How much force will be required to remove the lid?

  • Thread starter Thread starter iseidthat
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
To remove the lid of an airtight box, the force required varies with atmospheric pressure changes. At a mountain altitude with a pressure of 0.81·10^5 Pa, the force calculated is 3240 N. When taken to sea level, where the pressure is 1.013·10^5 Pa, the force increases to 4052 N. The key factor in determining the force is the pressure difference between the inside and outside of the box. Understanding this pressure differential is crucial for calculating the force needed to remove the lid.
iseidthat
Messages
15
Reaction score
0

Homework Statement


An airtight box has a removable (massless) lid of area 4·10-2 m2. A lid is placed on it while it is on top of a mountain (where Patm=0.81·105 Pa). It is then taken to sea level, where Patm = 1.013·105 Pa. How much force will be required to remove the lid?

Homework Equations


F=PA

The Attempt at a Solution


For the first part:
F=PA=.04 x 81000=3240 N
For the second part:
F=PA= .04 x 101200=4052 N

is there a change in pressure type of equation?
 
Physics news on Phys.org


You are only interested in the change of pressure. - this gives you the pressure differnce between the inside and outside of the box and so the force to open it
 
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 '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