Fluids Homework Help: Pascal's Barrel Burst Problem

  • Thread starter Thread starter ouma serumula
  • Start date Start date
  • Tags Tags
    Fluids Homework
AI Thread Summary
The discussion focuses on Pascal's principle, illustrating how fluid pressure can amplify force, using a scenario involving a wine barrel and a vertical tube. To calculate the mass of fluid in the tube, one must find the area of a circle with a 0.3 cm radius, multiply it by the height of 12 m (1200 cm) to determine the volume in cm³, and then use the density of water (1 g/cm³) to find the mass. The net force exerted by the water on the barrel lid is calculated using the formula F = M × g, where M is the mass of water in grams and g is the acceleration due to gravity (9.81 m/s²). The relationship between the forces acting on different areas is explained through the hydraulic principle, allowing for the calculation of the force on the lid based on the areas of the tube and barrel. This discussion effectively demonstrates the application of fluid dynamics principles in practical scenarios.
ouma serumula
Messages
1
Reaction score
0
help

in working out his principal,pascal showed dramatically how force can be multiplied with fluid pressure.he placed a long thin tube of 0.30cm radious vertically into a 20cm radious wine barrel.He found that when the barrel was filled with water and the tube filled to a height of 12m.the barrel burst.calculate the mass of fluid in the tube and the net force exerted by the water in the barrel on the lid just before ruptured.
PLZ EXPLAINE IN DETAILS
 
Physics news on Phys.org
For the mass, first you need to calculate the area of a circle with 0.3 cm radius, then multiply it with 12m = 1200cm to get the volume in cm3. Then multiply volume with the density of water, which is 1g/cm3 (see this link). So mass (in grams) = volume (in cm3) for water.
 
Now force is mass times gravitational acceleration: F = M\times g = M\times 9.81 m/s^2.

In your case mass = volume. Suppose the volume of the water in the tube is V cm3 (V is the volume that you are supposed to calculate in the first step above, you should put in the number you calculated for volume). Then its mass is V grams. So the force exerted at the bottom of the tube is f = (V grams) times (9.81 m/s2) or approximately 10V g m/s2.

Show the force on the lid as F. You need to calculate the area of a circle with a 20cm radius. Let's say this area is A. Also say that the area of the circle with the 0.3cm radius is a. Hydraulic principle* says that F/f = A/a, or F = f A/a. Since you have calculated f and also calculated A and a, you can calculate F.

* "A force acting on a small area can create a much larger force by acting on a larger area by virtue of hydrostatic pressure," as explained under Hydraulic machinery in Wikipedia, the free encyclopedia.
 
Last edited:
Check out http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html

The force on the bottle cap is similar to the force exerted by water in the tube in your example. The force acting on the bottom of the bottle is similar to the force acting on the lid in your example.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top