# Tension and Force in pulley system

• Plasmosis1
Your FBD looks good. Continue with the *hint* regarding how to orient your coordinate system for your FBD.

## Homework Statement

Due to settlement of soil, a recently planted tree has started to lean. To straighten it, the cable system shown is used, where a turnbuckle on cable AB is periodically tightened to keep the cable taut as the tree gradually straightens. If the force in cable AB is 395 N, determine the force in cable CBD.
(see attached diagram)

Fx=F*cos(θ)
Fy=F*sin(θ)

## The Attempt at a Solution

*hint* A FBD of the pulley at B is recommended to begin this problem. Isolating pulley B from its surrounding requires a closed surface passing through cables AB and BCD (along both segments BC and BD). The problem has a symmetry that becomes apparent with an orthogonal coordinate system having one axis aligned with AB.

What does the hint mean by "closed surface passing through cables AB and BCD"?
I don't know what the question means by the force in cable CBD. Is it referring to total tension or something?

#### Attachments

• IEA1.PNG
7.9 KB · Views: 1,997
• IEA2.png
2.9 KB · Views: 1,454
Last edited:
Plasmosis1 said:
What does the hint mean by "closed surface passing through cables AB and BCD"?
Imagine drawing a box around the pulley. Your "system" is everything in the box, which of course includes the pulley itself. The forces on the box are due to the various tensions in the cables attached to it.

I don't know what the question means by the force in cable CBD. Is it referring to total tension or something?
Just the tension in that cable. (Not sure what "total" tension would mean.)

Plasmosis1 said:
What does the hint mean by "closed surface passing through cables AB and BCD"?

It's just a way of visually isolating the pulley. As long as you can draw a FBD for the pulley, you can forget about the closed surface.

I don't know what the question means by the force in cable CBD. Is it referring to total tension or something?

The "force in cable CBD" is the tension in the cable. I'm not sure what you mean by "total" tension. The cable has a tension which you can assume has the same value at each point of the cable.

[EDIT: I see Doc Al already replied while I was still typing.]

Ok but that still doesn't help me with solving for the force in CBD. Can someone walk me through the steps please?

Plasmosis1 said:
Ok but that still doesn't help me with solving for the force in CBD.
Did you draw a FBD for the pulley? What forces act on it? What are their directions?

I just attached a FBD of the pulley but it doesn't really help much. I don't know any of the angles for BD so I don't know how to account for the forces with it.

Your FBD looks good. Continue with the *hint* regarding how to orient your coordinate system for your FBD.

Ok so AB bisects angle CBD. If you make AB the x-axis CB and DB will be 28-11=17 degrees above and below the x-axis. The two x-component of the two will add up to 395 so the x component of one of them is 395/2=197.5. So the tension is 197.5/cos(17)=207N.
Thanks!