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Summing the forces for an object hanging from a spring.

  1. Nov 9, 2016 #1
    1. The problem statement, all variables and given/known data
    "A 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length?"

    2. Relevant equations
    F = -kx
    W = mg
    The teacher said to call the force the spring pulls up with FT.

    3. The attempt at a solution
    Hello, I was wondering if somebody could help me understand my errors. I started the word problem by creating this diagram:
    upload_2016-11-9_18-11-18.png
    I then proceeded to sum the forces in the y-axis:
    ∑Fy: FT + W = ma
    and substituted to:
    -kx + mg = ma
    a = 0 so I re-arranged the equation to say
    ∑Fy: -kx = -mg
    I plugged the numbers from the problem into the equation as follows:
    -(250)(x) = -((1.5)(-9.8))
    I solved for x and arrived at the solution of
    x = -0.0588 m

    My teacher told me that my solution was correct, but my work was wrong. He said the work should look like this:
    ∑Fy: FT - W = ma
    -kx - mg = ma
    a = 0
    ∑Fy: -kx = mg
    -(250)(x) = (1.5)(9.8)
    x = -0.0588 m

    He explained that because the weight is in the negative (downward) direction I should subtract mg from -kx, and that I should plug in 9.8 for g instead of -9.8. He says that my way of adding together FT and W while plugging in -9.8 for g is incorrect because it "does not show the direction". I am confused about why I can not add together the two forces and plug in -9.8 for g, I must be missing something. He also explained that "The direction is taken care of in the free body diagram" what exactly does this mean? If direction is already taken care of in the FBD then how come we are even subtracting weight in the first place? I was wondering what exactly the free body diagram does for me and also why I am wrong in making g negative. Would the negative direction of the acceleration (g) not mean that weight is also negative? I guess what I am trying to say is, what makes my incorrect method different from the correct method, and why would I not plug in a = -9.8? Thank you for your help.
     
    Last edited: Nov 9, 2016
  2. jcsd
  3. Nov 9, 2016 #2
    Your teacher's and your solution are different. Both are can't be correct.

    You should take 9.8 as g because in your equation, ##mg -kx = 0##, You took negative y axis in Cartesian system as your positive direction and vice versa.
     
  4. Nov 9, 2016 #3

    haruspex

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    The answers only differ because you made a numerical error in the final step.

    In your method, you have taken up as positive, quite consistently, including that 'g' is the acceleration due to gravity, and this will take a negative value.
    That's all good.

    Your teacher's method, as you copied it, contains a sign error. It went from
    -kx + mg = ma
    to
    ∑Fy: -kx = mg
    So I am guessing what was intended was
    ∑Fy: FT - W = ma
    -kx - mg = ma
    [note the flipped sign]
    a = 0
    ∑Fy: -kx = mg


    This is ok, with the conventions that x is positive up, W is positive down, and g takes a positive value.
    Frankly, I prefer your choice of sign convention.
    (Ideally, you should always specify your sign conventions at the start. It makes it much easier for others to follow, and might even prevent mistakes.)
     
  5. Nov 9, 2016 #4
    I'm so sorry! That was a typo, we both arrived at a solution of -0.0588.
    I am not sure I understand what you are trying to say here, I thought I should mark the negative y direction as negative, and so I would assume (apparently incorrectly) that acceleration due to gravity is pointing in the negative y direction and so I should mark it as negative, why would I not do this?
     
  6. Nov 9, 2016 #5

    haruspex

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    I see no evidence for that. Am I missing something?
     
  7. Nov 9, 2016 #6
    My teacher did not tell me that, I have once again failed to catch my copying errors, he actually said that -kx - mg = ma where g = 9.8. I am sorry for being so careless, I will try much harder to catch these mistakes in the future. My confusion is why it would be incorrect for me to use -kx + mg = ma and make g = -9.8.
     
  8. Nov 9, 2016 #7

    haruspex

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    As I posted, there is nothing wrong with that. Indeed, I consider it the best style.
     
  9. Nov 9, 2016 #8
    I think you just blew my mind, how can both be correct? Also, what role does the free body diagram play in determining the sign convention? I will continue working these problems with my teacher's method (because hes grading). Is it the same to subtract the mg from FT with g = 9.8 as it is to add mg to FT with g = -9.8? The way he made it sound, it was not the same thing and it seemed to me like he was saying "the direction was taken care of in the FBD" what could this mean? I am having a lot of trouble understanding why using positive 9.8 is the only "correct" way to do this. Actually I am having trouble understanding the sign convention in general.

    I would also like to add that he told me that although the equation I wrote gives the correct numerical answers, it "does not show the problem correctly" so "it is wrong".

    So if hypothetically my method is correct, that's fine, but I am not allowed to use it. So because I will be working with my teachers method, I feel it is important to understand his logic. I currently do not understand the way he wants me to do it, or to put it in other words, how is his method is correct? (Specifically, how can he plug in a positive g?)
     
    Last edited: Nov 9, 2016
  10. Nov 9, 2016 #9

    haruspex

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    That's a good question.
    Arguably, the fact that you drew a down arrow and labelled it mg means you are taking mg to be positive down. On the other hand, it would look rather strange to draw the arrow pointing up. Perhaps you could label it -mg, and note in the working that up is positive everywhere.
    Your teacher's approach is to take each force as positive in the direction in which it is known to act. In more complicated problems, you sometimes do not know in advance which way some forces act, so you just have to adopt an arbitrary convention. As long as you are consistent it should all work out.
    If you adopt the convention that the force from the spring is positive up, the force of gravity is positive down, and that any acceleration of the mass is positive up, you get ma=T-mg. Yet again, you could make acceleration positive down, giving ma=mg-T. The bottom line is that the conventions ought to be stated. If using the FBD to indicate that, you could show the acceleration variable with an arrow there too.
     
  11. Nov 9, 2016 #10
    Thank you so much for helping me Haruspex, I think I understand it now. I really appreciate the time you took to explain this to me. Based on what you said about "more complicated problems", perhaps my teacher is trying to get us used to this method before we start adding in more forces to problems, that would make sense for the long run. Once again, I can not thank you enough!
     
  12. Nov 10, 2016 #11
    Hoophy took spring force as ##-kx##, which is also acting upwards in Cartesian system. So he took +ve y-axis as his negative axis. If he would have taken otherwise, then it would be ##kx##.
    In first case g should be positive and in second it should be negative.

    This was my reasoning, but I don't think you are wrong. Please help me by suggesting the mistake ?
     
  13. Nov 10, 2016 #12

    haruspex

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    A spring force acts in the opposite direction to the length change. Whether up is positive for both the force and the displacement, or negative for both, the force is -kx.
     
  14. Nov 10, 2016 #13
    I thought ##kx## is the magnitude and sign is the direction, which we need to assign depending on the situation.

    So is ##-kx## magnitude ?
     
  15. Nov 10, 2016 #14

    haruspex

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    A magnitude is by definition non-negative, so that would be |kx|.
    Even when dealing with scalars signs are important. If you define up as positive for everything (displacements, forces, velocities, accelerations, ....) then for a spring F=-kx. That says that the force acts oppositely to the displacement. As a differential equation, that leads to ##m\ddot x=-kx##, which has the usual SHM solution.
     
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