Finding weight of a beam using torque & equilibrium?

[chmergatroyd]

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
The actual answer is 52N[/B]

 

[TSny]

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.

 

[EMMAYAH]

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
The actual answer is 52N[/B

 

[EMMAYAH]

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

 

[got]

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
The actual answer is 52N[/B]

How will I get 52N

 

[haruspex]

How will I get 52N

Please post your attempt so that we can guide you.

 

[Mister T]

(28×2)+2.5F=10×3

Where does the 2.5 come from?

 

[berkeman]

How will I get 52N

You are replying to a 7 year old thread. If you have a new question or want to discuss the old problem, please start a new thread and show your work. Thank you. This old thread is now closed.