How Does String Tension Vary with Torque in a Triangular Rod System?

[pc100]

Any help on thinking about this general scenario involving string tension?

Homework Statement

Two rods A and B are connected by a joint (A, B are different lengths. Also this is a planar problem). The ends of the rods are connected by a string so the system is a triangle. A constant torque is applied between the rods at their joint. What's the tension (force) in the string?
Tension can be found by calculating forces at either end since it should be identical, but they don't match for me. Visually the vector component breakdown doesn't look correct either.

Homework Equations

perpendicular force F at end of rod A = half of torque / length = (T/2) / lengthA
perpendicular force F at end of rod B = half of torque / length = (T/2) / lengthB

The Attempt at a Solution

Here are extreme cases first. If the rods are the same length, and they're folded on top of each other, so the string is almost zero length, the tension will be just the normal forces (from the torque) at the rod ends. If the rods are swung out almost completely in line--and the string is slightly shorter so they can't swing out straight all the way--then the tension will be near infinite regardless of the torque, right? Or does it become very small?

If it becomes very small then my answers are the same for both ends of the string, but intuitively it doesn't make sense when drawing a vector diagram of the forces. If it goes to infinity, it makes sense in the vector diagram, but the answers are different for each end.

 

[tiny-tim]

Welcome to PF!

perpendicular force F at end of rod A = half of torque / length = (T/2) / lengthA
perpendicular force F at end of rod B = half of torque / length = (T/2) / lengthB

Hi pc100! Welcome to PF!

No, there's no perpendicular force, it's perpendicular distance …

torque = r x F,

so the magnitude of the torque is Frsinθ, where θ is the angle between r and F

And Frsinθ = Fd, where d = rsinθ (the "lever arm") is the perpendicular distance from (in this case) the pivot to the line of the string.

(and when the angle is 180º, the tension due to the torque should be zero, since it isn't even trying to make the string any longer )

 

[pc100]

Thanks tiny-tim. Why is the tension zero? See the attachment with two situations I drew. Don't they both lead to the same tension in the string?

 

[tiny-tim]

Sorry, but I don't understand either of your diagrams.

 

[pc100]

Sorry, but I don't understand either of your diagrams.

Ah, okay:D

Situation 1 has a rod (in red) on a joint connected to the ground. A torque is applied to the rod. A string (in blue) is slightly longer than the rod, and connected to the ground and the end of the rod. As a result of the torque on the rod about the joint, there must be a force F at the end of the rod equivalent to T / L. (Ah! I see, I had a * instead of / in the drawing. I've changed that now, sorry about that)

Situation 2 has two strings connected together. Each string has the same length as the one in situation 1. A force F is applied between the two strings, and since the angle between the strings is almost 180 degrees, there is a huge tension created in the strings (if it was 180 exactly, the tension would be infinite)

 

[tiny-tim]

ah, i see now … the T*L and the absence of the second rod was confusing me.

hmm … perhaps I've been misunderstanding how this torque is applied.

Yes, as drawn, the tension T becomes extremely large as θ -> 180º.

But the rest of my first post still stands …

the torque of the tension T (about the pivot) is Trsinθ, where θ is the angle between one rod and the string.

You've introduced two imaginary forces perpendicular to, and at the end of, each rod (in equilibrium, that wouldn't work, because the resultant force would move the whole assembly sideways but anyway …)

The torques of those two imaginary (unequal) forces must be equal and opposite (otherwise the whole system would rotate), and each individually must be equal and opposite to the torque of the tension … if you work it all out, there's no contradiction.

 

[pc100]

You've introduced two imaginary forces perpendicular to, and at the end of, each rod (in equilibrium, that wouldn't work, because the resultant force would move the whole assembly sideways but anyway …)

The torques of those two imaginary (unequal) forces must be equal and opposite (otherwise the whole system would rotate), and each individually must be equal and opposite to the torque of the tension … if you work it all out, there's no contradiction.

Okay, that makes sense. So then I think the attached diagram (for the full problem) looks okay. I will do this for torque as well soon to verify that it works, but for now it uses the imaginary forces at the rod ends.

And just making sure that it's half of the forces that I drew. So not Fa=T/a and Fb=T/b, but (1/2) (T/a) and (1/2) (T/b). This is because the torque is referenced to one of the rods, since the torque (motor?) is between the rods.

 

[tiny-tim]

(just got up :zzz: …)

Okay, that makes sense. So then I think the attached diagram (for the full problem) looks okay. I will do this for torque as well soon to verify that it works, but for now it uses the imaginary forces at the rod ends.

And just making sure that it's half of the forces that I drew. So not Fa=T/a and Fb=T/b, but (1/2) (T/a) and (1/2) (T/b). This is because the torque is referenced to one of the rods, since the torque (motor?) is between the rods.

(nice diagrams, btw )

hmm … it's better just to use torques, not imaginary forces.

Consider the two rods together, and each rod separately.

If rod A experiences an "external" torque t_A, and rod B experiences an "external" torque t_B (plus the torque from the tension T, of course),

then t_A + t_B = 0 (because the whole system does not rotate, and T is an internal force for the whole system) …

carry on from there. ​

 

[pc100]

This is helpful, thanks. I added a torque method to the bottom of the drawing. Is this what you were thinking of?

Since the system is static, T1_reaction + T2_reaction + Ta + Tb = 0

The torques Ta and Tb on each rod should be *half* the torque applied to the joint by the motor (or whatever is at the joint), right?

 

[tiny-tim]

Hi pc100!

(have a tau: τ )

This is helpful, thanks. I added a torque method to the bottom of the drawing. Is this what you were thinking of?

Yes.

(But why are you calling it "reaction torque"? The tension from the string is a force on the rod, and its torque is a torque on the rod, not a "reaction". )

The important thing to notice is that (from the geometry) asinα₁ = bsinα₂, confirming that the torques are the same.

Better still, you should get used to using vectors (not broken into coordinates, but as themselves) …

the torques are a x T and b x -T,

and they are obviously the same because if you subtract them you get (a - b) x T, which is zero!

The torques Ta and Tb on each rod should be *half* the torque applied to the joint by the motor (or whatever is at the joint), right?

Yes.

 

[pc100]

Thank you! You were very helpful, tiny-tim. And thanks for the τ for future reference.

(Good point about (a - b) x T = 0, that makes sense since a-b is a vector parallel to the tension T, so the cross product is zero.)