×2)+2.5F=10×3
56+2.5F=30
2.5F=-26
F=-10.4N