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Homework Help: Hooks Law, did I think correctly?

  1. Jan 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Hello there!

    I'm having a big issue with understanding my books explanation. The image is taken from my book. The book is trying to prove the formula F = -k*y. This is the chapter ( Simple Harmonic motion)

    They got the following equations(Explanation from book):
    ks0 = mg
    k(s0+s1) = ks0 + ks1
    F = ks0 + ks1 - mg = ks1, because ks0=mg
    ''But the y-coordinate is negative, and equal to -s1. Also F can be written as F = -ky''



    2. Relevant equations

    I can't understand this at all. IF I define positive direction as upwards, and negative as downwards, just like they have done I suppose, how isn't the s0 and s1 negative because the spring is obviously stretched downwards?

    Shouldn't s0 be a negative value? Because for example if the starting position is 10cm, and it moves to the ''Jämviktsläge=zero level'') then it should move 0-10 = -10cm from its starting position, in other words; it's stretched -10cm?
    The same with s1, if negative direction is down, then it should move -2-(0) from the resting position is this correct?

    However I get totally wrong equations, unlike theirs.. I've been trying for 4 hours now..

    Best regards,
  2. jcsd
  3. Jan 5, 2014 #2

    Doc Al

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    Staff: Mentor

    s0 and s1 are just distances and thus positive. The displacement from equilibrium would be negative.
  4. Jan 5, 2014 #3
    F = -kx
    The negative means it's a restoring force.

    Consider a spring which is horizontal, and a mass attached to the right side of it.
    If you move it to the right (positive), the force is negative (so it points to the left)
    If you move it to the left (negative) the force is positive (so it points to the right).

    And Hooke's law is based on observation (take a spring and test it).
  5. Jan 5, 2014 #4

    I can't understand why they are positive in that case. I assume.
    For example, always when I've calculated I've thought of it as negative in another chapter (Movement in 2D)

    Which is driving me insane..
    If I have a water-level that is defined as my zero-point and I stand 20m above, at a cliff and drop a stone, then it would travel -20m. If I throw it up from the water-level it would go +20m.

    What I'm trying to say is what is the difference here? :(

    Thank you for the fast response!

    Best regards,
  6. Jan 5, 2014 #5
    I do know this, but I'm trying to understand how they have explained the equation, which I obviously can't understand for some reason! :(

    Best regards,
  7. Jan 5, 2014 #6

    Doc Al

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    Staff: Mentor

    All they are concerned with here are the magnitudes of the forces.

    The minus sign in Hooke's law just says that the spring force is always opposite to the displacement.

    That's perfectly OK too. Here they are using the sign convention and the net force F equals -ky, which is an upward force since y is negative.
  8. Jan 5, 2014 #7

    So I would do like I was trying to, (like in my original post, and think in the way like I did in the second post of mine..) if I wanted to figure out the directions of the forces or..? :s
    Because if I understood you correctly the book has simply assumed that mg goes downwards, ks goes upwards and therfore not paid any attention to the direction of the displacement?

    It seems like I've almost understood.
    Best regards,
  9. Jan 5, 2014 #8

    Doc Al

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    I think you're getting it.

    The book indicates the direction of the forces by the vectors drawn in the diagram. The magnitudes of those forces are given by ks0, ks1, and mg.
  10. Jan 5, 2014 #9

    So lets say if I wanted to calculate the direction of the vectors(forces), would I then pay attention to the direction of the s0 and s1?

    If not then, when would I think like I was trying to do in my original post?

    For example if I've the equation on its final form: F = -ky. If I then assume that up is equal to the positive direction, and down is equal to the negative one. If I then use the equation F = -k * 2 I would assume that the spring has stretched 2m inwards, and -2 that the spring has stretched 2m downwards?

    When they wrote the y-coordinate is negative and equal to -s1, I assume that they then mean that the negative direction is downwards, positive upwards?

    Best regards,
  11. Jan 5, 2014 #10

    Doc Al

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    Staff: Mentor

    You always pay attention to the direction of the forces and displacement. Your book did! That's how they knew where to draw the arrows.

    If you want to use the equation to tell you the direction of the force, then use F = -ky. When y is negative, which is a downward displacement, then F is positive and upward. The reverse when y is positive.

