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What should be squared?

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A hydraulic jack is used to lift a load of weight 1000N. The load-bearing cylinder has a radius of 4cm and the other piston has a radius of 1cm. What force must be exerted on the smaller piston to support the load?

    f2/f1 = A2/A1

    My question is:

    I have this calculated already but why the A2 must be squared and be 16 whereas I don´t need to square f2?

    When do I have to square and what? I thought I was supposed to square F2 and A2.

    Now, check the following problem:

    One section of a pipe has a cross-sectional area of 20cm^2. In this section water has flow velocity of 1.0m/s. Another section of this pipe has a constricted cross-sectional area of 5cm^2. If flow is steady, what is the water velocity in the constricted section?

    A2v2 = v1a1

    The A2 in this problem is not squared. Why?
    I thought A2 or V2 was supposed to be squared.

    Could anyone tell me exactly what is supposed to be squared when I am using any of these formulas?
     
    Last edited: Apr 19, 2012
  2. jcsd
  3. Apr 19, 2012 #2
    I am sorry, guys, but is there anything wrong with the question?
    I am not following the template simply because I do not have a question to be solved here. The answer is already on my notes. I just would like to understand the mathematics of it.
     
    Last edited: Apr 19, 2012
  4. Apr 19, 2012 #3
    I have an exam and I really need to have that clear to me. Could anyone just clarify this? Please?
     
  5. Apr 19, 2012 #4
    [itex]IMA=\frac {distance \ moved \ by \ input \ force}{distance\ moved\ by\ \ load}[/itex]

    What is AMA?
    What is the efficiency of a machine?
     
  6. Apr 20, 2012 #5
    My quizzes cover mostly the mathematical aspects of it instead of conceptual´s.
    That´s why I want to know specifically whether to square certain numbers or not.
     
    Last edited: Apr 20, 2012
  7. Apr 20, 2012 #6
    You can look at it from a pressure-area standpoint. The pressure is the same in both cylinders. Force is pressure multiplied by area. In one case the radius is 1, in the other it's 4. You square each of them to get the areas. 1 squared is still 1. 4 squared is 16.

    A bit of advice: Don't try to memorize formulas because in the end there will be far too many. You need to reason things out in the sciences. Understanding rather than memorization is the key.
     
  8. Apr 20, 2012 #7

    Integral

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    A is the area of the piston, you are given a radius of 4cm so now how is the area computed? The other piston's radius is also squared, but what is 1 squared?
     
  9. Apr 20, 2012 #8
    1 and 16.

    I think I am understanding. Thanks to both of you!

    See if I am correct:

    In the first problem, I am given the radius and to get the area, I should square it.

    In the second problem, the area is already given and that´s why it´s not squared.

    Would that be correct?
     
  10. Apr 20, 2012 #9
  11. Apr 20, 2012 #10
    Could you clarify another doubt? It´s about the exact same topic.

    It might be a little stupid but here it is:

    For some reason, I had in mind that every time I saw A2 or F2, I should square things.
    That´s because I read:

    A gas initially is at a pressure of 100kpa. If its volume is halved while the absolute temperature is kept constant the new pressure will be 200kpa.

    If that is Boyle´s Law p1v1=p2v2, I thought the answer was because 100 = 200.
     
  12. Apr 20, 2012 #11
    P1*V1 = P2*V2 so P2 = P1 * (V1/V2)

    But V2=V1/2

    Then

    P2 = P1 *(2) = 2P1

    If you halve the volume while the temperature is constant, the pressure doubles.
     
  13. Apr 20, 2012 #12
    Thanks a lot! Now I am able to answer the other questions by myself. :)
     
    Last edited: Apr 20, 2012
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