How Much Force Is Needed for One Million Atmospheres with Diamond Tip Presses?

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Diamond tip presses can generate pressures exceeding one million atmospheres, and calculating the required force involves using the formula P = F/A. For a circular tip with a diameter of one millimeter, the pressure converts to approximately 1.47 x 10^7 lb/in^2, leading to a calculated force of about 17,900 pounds. However, there is a discussion about the importance of using consistent units, particularly converting to Newtons since the diameter is given in millimeters. While some prefer pounds for a tangible sense of weight, using Newtons aligns with the SI system and avoids unit mixing. Ultimately, clarity in unit usage is crucial for accurate calculations in physics problems.
JWHooper

Homework Statement


Diamond tip presses are now capable of creating pressures in excess of one million atmospheres. Suppose the tips of the presses were circles with an diameter of one millimeter. How much force must be applied to the tips to create a pressure of one million atmospheres?


Homework Equations


P = F/A, where P is the pressure, F is the force and A is the area.
r is the radius in my solution.


The Attempt at a Solution


P = 10^6 atm, r = 1.97 x 10^-2 in
P = (10^6)(14.7 lb/in^2) = 1.47 x 10^7 lb/in^2 (unit conversion)
A = (pi)(1.97 x 10^(-2))^2 = 1.22 x 10^(-3) in^2

F = PA = (1.47 x 10^7 lb/in^2)(1.22 x 10^(-3) in^2) = 1.79 x 10^4 lb
Did I do anything wrong? I just want to make sure.

J.
 
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JWHooper said:

Homework Statement


Diamond tip presses are now capable of creating pressures in excess of one million atmospheres. Suppose the tips of the presses were circles with an diameter of one millimeter. How much force must be applied to the tips to create a pressure of one million atmospheres?


Homework Equations


P = F/A, where P is the pressure, F is the force and A is the area.
r is the radius in my solution.


The Attempt at a Solution


P = 10^6 atm, r = 1.97 x 10^-2 in
P = (10^6)(14.7 lb/in^2) = 1.47 x 10^7 lb/in^2 (unit conversion)
A = (pi)(1.97 x 10^(-2))^2 = 1.22 x 10^(-3) in^2

F = PA = (1.47 x 10^7 lb/in^2)(1.22 x 10^(-3) in^2) = 1.79 x 10^4 lb
Did I do anything wrong? I just want to make sure.

J.
Looks good. But you might want to convert the force to Newtons since the diameter was given in millimeters (personally, I like it in pounds).
 
PhanthomJay said:
Looks good. But you might want to convert the force to Newtons since the diameter was given in millimeters (personally, I like it in pounds).
So, it doesn't really matter if pounds is used or Newtons is used?
 
JWHooper said:
So, it doesn't really matter if pounds is used or Newtons is used?
The USA, Myanmar, and Liberia use pounds as the standard unit of force. Everywhere else in the world, the Newton (SI system) is the standard unit of force. Since the problem noted the diameter in mm, it may be looking for Newtons of force, not pounds, because it's bad enough working with 2 systems, let alone mixing the two. Being from the States, I like pounds, because I get a good 'feel' for how much 18000 pounds is. When you convert pounds to Newtons, 18000 pounds is about 80000 Newtons. I wouldn't have a good feel for how much 80000N is without first converting it to pounds. If this is a Physics problem where you are asked to use SI, use Newtons.
 
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