  12. Jan 5, 2014 #11
    I'm with you on all points except the first one, Al.

    Mg is drawn downwards, Ks0 is drawn upwards. But! If s0 is + or - direction, I can't see where that's mentioned. Because if I would of defined positive direction upwards, and at b) We can see that the spring has a downward displacement. Would not that mean that s0 infact is a negative value? That's the part that's confusing me.

    Then it would of been like ks*-s0 if I determined positive direction upwards, and ks*+s0 if I determined it downwards, according to the above.

    (I would do this, if I wanted to know the direction of the forces according to me), if we already know the vector direction of the force, then we just take absolute value of the displacement.
  13. Jan 5, 2014 #12

    Doc Al

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    Again, s0 is just a distance, a magnitude. You can see from the diagram that the displacement is negative (downwards).

    If the spring is displaced a distance s0 downwards, the (signed) displacement is y = -s0.

    s0 and s1 are just positive numbers.
  14. Jan 5, 2014 #13
    Ok, so if I understood you correctly.

    Because we know the direction of the forces, we take the absolute value of the displacements s0 and s1, because Fspring = k * (Displacement), and we already know which direction the force is going into, so it's to no interest to us to take the signed displacement into consideration?

    ''If the spring is displaced a distance s0 downwards, the (signed) displacement is y = -s0.''

    So you mean that if it displaces anywhere downwards, the signed displacement is y = -s0, it can be both above and below the zero-level (Jämviktslaget)?

    Best regards,
  15. Jan 5, 2014 #14

    Doc Al

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    When the direction is obvious, all you need is the magnitude.

    I'm not sure what you mean by 'zero-level'. Do you mean the equilibrium point, where the net force on the mass is zero?

    In your diagram, s0 is the distance between the unstretched position of the spring and the equilibrium point once the mass is hung from it. When the mass is pulled down an additional distance of s1 below the equilibrium point, the spring is now stretched a total distance of s0+s1.

    When the mass is released, it will oscillate up and down about the equilibrium point. It will go from a distance of s1 below equilibrium to a distance of s1 above equilibrium.
  16. Jan 5, 2014 #15
    I'm not sure what you mean by 'zero-level'. Do you mean the equilibrium point, where the net force on the mass is zero?
    Yes exactly.

    I'll re-write my question a bit perhaps was a bit unclear.

    So you mean that if the spring displaces anywhere downwards, the signed displacement is y = -s0, can it be both above and below the equibrilium point (Jämviktslaget), as long as the direction is downwards?

    For example, imagine that the spring moves only a tiny bit down, (it's still above the equibrilium point), is the displacement still y = -s0?

    On the other hand if it displaces anywhere upwards, the signed displacement is y = +s0?

    ( Based on that the Positive direction defined up, negative down)

  17. Jan 5, 2014 #16
    This is an image of what I ment;

    http://s27.postimg.org/z5uaeoww3/thisedit.png [Broken]

    I mean that when the spring goes downwards, but it's actually still above the equibrilium point, is the displacement then -S3, if we name the displacenment S3? And is the sign then - if we wish to take the direction in consideration and not only the magnitude. (The displacement with respect to it's sign).

    The black line is the equibrilium level.

    Sorry for my bad english.

    Best regards,
    Last edited by a moderator: May 6, 2017
  18. Jan 5, 2014 #17

    Doc Al

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    Staff: Mentor

    Careful: s0 is a fixed distance = mg/k.

    Measuring the displacement of the spring from its unstretched position, anytime it is stretched downward you can call the displacement negative (regardless of whether it is above or below the equilibrium point). If it is compressed (displaced upward) then the displacement is positive.

    Note that it is common practice, when describing the motion of a mass hanging from a spring to measure displacement from the equilibrium position.
  19. Jan 5, 2014 #18
    Ok, I'm with you on all points however there's just a final thing I'm a bit unsure of.

    If we're only looking at the magnitude of the displacement, then in our formula; F = -ky, why do we pay attention to the displacement when we're going below our equibrilium, for example if we go 3 units below we insert: y = -3, then we've paid attention to the direction of the displacement and not only the magnitude, because then it would be y = +3?

    Best regards, and thank you for all of the help. Very good explanations!

    Edit; Or do we now pay attention to the direction, because we want to calculate both the magnitude and the direction of the force and not only the magnitude?
    Last edited: Jan 5, 2014
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