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Question related to SHM and free body diagrams

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Draw three free-body diagrams showing the forces acting on a toy suspended by a vertical spring. The diagrams should show the situations when the toy is at its equilibrium position, and above and below the equilibrium position.


    2. Relevant equations



    3. The attempt at a solution
    At eq. it is fine.. Mg=F restoring (though i don't belive that restoring force acts in the eq. postion cs F=-kx)... above eqm, mg + air resistance > k(x - d) .... (how is this possible).. when it is above F restoring should be inline with mg... my question is what gives the toy its upward motion.... ; below eqm, mg < air resistance + k(x + d). (again this time mg dominates) the reason it actually moves down... These are the answers given by my textbook... i feel that they r the other way round.. I need help in understanding the concept URGENTLY and thanks to whoever gives me a GOOD REPLY !!!
     
  2. jcsd
  3. Jun 12, 2012 #2

    Doc Al

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    Forget about air resistance for the moment. Just compare the weight (mg) with the restoring force. Note that the restoring force depends on the amount of stretch in the spring--measured from its unstretched length.
     
  4. Jun 12, 2012 #3
    Let us take above equillibrium position.. Where's mg pointing.. Always downwards.. What bout F.. again downwards... No what gives the upward motion.. I am confused...
     
  5. Jun 12, 2012 #4

    Doc Al

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    Why do you think the restoring force acts downwards?
     
  6. Jun 13, 2012 #5
    Restoring force acts towards the equillibrium position.. Isn;t it?
     
  7. Jun 13, 2012 #6

    Doc Al

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    No, the force of the spring always acts to pull back towards the unstretched position. In the equilibrium position (once there's a mass hanging from the spring) the spring is stretched and thus is exerting a force.

    (Note: The net force on the mass--the sum of the weight plus spring force--is towards the equilibrium position.)
     
  8. Jun 13, 2012 #7
    Pardon my dumbness!!! .... U said the force pulls back the unstretched spring towards equillibrium... So my words were right!!!
     
  9. Jun 13, 2012 #8

    Doc Al

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    The net force, not the spring force. But you need to understand where the net force comes from and why it always acts towards the equilibrium point (which is not the unstretched position).

    You want to show the individual forces on your free body diagram.
     
  10. Jun 13, 2012 #9
    Oh! Why do i feel that i am the dumbest person in this world... Dude, I guess i;ve got confused between Restoring force, Force exerted by spring and position of eq. postion... Please clarify cs i understand nothing if u please! Thanks
     
  11. Jun 13, 2012 #10

    Doc Al

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    Only two forces (ignoring air resistance for the moment) act on the mass: The weight, which you know acts down, and the force from the spring, which acts up.

    Answer these questions:

    Imagine the spring is just hanging there, unstretched. What force does it exert?

    Now hang the mass on the spring and bring the mass to the new equilibrium position. What force does the spring exert? What's the net force on the mass?

    If the mass moves above the equilibrium point, does the spring force increase or decrease? What happens to the net force on the mass?

    If the mass moves below the equilibrium point, does the spring force increase or decrease? What happens to the net force on the mass?
     
  12. Jun 16, 2012 #11
    Allright, let me take each sitauation. During eqillibrium point.. Net force should be 0, so mg=kx... Now, the sitaution below eqm, Net force is downwards, so y not mg>kx???? According to ur question, i really have no idea cs i messed between restoring force and the force of the spring. Btw, this was not a hw task, more than it was a imple question and i needed to know the concept so i would like a straight reply next time .. thanks|!
     
  13. Jun 16, 2012 #12

    Doc Al

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    OK.
    Why do you think the net force is downward?

    Only two forces act. The weight is the same. What happens to the spring force as the spring is stretched further downward? Does it increase or decrease? What direction does it act?
     
  14. Jun 16, 2012 #13
    Whether below or above, the restoring force should be the SAME!! beacuase F=-kX... ???
     
  15. Jun 16, 2012 #14

    Doc Al

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    You say the spring force is the same yet it is also proportional to x?

    When you stretch the spring downward, which way does the spring pull back on you?

    If you stretch it even further downward, how does the spring force change?
     
  16. Jun 16, 2012 #15
    well, the same whether the same distance x below or above.. Look!! I gotta anoither doubt which may make this simple.. What is K(x-d) what is d??
     
  17. Jun 16, 2012 #16

    Doc Al

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    x = amount by which spring is stretched.
    d = the position of the equilibrium position below the unstretched position; d = mg/k

    The two forces on the mass are:
    Upward force of the spring = kx
    Downward force of gravity = mg = kd

    Net upward force = kx - kd = k(x-d)

    Note that when the mass is above the equilibrium point x < d and thus the net force is downward; when below equilibrium, x > d and thus the net force is upward.
     
  18. Jun 16, 2012 #17
    I really appreciate your effort in making me understand but i am afraid, i am more confused now. To settle this down, am gonna give my own words and correct me if i am wrong.. (Please mark which situation you are talking about.. like for example, equillibrium position with /. wothout load)
    Anyway, when it is in the new eq. position with the load fx=mg... because the net force has to be zero. I agree with this... Now, mg > k(x-d)... I know that when in motion, the force tries to bring the toy back to the eq. postion.. That's y mg dominates. But what does it actually dominate over.. U said x= amount by which spring is stretched... What do u mean, which point was it stretched from?? Original eq. position or the new one.. In my situation here, when above eq. point, what is x??? Then u mentioned d "the position of the equilibrium position below the unstretched position; d = mg/k" Do u meand is the amplitude of the the new system when load is attached... And how did u come to know the formula... d=mg/k.. Thanks a lot by the way for bearing me ( I think am getting more stupid).. :(
     
  19. Jun 16, 2012 #18

    Doc Al

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    Please stop using 'text speak'--it's annoying.
    That's because at that point the spring force equals the weight. To find where that point is, just set kx = mg and solve for x. That's the distance we call 'd'.
    This is confusing. I assume you mean mg > kx, which is true when the mass is above the equilibrium point.
    There are only two forces acting. So it must be the spring force!
    x is the amount the spring is stretched, which is measured from the unstretched position. When the mass is at the equilibrium position, the spring is stretched by an amount 'd'. You can easily solve for 'd' as I indicated above.
     
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