Tension in a deformed cylindrical bag on a surface

[phantomvommand]

Homework Statement

See image below

Relevant Equations

Force balance (Alternatively, virtual work)

A cylindrical bag is made from a freely deformable fabric, impermeable for air, which has surface mass density σ; its perimeter L is much less than its length $l$. If this bag is filled with air, it resembles a sausage. The bag is laid on a horizontal smooth floor (coefficient of friction µ = 0). The excess pressure inside the bag is p, the free fall acceleration g. The density of air is negligible.

Prove that the tension in the bag's fabric T = αx + β, where x is the height of the given point P above the floor, and find the coefficient α. Hint: consider the force balance for a small piece of fabric (using the vertical cut shown infigure below).

Remark: fabric's tension is (loosely speaking) the force per unit length. Actually, fabric's tension is not as simple concept as a rope's tension, because different force directions are possible; here, however, we neglect the force component along the sausage's axis. Thus, we can describe the fabric's tension here with a single number, the force per unit length T acting across a horizontal cut of the sausage.

There is a virtual work solution which shows that $T = \sigma gx + T_0$, where $T_0$ is the tension of the fabric where it contacts the surface, ie tension along part 'c' of the above figure. I am however unsure as to how to use force balance to solve this. My attempt is as such:

Knowing that $T_0$ is involved, I consider a portion of the fabric from the contact surface up till a certain height x. Let the end of this portion form an angle $\theta$ with respect to the vertical. Note: In the above diagram, $\theta$ will be an *obtuse* angle.

(1) Horizontal pressures must balance, so $T cos \theta + p(x) = T_0$.
(2) Assuming that the deformation is minimal, the cylinder largely resembles a circle, so $cos \theta = \frac {r - x} {r}$, where $r$ is the radius of the circle.
(3) Vertical force balance shows that: $T sin \theta$ = Weight + pressure downwards, so $T sin \theta = \sigma gr\theta + p\sqrt {r^2 - (r - x)^2}$
(4) How does everything reduce to the solution $T = \sigma gx + T_0$, or is there an alternative, simpler method (which I think there should be)?

Thanks

 

[haruspex]

If $\theta$ is an angle to the vertical then the horizontal component would be $\sin(\theta)$, no?

 

[phantomvommand]

If $\theta$ is an angle to the vertical then the horizontal component would be $\sin(\theta)$, no?

$\theta$ is the angle swept out from the base of the cylinder to the fabric at height x. Under my definition of $\theta$, $\theta$ will be an obtuse angle in the provided diagram.

 

[Chestermiller]

The horizontal force balance seems incorrect to me. I get $$T_0 \cos{\theta}+T_0=px$$where p is uniform inside the fabric, and independent of x.

 

[phantomvommand]

The horizontal force balance seems incorrect to me. I get $$T_0 \cos{\theta}+T_0=px$$where p is uniform inside the fabric, and independent of x.

This has to do with your definition of $\theta$. I defined $\theta$ to be the angle swept out from the bottom of the cylinder, so under my definition, in the diagram in #1, $\theta$ is obtuse.

 

[Chestermiller]

This has to do with your definition of $\theta$. I defined $\theta$ to be the angle swept out from the bottom of the cylinder, so under my definition, in the diagram in #1, $\theta$ is obtuse.

I defined theta as the angle the fabric makes with the horizontal. Do you agree that p is uniform within the fabric cylinder?

 

[phantomvommand]

I defined theta as the angle the fabric makes with the horizontal. Do you agree that p is uniform within the fabric cylinder?

Yes, p is uniform. but I do not see how the equations can be reduced to the text’s solution.

 

[haruspex]

The horizontal force balance seems incorrect to me. I get $$T_0 \cos{\theta}+T_0=px$$where p is uniform inside the fabric, and independent of x.

I think @phantomvommand intended p(x) to indicate a product, not a function.

 

[phantomvommand]

I think @phantomvommand intended p(x) to indicate a product, not a function.

@Chestermiller yep, I meant px as a product Your equation is correct, though of course under a different definition of theta.

 

[haruspex]

I do not see how the equations can be reduced to the text’s solution.

Combining your equation with the target equation produces the relationship between x and θ, so in principle determines the shape of the cylinder. Therefore it is not possible to obtain the target equation from yours. If it can be done at all, it must be from some other starting point.

 

[haruspex]

The first step in this problem should be to show that the curved section of the fabric is a perfect circle.

But is it?

 

[Chestermiller]

But is it?

No. My mistake.

