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Question about the Hooke's Law

  1. Nov 11, 2006 #1
    Hey, I am confused about the Hooke's Law about the spring. If I am holding a spring vertically and hanging weights on the bottom of the spring so that my spring will be stretched longer, do I get a positive delta x? The equation of Hooke's Law is F = kx. If I do get a positive delta x, do I get a negative tension force or k? ie, say the spring was 46 cm, and it was stretched to 55 cm, and the weight put on the spring was 100g. Then in this case, to calculate k, the tension force would be .1 * 9.8 = 0.98 and delta x = final x - initial x = 55-46 = 9cm. So then k would be -0.98/9 = -0.11? How can you ever have a negative k? This is the part I don't understand
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  3. Nov 11, 2006 #2

    Andrew Mason

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    Hooke's law is:

    [tex]\vec{F} = -k\vec{x}[/tex]

    The force is in the opposite direction of the displacement.

  4. Nov 11, 2006 #3
    So in the example I just gave, would delta x be 9 or -9?
  5. Nov 11, 2006 #4
    You need to think about the orientation of your coordinate axes.
  6. Nov 11, 2006 #5
    So if displacement is pointing the opposite of the force, then that means the coordinate system is pointing up. If so, delta x would be -9 cm. Can someone confirm this? My test is coming up tommorow (it's an online class) and I am frantically trying to understand this.
  7. Nov 11, 2006 #6
    ohh... it's an online, (multiple choice?), means you can't explain how you define your coordinates.... but BREATHE.... RELAX.... in those cases they usually ask for magnitude and direction, instead of asking positive, negative. The spring constant k is usually DEFINED to be a POSITIVE constant. So knowing the spring stretched 9cm down, at equilibrium the spring pulls up on the mass with a force of F and the mass pulls down on the spring with F, and the magnitude of the Force is 0.98 N, and the spring constant is 0.98N and the spring constant is 0.11 N/cm would be okay. Best advice I'd actually think of given your initial question wording is WATCH UNITS on things like forces, constants. Those probably will give you more insight on multiple choice exams.
  8. Nov 11, 2006 #7

    Andrew Mason

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    It does not matter whether you have greater or less than zero. It is an arbitrary choice. You just have to recognize that the force of the spring is in the opposite direction to the displacement of the spring. The spring force resists further extension. If down is + then x > 0 and F is - (up). If down is - then x < 0 and force is + (up).

  9. Nov 12, 2006 #8
    Thanks, alright I'm getting this. But in my textbook, I am seeing another equation, F = kx. and it says x is the compressed amount. What is this about. Are there two equations?
  10. Nov 12, 2006 #9
    That one probably just relates about the magnitude of the force to the magnitude of the change in x... not including the fact that these are vectors and have direction.

    In general, I don't think your exam will try to trick you up on things like negative signs. In questions they would, I'd think, usually put a picture there and ask for particular directions (up or towards the top of the page, down or towards the bottom of the page, left, or right). If they ask for + or -, I would think they would put a coordinate system in the picture for you. Has your instructor put any practice questions on the web for you to try? That would help you get in idea of what to expect.
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