Difference in tension introduced by friction

[Ithilrandir]

Homework Statement

A cord moving at a low speed v rubs against a round post and deviates from a straight line by a small angle Δθ << 1 radian, as shown in the figure. If the tension on one side of the post is T + ΔT and on the other side is T, what is the difference ΔT introduced by the friction?

Relevant Equations

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I have no clue how to do this, so I did my best guess of it.

Friction is μN, N being the normal force.

The normal force when it is in contract with the pole should be V²/R, R being the radius of the pole.

So ΔT = μV²/R

The answer provided is μTΔθ

 

[jbriggs444]

The normal force when it is in contract with the pole should be V2/R, R being the radius of the pole.

In a typical exercise involving cords, pulleys and the like, the cord is approximated as having negligible mass.

There is a different way to determine the normal force based on the givens of the problem together with the assumption that the cord has negligible mass.

There is also a problem with the formula: $F=\frac{v^2}{r}$. What are the units on the left. What are the units on the right?

 

[Ithilrandir]

In a typical exercise involving cords, pulleys and the like, the cord is approximated as having negligible mass.

There is a different way to determine the normal force based on the givens of the problem together with the assumption that the cord has negligible mass.

There is also a problem with the formula: $F=\frac{v^2}{r}$. What are the units on the left. What are the units on the right?

I can't see anything wrong with the formula as the force experienced by the cord is by turning a small degree against the post.

 

[jbriggs444]

I can't see anything wrong with the formula as the force experienced by the cord is by turning a small degree against the post.

Again, answer the question: What are the units on the left? What are the units on the right?

Or, perhaps more obviously: What is the correct formula for centripetal force?

Additionally:

In what direction does a hypothetical centripetal force act on a segment of string?
In what direction does the normal force from the cylinder act on the segment of string?
Do you see that your formula has the wrong sign?!
The faster a [massive] cord moves around this cylinder, the smaller the normal force must become. Which is why an assumption that the cord is massless is so very handy.

 

[haruspex]

I can't see anything wrong with the formula as the force experienced by the cord is by turning a small degree against the post.

Centripetal force has almost nothing to do with the force between the cord and the post.
True, there will be some centripetal force required to coax the cord around the post, but it would be very small and anyway we are not given any mass or density for the cord so it is unknowable.
What centripetal force there is is provided by the tension in the cord, thereby slightly reducing the normal force.
So a) your equation for centripetal force is wrong because the expression on the right is not a force;
b) centripetal force is irrelevant to the question anyway.

Consider the short segment of cord in contact with the post. What are the forces on it?

 

[Ithilrandir]

Centripetal force has almost nothing to do with the force between the cord and the post.
True, there will be some centripetal force required to coax the cord around the post, but it would be very small and anyway we are not given any mass or density for the cord so it is unknowable.
What centripetal force there is is provided by the tension in the cord, thereby slightly reducing the normal force.
So a) your equation for centripetal force is wrong because the expression on the right is not a force;
b) centripetal force is irrelevant to the question anyway.

Consider the short segment of cord in contact with the post. What are the forces on it?

Right I see. The forces on the segment should be the friction, the normal force and the tension from both sides.

 

[Ithilrandir]

Right I see. The forces on the segment should be the friction, the normal force and the tension from both sides.

OK I did a force body diagram, with T to the right, F to the left, N upwards and T + ΔT at an angle Δθ to the horizontal.

Using small angle approximation,

(T + ΔT)Δθ = N,

μ(T + ΔT)Δθ = F

(T + ΔT)cosΔθ = T - F, small angle approx,

T + ΔT = T - F

ΔT = F

μ(T + ΔT)Δθ = ΔT

If the μΔTΔθ part of the above is gone I would have the correct answer, but I don't see how to do it from here.

 

[jbriggs444]

(T + ΔT)Δθ = N

I think that you are letting the $\Delta T$ distract you here. Take one step at a time. You are trying to first compute the normal force $N$. Then you can use that to figure out $\Delta T$.

Since $\Delta T$ is small relative to $T$, one can simplify this to $N = T \Delta \theta$. This is a simpler and less distracting starting point.

 

[Ithilrandir]

I think that you are letting the $\Delta T$ distract you here. Take one step at a time. You are trying to first compute the normal force $N$. Then you can use that to figure out $\Delta T$.

Since $\Delta T$ is small relative to $T$, one can simplify this to $N = T \Delta \theta$. This is a simpler and less distracting starting point.

Right. Thanks, I tend to overlook simplifying things involving small values unless it's an angle.