Calculating Tension in a horizontal cable

[rml4589]

Homework Statement

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.

Homework Equations

sum forces = 0
sum torqes = 0

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

 

[Doc Al]

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.

 

[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?

 

[Doc Al]

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

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

if i chose the pivot point being the point where the beam touches the wall,

Good choice.

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?

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

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

 

[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?

 

[Doc Al]

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

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.

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

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

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

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.)

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

Just a little.

 

[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, F_beam * r_beamsin(theta) = -F_cable * r_cable


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 F_cable and that should be the tension.

EDIT:
would it be this equation instead?

F_beam*r_beamsin41 = F_cable*r_cable*sin49?

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

 

[Doc Al]

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.

Good!

since tension is just a force, and force is just ma, in this case a = 9.8,

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

F_beam * r_beamsin(theta) = -F_cable * r_cable


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

Why do you think the angle is 90 degrees?

Is that equation right so far?

Put that sinθ back for the tension force.

i would just solve for F_cable and that should be the tension.

Yes, once you fix the equation you can solve for the tension.

 

[Doc Al]

EDIT:
would it be this equation instead?

F_beam*r_beamsin41 = F_cable*r_cable*sin49?

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

Much better! (What are r_beam and r_cable equal to?)

 

[rml4589]

R_beam = 2m
R_cable = 3.02/sin41 = 4.6 / 2 = 2.3

so it would be, with everything plugged in

F_beam*R_beam*sin(41) = -F_cable*R_cable*sin(49)

and therefore

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

solving for F_cable

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

F_cable = -103.7 N

or would it be positive?

 

[Doc Al]

R_beam = 2m

Good. The weight acts in the center of the beam.

R_cable = 3.02/sin41 = 4.6 / 2 = 2.3

No. The tension acts at the end of the beam, so R_cable should equal the full length of the beam.

so it would be, with everything plugged in

F_beam*R_beam*sin(41) = -F_cable*R_cable*sin(49)

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).

 

[rml4589]

Good. The weight acts in the center of the beam.

No. The tension acts at the end of the beam, so R_cable 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).

ok so it would be

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

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

 

[Doc Al]

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

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

 

[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?

 

[Doc Al]

4.6 was the length of the cable not the beam.

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, R_weight = 2m (half the length of the beam). The tension acts at the end of the beam, so what must R_tension equal?

Note: It might be less confusing if you used F_weight instead of F_beam, since both forces act on the beam. Similarly, use R_weight instead of R_beam.

 

[rml4589]

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, R_weight = 2m (half the length of the beam). The tension acts at the end of the beam, so what must R_tension equal?

Note: It might be less confusing if you used F_weight instead of F_beam, since both forces act on the beam. Similarly, use R_weight instead of R_beam.

ok you lost me :(

so R_weight is equal for both sides of the equation as well as F_weight?

or is R_tension = R_beam?

i'm confusing myself i think

 

[Doc Al]

or is R_tension = R_beam?

No, R_weight is equivalent to R_beam.

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.

 

[rml4589]

No, R_weight is equivalent to R_beam.

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.

ok, so using your suggestion, the formula is:

F_weight*R_weight* sin41 = F_tension*R_weight * sin49?

are the R_weight supposed to cancel?

if so,
F_weight * sin41 = F_tension*sin49
137.2 * sin41 = F_tension*sin49
F_tension = 119.3 N?

 

[Doc Al]

ok, so using your suggestion, the formula is:

F_weight*R_weight* sin41 = F_tension*R_weight * sin49?

are the R_weight supposed to cancel?

No, the formula should be:

F_weight*R_weight* sin41 = F_tension*R_tension * 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 R_tension (which you called R_cable)? R_tension and R_weight are distances that you should know.

 

[rml4589]

No, the formula should be:

F_weight*R_weight* sin41 = F_tension*R_tension * 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 R_tension (which you called R_cable)? R_tension and R_weight are distances that you should know.

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

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

as for R_tension, 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 L_tension?

just to make sure, the unknown is still F_tension right? and R_tension = 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?

 

[Doc Al]

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

Right.

as for R_tension, 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 L_tension?

It's the distance from pivot point (at one end of the beam) to the other end of the beam. (I don't know what you mean by L_tension.)

just to make sure, the unknown is still F_tension right?

Right.

and R_tension = distance from pivot point to the point of application of force.

Exactly.

Wouldn't that distance be the length of the cable?

No. The length of the cable is the distance from one end of the cable to the other. That has nothing to do with R_tension.

or is the point of application of force NOT where the cable and the beam meet?

No, that's exactly where the point of application of the tension force is.

Do this: On your diagram, put a mark at the pivot and a second mark at the end of the beam where the cable is attached. The distance between those two points is R_tension. (What is that distance? Hint: It's a really easy question!)

 

[rml4589]

so R_tension = length of the BEAM not the cable?

and it doesn't cancel out because R_weight = 2R_tension.

SO

137.2N * 2m * .656 deg. = F_tension * 4m * .755
and
F_tension = 59.6 N?

please tell me I'm finally right haha

 

[Doc Al]

so R_tension = length of the BEAM not the cable?

Yes!

and it doesn't cancel out because R_weight = 2R_tension.

Well, you have those swapped, but yes, that's the idea. (2R_weight = R_tension.)

SO

137.2N * 2m * .656 deg. = F_tension * 4m * .755
and
F_tension = 59.6 N?

please tell me I'm finally right haha

Yay!

 

[rml4589]

WOOO! Thank you SO much lol. I've seriously been working on that problem for 3 days now. it was the last of my pre-final practice test. It was only worth like 1/1000 of my grade, but it will definitely help me on the final tomorrow. We got snowed in here in Mass so they moved the final to tomorrow. It was supposed to be this morning but they postponed it to tomorrow lol. There had BETTER be a question like that on the test tomorrow now that i actually know how to do it haha. thanks again!