# Calculating Tension in a horizontal cable

1. Dec 19, 2008

### rml4589

1. The problem statement, all variables and given/known data
The beam is 4.0 m long and has a mass of 14.0 kg. If the beam makes an angle of 41.0 degrees with the wall, what is the tension in the cable?

The attachment is the picture to go along with visualizing the problem.

2. Relevant equations

sum forces = 0
sum torqes = 0

3. The attempt at a solution

ok so since the object is not moving, both the net force and net torque = 0. I drew a FBD with the arrows indicated in the attachment.

would I have to figure out the horizontal force from the beam and subtract it from the Tension of the cable which would be something like

cos49 = adjacent / (3.02 / cos41) = 2.63

That would be the length of the cable.

how can i calculate the tensions? I'm kinda lost at this point. thanks for your help

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Last edited: Dec 19, 2008
2. Dec 19, 2008

### Staff: Mentor

Not quite sure what you're doing there.

Hint: If you choose your pivot point wisely, you can solve for the tension in the cable in one step by setting net torque = 0.

3. Dec 19, 2008

### rml4589

i was trying to figure it out using the net forces = 0.

if i chose the pivot point being the point where the beam touches the wall, would i have to figure it out in the x and y directions? or could i just take the d to be the length of the beam?

4. Dec 19, 2008

### Staff: Mentor

That's not enough. Whenever you have an extended body (like a beam), be ready to use torques.

Good choice.
I don't understand what you're saying here. (What's d?)

About that pivot point, what torques act on the beam?

5. Dec 19, 2008

### rml4589

the only torque about that point would be the torque of the beam right?

so the torque of the beam would be offset by the tension in the cable?

sooo if i fiind the torque of the beam, that will equal the tension in the cable

or am i making up my own laws of physics here?

6. Dec 19, 2008

### Staff: Mentor

By "torque of the beam" I think you mean the torque due to the weight of the beam. That's one torque. The tension creates another torque.

The torque due to the weight will be balanced by the torque due to the tension.

Torque and force (such as tension) are different things with different units, so a torque cannot equal a force. (Of course they are closely related.)

Just a little.

7. Dec 19, 2008

### rml4589

ok i think i know what you are saying.

the torque of the cable will be equal but opposite the torque of the beam, causing a net torque of 0. using the torque of the cable i should be able to figure out the tension of the cable. since tension is just a force, and force is just ma, in this case a = 9.8, Fbeam * rbeamsin(theta) = -Fcable * rcable

since sin(90) = 1, that doesn't need to be taken into accord for the cable.

Is that equation right so far?
i would just solve for Fcable and that should be the tension.

EDIT:
would it be this equation instead?

Fbeam*rbeamsin41 = Fcable*rcable*sin49?

i realized the angle is probably 49 and not 90. i was looking at the incorrect place

Last edited: Dec 19, 2008
8. Dec 19, 2008

### Staff: Mentor

Good!
Forget that last part about a = 9.8. Nothing's accelerating here.

Why do you think the angle is 90 degrees?

Put that sinθ back for the tension force.
Yes, once you fix the equation you can solve for the tension.

9. Dec 19, 2008

### Staff: Mentor

Much better! (What are rbeam and rcable equal to?)

10. Dec 19, 2008

### rml4589

Rbeam = 2m
Rcable = 3.02/sin41 = 4.6 / 2 = 2.3

so it would be, with everything plugged in

Fbeam*Rbeam*sin(41) = -Fcable*Rcable*sin(49)

and therefore

137.2N * 2m * .656 deg. = -Fcable * 2.3m * sin(49)

solving for Fcable

Fcable = -(137.2N * 2m * .656 deg.) / (2.3m * .755 deg.)

Fcable = -103.7 N

or would it be positive?

11. Dec 19, 2008

### Staff: Mentor

Good. The weight acts in the center of the beam.
No. The tension acts at the end of the beam, so Rcable should equal the full length of the beam.

Get rid of the minus sign. What you're really doing is setting clockwise torques (from the weight) equal to counter-clockwise torques (from tension).

12. Dec 19, 2008

### rml4589

ok so it would be

137.2N * 2m * .656 deg = Fcable * 4.6 * .755

Fcable = (137.2N*2m*.656) / 4.6 * .755
Fcable = 51.83N ?!?!?

13. Dec 19, 2008

### Staff: Mentor

How did you get 4.6? The beam is only 4 m long.

14. Dec 19, 2008

### rml4589

4.6 was the length of the cable not the beam.

since i know the angle between the wall and the beam, and i knew the length of the beam (the hypotenuse), i used sin41 = Lcable / 4....

which i just calculated to be 2.62 haha wow strange mistake there....

ok so 137.2N * 2m * .656 = Fcable * 2.62 * .755
Fcable = (137.2N*2m*.656) / (2.62 * .755) = 90.999 so 91 N?

15. Dec 19, 2008

### Staff: Mentor

The length of the cable is not what you need.

Since you are finding torques on the beam, all distances are along the beam. R is the distance from pivot to the point of application of the force on the beam. Since the weight acts at the middle of the beam, Rweight = 2m (half the length of the beam). The tension acts at the end of the beam, so what must Rtension equal?

Note: It might be less confusing if you used Fweight instead of Fbeam, since both forces act on the beam. Similarly, use Rweight instead of Rbeam.

16. Dec 19, 2008

### rml4589

ok you lost me :(

so Rweight is equal for both sides of the equation as well as Fweight?

or is Rtension = Rbeam?

i'm confusing myself i think

17. Dec 19, 2008

### Staff: Mentor

No, Rweight is equivalent to Rbeam.

All I'm suggesting here is to use the word "weight" (which describes the force) instead of the word "beam" when describing the force due to the weight of the beam.

18. Dec 19, 2008

### rml4589

ok, so using your suggestion, the formula is:

Fweight*Rweight* sin41 = Ftension*Rweight * sin49?

are the Rweight supposed to cancel?

if so,
Fweight * sin41 = Ftension*sin49
137.2 * sin41 = Ftension*sin49
Ftension = 119.3 N?

19. Dec 19, 2008

### Staff: Mentor

No, the formula should be:

Fweight*Rweight* sin41 = Ftension*Rtension * sin49

(I didn't realize that I changed "cable" to "tension" as well, so that might have confused you.)

The key thing is: What is Rtension (which you called Rcable)? Rtension and Rweight are distances that you should know.

20. Dec 19, 2008

### rml4589

as a pre-ps, thanks for putting up with me as long as you have haha

Rweight = 2 because the weight acts on the middle of the beam.

as for Rtension, like you said, acts on the end of the beam. so wouldn't it be the distance from the wall to the end of the beam, which happens to be Ltension?

just to make sure, the unknown is still Ftension right? and Rtension = distance from pivot point to the point of application of force. Wouldn't that distance be the length of the cable? or is the point of application of force NOT where the cable and the beam meet?