Plz help me?

Question

If a can begins empty, sitting at rest in a sea of pressurized air and the valve is opened, then what would happen to the can?
Wil the canl move to the left, right, or not move once the valve is opened and air is let inside

Make the following assumptions: Pressure outside can = P ; mass of can = m; Area of can ends(each) = A; area of valve hole = s.

Newton's second law :

m*a = P*s*e^(-bt) - C*v(t)

a = d^2x / dt

v = dx / dt

m*d^2x / dt = P*s*e^(-bt) - C*dx / dt

m*d^2x / dt + C*dx / dt = P*s*e^(-bt) >>> Differential equation

the solution for this differential eq is:
y’ = Ps/(mb-c) { e^ (- c/m * t) - e^ (- bt)}

^^ according to this solution, which is velocity, the can will move to the right, but how can i prove that? we need to find something in common between all these relationships that prove it moves to the right.........

according to this equation, which is velocity,
y’ = Ps/(mb-c) { e^ (- c/m * t) - e^ (- bt)}
all of the terms are constants and scalars, but how can i elaborate on that and prove it moves to the right

the valve is located on the right and that is why it is definitely going to move to the right, but how am i to prove that with the final velocity equation i got?

y’ = Ps/(mb-c) { e^ (- c/m * t) - e^ (- bt)}

How can i show that (mb-c) and { e^ (- c/m * t) - e^ (- bt)} are both positive?
granted, that we say positive=the right direction

Answer 1

With your equation, you can say that P is the pressure from outside on the left side area equal to the valve hole, which is directed to the right. Therefore, for the equations purpose it is a vector.

You do not, as you expressed, want to prove that mb-c and e^(-ct/m)-e^(-bt) are both positive. You simply want to prove that they are OPPOSITE, so that the overall equation is positive. This is relatively simple:

If e^... is to be positive, the first term must be greater than the second (since e^anything is always positive). Therefore, you need to prove that c/m is less than b (and therefore greater when multiplied by -t). Going from there:

c/m < b
c < mb (we know that m, c, and b are all positive)
0 < mb-c

So mb-c is positive also, and since the numerator and denominator are positive, the overall equation is positive.

However, if e^... is to be negative, the first term must be less than the second, and you need to prove that c/m is greater than b (and therefore less when multiplied by -t). Going from THERE:

c/m > b
c > mb
0 > mb-c

Therefore, the numerator and denominator are both negative, so the overall equation is still POSITIVE.

Since you know that the force acting on the can (from pressure) is directed to the right, you can prove that the velocity of the can is directed in the same direction.

Hope that helped. Have a good day.

Answer 2

Where is the valve in relation to the can; what is "left" "right" "up" etc. in relation to the can.

The air will be pushed into the can by P. Therefore, a mass of air will move into the can through the valve.

Since air is moved one way (mass of moved air * rate of movement), the can must move in exactly the opposite way (mass of can * rate).

There are many ways (all mathematically equivalent) to look at this problem.

For example:

P pushes onto all sides of the can and all the local pressures cancel themselves out. Once the valve is open, the tiny area of the valve no longer resists P. The sum of all the other local pressures will leave a net vector in the direction of the opening (the size of the vector will be a function of the area of the valve versus the area of the rest of the can)

A mass of air moves one way into the can, the can must move the other way to keep the total system in balance. Find out which way the mass of air moves (from the valve to the centre of the can) and you will have the direction the can will move (opposite).

Answer 3

if it is sitting at rest? nothing moves with out force , you know newtons law, if anything since it is sitting in a pressurized area, it's movement would be upon it self...another words )implosion) crushed!