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phantomvommand

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- 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

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

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