Deriving the captstan equation

[Mr Davis 97]

Homework Statement

A device called a capstan is used aboard ships in order to control a rope which is under great tension. The rope is wrapped around a fixed drum. The load on the rope pulls it with a force $T_A$ and the sailor holds it with a much smaller force $T_B$. Show that $T_A = T_B e^{\mu \theta}$, where $\mu$ is the coefficient of friction, and $\theta$ is the total angle subtended by the rope on the drum.

Homework Equations

Newton's second law

The Attempt at a Solution

So I think that I can get a solution. That is, I think that I can look at an infinitesimal section of the rope, apply Newton's second law in the horizontal and vertical directions, take the limit as $\theta$ goes to zero, and then integrate. However, I have some questions about which forces will be acting on the rope. My first question is, is the rope massless? If so, how would friction and a normal force be acting on the rope? Also, if the rope is massless, wouldn't the tension in the rope be uniform?

 

[kuruman]

The rope is not massless. The normal force and friction increase as you move from the low tension to the high tension end.

 

[Mr Davis 97]

The rope is not massless. The normal force and friction increase as you move from the low tension to the high tension end.

So if the rope is not massless, why don't we account for its weight?

 

[kuruman]

The weight of the rope is negligible compared against the normal force.

 

[Mr Davis 97]

The weight of the rope is negligible compared against the normal force.

Is the normal force much greater than the weight because of the two tensions pulling down?

 

[kuruman]

Yes, and the tensions increase as you go around the wrapped rope.

 

[Mr Davis 97]

Okay that makes sense. But how is the rope is equilibrium if the two forces on each side of the rope are so different? Shouldn't the rope accelerate in the direction of the tension caused by the load?

 

[kuruman]

The maximum force of static friction increases as you move around the wrapped rope because the normal force increases. Derive the capstan equation including the dependence of the normal force on the angle and you will see how this works.

 

[Mr Davis 97]

Alright , that makes sense. Another question. After deriving that $\Delta T = T \mu \Delta \theta$, we can have that $\displaystyle \int \frac{1}{T} dT = \int_0^\theta \mu d \theta$ However, I am not sure what the bounds of integration should be on the left integral. I think that it might look like $\displaystyle \int_{T_B}^{T_A}$, but am not sure why this would be the case.

 

[kuruman]

What is the tension when the angle is zero? That's the lower limit. What is the tension when the angle is θ? That's the upper limit.

 

[Charles Link]

One thing you are needing, and it would help to show you with a diagram is the normal force per unit length from a rope around a circle is given by $f_L=T/r$. A derivation is done using a segment $\Delta s$ where the angle changes by $\Delta \theta/2$ on each side. Meanwhile, the tension changes because of the frictional force which comes from the normal force. I see from a previous post that perhaps you already have the equation $f_L=T/r$ or a variation of it in terms of $\theta$.

 

[Mr Davis 97]

What is the tension when the angle is zero? That's the lower limit. What is the tension when the angle is θ? That's the upper limit.

Why would $T_B$ be the tension when the angle is zero, and $T_A$ be the tension when the angle is $\theta$?

 

[kuruman]

The capstan doesn't know the alphabet. Like I said, the lower limit of the integral on the left matches the lower limit of the integral on the right and similarly for the upper limits of the integrals.

 

[Charles Link]

I think in post #9 your differential equation for $\Delta T$ needs a minus sign. (The tension decreases due to the frictional force the more the thing gets wrapped around). Also you have an extra $\theta$ next to the $\mu$ that shouldn't be there. The $T_B$ is the force from the sailor, (should be the upper limit of the $T$ integral ), and $T_A$ is the force from the sail. (needed to edit because I read the original problem more carefully.)(equation should read $\Delta T=-\mu T \Delta \theta$).

 

[Mr Davis 97]

I think I corrected the equation; also, I don't see why there would need to be a minus sign.

However, I am just not seeing how the force on the sailor and the force that the sailor supplies would correspond to when $\theta$ is zero and when $\theta$ has a value.

 

[Charles Link]

I think I corrected the equation; also, I don't see why there would need to be a minus sign.

However, I am just not seeing how the force on the sailor and the force that the sailor supplies would correspond to when $\theta$ is zero and when $\theta$ has a value.

Please read my edited version of post #14.

 

[kuruman]

If we agree that T_A is the larger force, then T_A = T_Be^μθ. This can be inverted to give T_B = T_Ae^-μθ, but the latter is not the traditional capstan equation. Either equation gives T_A = T_B when the angle is zero.

 

[Charles Link]

If we agree that T_A is the larger force, then T_A = T_Be^μθ. This can be inverted to give T_B = T_Ae^-μθ, but the latter is not the traditional capstan equation. Either equation gives T_A = T_B when the angle is zero.

I think the derivation will give (from the differential equation ) $\$ $T_B=T_A e^{- \mu \theta}$. A little algebra gives the textbook result.