 

[Chestermiller]

Let $\phi$ be the angle that the fabric contour makes with the positive x axis and let ds represent the differential distance along the fabric contour. Then the unit tangent vector along the contour is $$\mathbf{i}_s=\mathbf{i_x}\cos\phi+\mathbf{i_y}\sin\phi$$and the vector force balance on the arc of fabric between s and $s+ds$ is: $$\frac{d(T\mathbf{i}_s)}{ds}-\alpha g\mathbf{i}_y-p\mathbf{i}_n=\mathbf{0}$$where $\mathbf{i}_n$ is the unit normal to the contour: $$\mathbf{i}_n=-\mathbf{i_x}\sin\phi+\mathbf{i_y}\cos\phi$$In terms of components tangential and normal to the contour, this equation becomes: $$\frac{dT}{ds}=\alpha g \sin\phi$$and$$T\frac{d\phi}{ds}-\alpha g \cos{\phi}=p$$

 

[phantomvommand]

Let $\phi$ be the angle that the fabric contour makes with the positive x axis and let ds represent the differential distance along the fabric contour. Then the unit tangent vector along the contour is $$\mathbf{i}_s=\mathbf{i_x}\cos\phi+\mathbf{i_y}\sin\phi$$and the vector force balance on the arc of fabric between s and $s+ds$ is: $$\frac{d(T\mathbf{i}_s)}{ds}-\alpha g\mathbf{i}_y-p\mathbf{i}_n=\mathbf{0}$$where $\mathbf{i}_n$ is the unit normal to the contour: $$\mathbf{i}_n=-\mathbf{i_x}\sin\phi+\mathbf{i_y}\cos\phi$$In terms of components tangential and normal to the contour, this equation becomes: $$\frac{dT}{ds}=\alpha g \sin\phi$$and$$T\frac{d\phi}{ds}-\alpha g \cos{\phi}=p$$

Thanks. I've used the tangential equation to obtain the text's solution, with the approximation $s \sim r\phi$. However, I am curious about how you obtained the normal equation. I get that the quantity $p + \sigma gcos\phi$ must be balanced, but I do not see how it is balanced by $T \frac {d\phi} {ds}$

 

[hutchphd]

What is the definition of T? That will tell you.

 

[Chestermiller]

Thanks. I've used the tangential equation to obtain the text's solution, with the approximation $s \sim r\phi$. However, I am curious about how you obtained the normal equation. I get that the quantity $p + \sigma gcos\phi$ must be balanced, but I do not see how it is balanced by $T \frac {d\phi} {ds}$

The tangential unit vector is changing direction along the fabric contour, and this results in a normal component of force on the fabric.

 

[phantomvommand]

The tangential unit vector is changing direction along the fabric contour, and this results in a normal component of force on the fabric.

I currently see $T \frac {d\phi} {ds}$ as normal force = $Td\phi$, and this is spread over ds. Is this how it should be seen, or is there a way to see it as $\frac {d\phi} {ds}$ multiplied by T? Also, I think the it balances $p\sigma + \sigma gcos\phi$

 

[TSny]

Let $\phi$ be the angle that the fabric contour makes with the positive x axis and let ds represent the differential distance along the fabric contour. Then the unit tangent vector along the contour is $$\mathbf{i}_s=\mathbf{i_x}\cos\phi+\mathbf{i_y}\sin\phi$$and the vector force balance on the arc of fabric between s and $s+ds$ is: $$\frac{d(T\mathbf{i}_s)}{ds}-\alpha g\mathbf{i}_y-p\mathbf{i}_n=\mathbf{0}$$where $\mathbf{i}_n$ is the unit normal to the contour: $$\mathbf{i}_n=-\mathbf{i_x}\sin\phi+\mathbf{i_y}\cos\phi$$In terms of components tangential and normal to the contour, this equation becomes: $$\frac{dT}{ds}=\alpha g \sin\phi$$and$$T\frac{d\phi}{ds}-\alpha g \cos{\phi}=p$$

This is very nice!

You used the symbol $\alpha$ for the mass density $\sigma$. So, your next to last equation can be written as $$\frac{dT}{ds}=\sigma g \sin\phi$$
It is easy to use this equation to derive the result for $T$ as a function of vertical height.

 

[Chestermiller]

I currently see $T \frac {d\phi} {ds}$ as normal force = $Td\phi$, and this is spread over ds. Is this how it should be seen, or is there a way to see it as $\frac {d\phi} {ds}$ multiplied by T? Also, I think the it balances $p\sigma + \sigma gcos\phi$

p is not multiplied by $\sigma$.