Problems about atmospheric pressur

  • #1
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Hi, i'm doing some physics exercises but some don't result correct, could you help me? thanks!

Homework Statement


1)Calculate the necessary force to open a 0.3 m lenght square metallic container with a 410 mmHg inside pressure knowing external pressure is 0,980 atm

2)There is a hollow cilinder with a 5 kg mass and 30 cm diameter cap in a side and hanging on the ceiling in the other side. By a pump you reduce the internal pressure in the way of keeping the cap adherent to cilinder. Knowing external pressure is the normal atmospheric one and internal pressure is 12.6 mmHg, calculate the mass you need to separare cap.

Homework Equations


p = F/A
P = m*g

The Attempt at a Solution



first problem)
I converted all into pascal, so 54662 and 99299.
Then:
A = 0.09 m2
99299 - 54662 = 44637 Pa
F = 44637 Pa * 0.09 m 2 = 4017 N but my book says 4020 N, i forgot somewhat or is my procedure correct?

second problem)
Converted internal pressure in 1680 Pa.
A = 0.282743 m2
Atm. force = 101300 *0.282743 = 28642 N
Internal force = 1680 * 0.282743 = 475 N
Cop force = 49 N
Total acting force = 28642-49-475 = 28118 N

To win a 28118 N force I need a 28118/9.8 = 2870 Kg but my book says only 470 Kg how is it possible? where did i wrong?


Thanks!
 

Answers and Replies

  • #2
gneill
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4020 is just 4017 rounded to three significant figures.

In the second problem, it looks like your cap area is about 4x too big. Did you forget to divide the diameter by two to find radius?
 
  • #3
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What stupid i'm :( Thanks!

When I immerse a 0.2 m3 object more dense then water in water for the half (0.1), buoyance will do a force that is the weight meant as m*g of water or simply the mass value of the water moved?
Why in my problems wasn't necessary calculare the weight (m*g) of the water?


Thanks!
 
  • #4
gneill
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Buoyant force is equal to the weight (m*g) of the water displaced. Buoyancy is a force, so its units are Newtons, just like weight.

If your equations are set up so that you've already divided through by g everywhere, then you're just left with the mass part of m*g.
 
  • #5
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in the test there was this problem:
"There is a 100g mass glass cilinder that is height 50 cm and the diameter is 25mm. On the bottom there is a 90g mass stopper that close the cilinder in the way it's perfectly adherent to cilinder by buoyance. Inside the cilinder there is no air and the part of cilijnder with stopper is immersed for half-height in oil. Calculate the mass of water to put in cilinder needed to separate the bottom and the height of mercury to do the same think."

I calculated mass of cilinder + mass of stopper but mass of cilinder was marked as unnecessary, but why if it contribuiteto weight-force?


Thanks!
 
  • #6
gneill
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Is the cylinder free to float, or is it held fixed in place? If it was fixed in place, was the density of the oil given?

P.S. You should really start a new thread for new questions.
 
  • #7
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what do you mean for "free to float"?
yes the density of oil was given, 960 kg/m3.
Thanks!

I prefer don't start new threads for every problem about the same think.
 
  • #8
181
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what do you mean for "free to float"?
yes the density of oil was given, 960 kg/m3.
Thanks!

I prefer don't start new threads for every problem about the same think.
 
  • #9
gneill
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what do you mean for "free to float"?
I mean is the cylinder being held in place, submerged to 1/2 its length in oil, or is it just floating and not being held fixed.

yes the density of oil was given, 960 kg/m3.
Well, I think that answers the fixed/float question. If they give the oil density then you don't need to determine it via a buoyancy method.

I prefer don't start new threads for every problem about the same think.
You will probably find that more people will view your posts if they are in new threads. They are unlikely to wade through all the older (probably solved) material to reach your new idea or problem.
 
  • #10
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yes it float in a place fixed but why isn't cilinder mass important to calculare the height of mercury to put in?

Sorry for the double posting before :(
 
  • #11
gneill
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The problem I have with the cylinder floating freely (not held still by some external mechanism) at the depth they give (25cm of it submersed), is that it doesn't displace enough oil to float. The weight of the stopper plus cylinder is 1.863 N. The weight of oil displaced is Area*25cm*ρoil*g = 1.155 N, where Area is the cross-sectional area of the cylinder. The cylinder should sink further into the oil.

Also, just taking the atmospheric pressure into account, the force holding the stopper in place in the empty cylinder would be:

f = Patm*Area = 49.738N

For the weight of the stopper plus water on top of it to equal that force, the height of the water column would have to be

(f/g - 90gm)/ρwater/Area = 10.15 meters

which doesn't make sense, given that the cylinder is only given as 50cm in length.

Are you sure that the cylinder has no air inside (it's a vacuum)?
 
  • #12
181
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perhaps the cilinder isn't immersed for half-height but i remember it floated...
 

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