Solve Pressure Problem: Find Plug Position

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AI Thread Summary
The problem involves a vertical cylindrical tank with a plug that falls inside, and the goal is to determine the position of the plug when it comes to rest. The atmospheric pressure is given, and the forces acting on the plug must be analyzed, including the downward force due to its weight and the upward force from the trapped air. The relationship between pressure and volume is crucial, utilizing the ideal gas law to connect the pressures and volumes before and after the plug is inserted. The discussion emphasizes calculating the balance of forces and pressures to find the equilibrium position of the plug. The solution is confirmed as correct by the participants involved.
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[SOLVED] Pressure Problem

SOLVED

Homework Statement



A vertical cylindrical tank is 42 cm tall, 2.00 cm in radius, and is open at the top. Atmospheric pressure is 1.013 x 10^5 N / m^2. A close-fitting cylindrical plug of mass 5 kg is inserted at the top, and is let fall inside. If the temperature of the trapped air does not change, how far from the top of the cylinder is the base of the plug when it comes to rest?

(This is the exact problem, word for word)

Homework Equations



(p1*v1)/t1=(p2*v2)/t2

The Attempt at a Solution



Should I compare the pressure caused by the plug to the atmospheric pressure? ((5*9.8*x)/((.02)^2 * pi))((.02)^2 * pi * x)=1.013 * 10^5 * (.02)^2 * pi * .42 ...or, in terms of variables... m*g*h/(pi*r^2) * (pi*r^2*h)=atmospheric pressure * (pi*r^2*height_of_cylinder)

(pi*r^2*height_of_cylinder) is the volume of the cylinder, (pi*r^2*h) is the volume under the plug.

EDIT: I think this is right. Thanks!
 
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bfr said:
Should I compare the pressure caused by the plug to the atmospheric pressure? ((5*9.8*x)/((.02)^2 * pi))((.02)^2 * pi * x)=1.013 * 10^5 * (.02)^2 * pi * .42 ...or, in terms of variables... m*g*h/(pi*r^2) * (pi*r^2*h)=atmospheric pressure * (pi*r^2*height_of_cylinder)

(pi*r^2*height_of_cylinder) is the volume of the cylinder, (pi*r^2*h) is the volume under the plug.
What is the downward force on the plug? What is the upward force on the plug? How are the two related? (ie. does the plug move?). Translate the downward force into pressure and apply the ideal gas law: P1V1 = nRT1 = nRT2 = P2V2

AM
 
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