 

[Mr Davis 97]

If we agree that T_A is the larger force, then T_A = T_Be^μθ. This can be inverted to give T_B = T_Ae^-μθ, but the latter is not the traditional capstan equation. Either equation gives T_A = T_B when the angle is zero.

So why does $T_B$, the smaller force, correspond to $\theta = 0$, and $T_A$, the larger force, correspond to the $\theta$ when it is nonzero? I'm just trying to understand why we use those limits of integration

 

[kuruman]

Integration is a summation. You get a negative exponential if you add dT/T on one side and - μdθ on the other starting at the high tension end or add dT/T on one side and μdθ on the other starting at the low tension end. What I am saying here is that you can change the sign of the integral on the left and swap the limits of integration without changing anything.

 

[Charles Link]

The sailor problem complicates the matter somewhat by calling it by the fancy names. Suppose instead that you lasso a horse. You would have a hard time hanging on to the horse, but if you could wrap the rope a couple of times around a telephone pole, you would have a much easier job of hanging onto the horse. $T_A$ is the force from the horse, and $T_B$ is the force you need to hold the horse.

 

[Mr Davis 97]

So physically, in the limit that $\theta$ approaches zero, we have low tension with rope being in equilibrium (which is $T_B$), and in the limit where we have some higher angle $\theta$, we have some greater tension $T_A$?

 

[Charles Link]

$\theta$ is measured in radians. You get $2 \pi$ increase in $\theta$ for every revolution. If you just pull the rope partly around the telephone pole, $\theta$ will be less than $2 \pi$. The problem can alternatively be worked using distance around the pole and force per unit length, etc. and tenstion as a function of distance around the pole.

 

[kuruman]

So why does TBT_B, the smaller force, correspond to θ=0\theta = 0, and TAT_A, the larger force, correspond to the θ\theta when it is nonzero? I'm just trying to understand why we use those limits of integration

Perhaps it should be clearer to you if you think of the capstan equation as a function of θ, namely
T_sail(θ) = T_sailore^μθ.
This is the condition for the rope to be at the threshold of sliding when the rope is wrapped around by angle θ. When θ = 0, i.e. no rope is wrapped around the capstan, the sailor needs to match the pull of the sail and nothing moves. If the sail pulls harder and the sailor is unable to provide the necessary force directly, the rope needs to be wrapped around the capstan to an appropriate angle θ in order to multiply the maximum force that the sailor can provide by a factor e^μθ.

 

[Charles Link]

Perhaps it should be clearer to you if you think of the capstan equation as a function of θ, namely
T_sail(θ) = T_sailore^μθ.
This is the condition for the rope to be at the threshold of sliding when the rope is wrapped around by angle θ. When θ = 0, i.e. no rope is wrapped around the capstan, the sailor needs to match the pull of the sail and nothing moves. If the sail pulls harder and the sailor is unable to provide the necessary force directly, the rope needs to be wrapped around the capstan to an appropriate angle θ in order to multiply the maximum force that the sailor can provide by a factor e^μθ.

It should be noted, the capstan holds the sail still. It does not provide extra force for moving it like a block and tackle (pulley ) system would.

 

[kuruman]

It should be noted, the capstan holds the sail still. It does not provide extra force for moving it like a block and tackle (pulley ) system would.

I am not a sailor, but I would think that if the rope is wrapped just right, the sailor can relax or increase his/her pull to allow the sail to change its angle one way or the other.

 

[Charles Link]

I am not a sailor, but I would think that if the rope is wrapped just right, the sailor can relax or increase his/her pull to allow the sail to change its angle one way or the other.

No. The sailor cannot move the sail with it. Just like with wrapping a rope around a telephone pole will not allow you to move the horse. It does allow you to use much less force to hold the sail steady. One question that might arise is, what happens if you pull harder than what is necessary: Does the sail move? And the answer is no. Friction works in such a way that $F_{friction}=\mu F_{normal}$ is the maximum force friction will supply. e.g. a block sitting on a level surface has zero frictional force until you try to move the block in a horizontal direction. The frictional force will then be equal and opposite the horizontal force you supply as long as the block remains motionless.

 

[Mr Davis 97]

Let's just say that I integrate with no bounds of integration, then I would have $T = e^{\mu \theta + C}$ When $\theta = 0$, we have that $T_0 = e^C$, so then we have $T = T_0 e^{\mu \theta}$. From this point of view, it seems as though we are finding the tension of the whole rope for angles as a multiple of the initial tension in the rope. How do I go to say that $T$ and $T_0$ are somehow related to the tensions $T_{sail}$ and $T_{sailor}$?

 

[Charles Link]

Let's just say that I integrate with no bounds of integration, then I would have $T = e^{\mu \theta + C}$ When $\theta = 0$, we have that $T_0 = e^C$, so then we have $T = T_0 e^{\mu \theta}$. From this point of view, it seems as though we are finding the tension of the whole rope for angles as a multiple of the initial tension in the rope. How do I go to say that $T$ and $T_0$ are somehow related to the tensions $T_{sail}$ and $T_{sailor}$?

This one is a definite integral on both sides with no constant of integration. The result is $\ln (T_B/T_A)=-\mu \theta$. $\\$ ... $\$ Some finer detail: $\theta$ (the upper limit in the $d \theta$ integration) is angle of rotation in radians of how far in angle the rope wrapped around the pole or cylindrical piece (and then it leaves the cylindrical piece and goes to the hand of the sailor.) The $\theta=0$ (lower limit in the $d \theta$ integration) is defined as the point where the rope from the sail first makes contact across the pole as it wraps around it. $\$ ... $\\$ This gives $T_B=T_A e^{-\mu \theta}$ which is equivalent to $T_A=T_B e^{\mu \theta}$ . $T_A$ is the force from the wind on the sail which we have no control over. The form the equation is presented is actually rather confusing because $T_B=T_A e^{-\mu \theta }$ really is a more logical presentation of it. $\\$ One question they could ask is if $T_A=200$ pounds, what does $\theta$ (in radians) need to be to require only $T_B=10$ pounds of pulling force from the sailor if $\mu=.25$? (And then also compute how many revolutions of the rope (around the capstan) are required corresponding to the $\theta$.) $\\$ Also, please read my edited # 27.

 

[Mr Davis 97]

This one is a definite integral on both sides with no constant of integration. The result is $\ln (T_B/T_A)=-\mu \theta$. $\\$ ... $\$ Some finer detail: $\theta$ (the upper limit in the $d \theta$ integration) is angle of rotation in radians of how far in angle the rope wrapped around the pole or cylindrical piece (and then it leaves the cylindrical piece and goes to the hand of the sailor.) The $\theta=0$ (lower limit in the $d \theta$ integration) is defined as the point where the rope from the sail first makes contact across the pole as it wraps around it. $\$ ... $\\$ This gives $T_B=T_A e^{-\mu \theta}$ which is equivalent to $T_A=T_B e^{\mu \theta}$ . $T_A$ is the force from the wind on the sail which we have no control over. The form the equation is presented is actually rather confusing because $T_B=T_A e^{-\mu \theta }$ really is a more logical presentation of it. $\\$ One question they could ask is if $T_A=200$ pounds, what does $\theta$ (in radians) need to be to require only $T_B=10$ pounds of pulling force from the sailor if $\mu=.25$? (And then also compute how many revolutions of the rope (around the capstan) are required corresponding to the $\theta$.) $\\$ Also, please read my edited # 27.

But why are $T_B$ and $T_A$ the bounds of integration that correspond to the lower and upper limit of $\theta$?

 

[Charles Link]

But why are $T_B$ and $T_A$ the bounds of integration that correspond to the lower and upper limit of $\theta$?

It is only between these points that your differential equation applies. From the sail to the point of contact of the rope the tension is constant at $T=T_A$ and from the point where the rope leaves the cylinder to go to the sailor's hand it is constant at $T_B$. Between these two points, the tension decreases steadily as it wraps around the pole because of the frictional force. You have your equation $\Delta T=-\mu T \Delta \theta$. Perhaps if you would go over its derivation in more detail, you would see how the tension $T$ is changing from the frictional force. Meanwhile $\theta$ has limits which we call $\theta=0$ and $\theta$. The upper limit we could have called $\theta_B$, but in the final formula, $\theta$ is sufficient. As previously mentioned, this problem can be computed using length $ds$ instead of angle, but I think you can recognize that $ds=r d \theta$. e.g. you could find how the tension in the rope changes with distance around the pole.

 

[Mr Davis 97]

So does $\displaystyle \frac{dT}{d \theta} = - \mu T$ represent how to tension changes over the contact area of the rope and capstan?

 

[Charles Link]

So does $\displaystyle \frac{dT}{d \theta} = - \mu T$ represent how to tension changes over the contact area of the rope and capstan?

The capstan is a cylinder. The rope is wrapped around the cylinder as many times as necessary to reduce the force needed to keep the rope (sail end) still. Suggestion: Try it with a friend with a lamppost=have them hold a rope and pull on it and start wrapping it around the pole. You wrap it enough times and you won't really need to pull on it at all...Even one loop and they won't be able to win a tug-o-war... You won't be able to pull them either though=the same equations apply going the other way=i.e. if they stay fixed with a light steady tension and you tug... And yes, the differential equation only computes $T$ for the region of contact with the pole. Outside of this region $T$ is constant. (On one side it is equal to $T_A$, and on the other side $T_B$. If it is 10 ft. from the sail to the pole, $T=T_A$ stays constant for 10 ft.)