1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measure of spring in system with roller

  1. Jun 11, 2017 #1
    1. The problem statement, all variables and given/known data
    m1= 0,05 kg, m2= 0,03 kg, mass of spring is omitted. We look at movement after spring stopped vibrate. How long does the spring measure during the movement if not stretched spring measure 0,1 meter and if we use force 0,1N elongation measures 0,02 meter.
    (sorry, English isn't my main language, I've never learn scientific words so I'm not quite sure if my translation is right)

    I know the answer is 0,174 meter.
    2. Relevant equations

    3. The attempt at a solution
    m1*a = m1*g - Ft (Ft is thread tension)
    m2*a = Ft + Fs - m2*g

    0,1 = k*0,02

    Don't know what next and if it's even right.

    Attached Files:

    • nn.jpg
      File size:
      24.2 KB
  2. jcsd
  3. Jun 11, 2017 #2
  4. Jun 11, 2017 #3
    Hym, I can solve problems when there is no spring. That spring is real problem since I don't know how to mark forces at that device. Is Fs=Fg? My friend tried to solve that without Fs but I'm not sure...
  5. Jun 11, 2017 #4
    Why would Fs be Fg?
    Is the spring applying force in either direction? what does Newton's third law tell us?
    Last edited: Jun 11, 2017
  6. Jun 11, 2017 #5
    I'm not sure but aren't second mass applying force? Can't Fg be force of action and Fs force of reaction?
  7. Jun 11, 2017 #6
    Is there any effect from mass 1?
  8. Jun 11, 2017 #7
    Well, because of mass 1 that system is moving to the left.
  9. Jun 11, 2017 #8
    I think you have the right idea. Translation may be a little off. Because of mass 1, the force on mass 2 does not equal the the force of gravity on mass 2.
    With that being said, because the spring is without mass it does not contribute to the acceleration of mass 1. The tension in the string, Ft , is therefore also the same force acting on the spring, Fs .
  10. Jun 11, 2017 #9
    0,1 = k*0,02

    m1*a = m1*g - Ft
    m2*a = Ft - m2*g

    a(m1 + m2) = g(m1 - m2)
    a = 0,2/0,08 = 2,5 m/s^2

    Ft = m2(a + g) = 0,03*12,5 = 0,375 = Fs

    0,375 = 5*x
    x = 0,075 m

    x+0,1 = 0,175 m = 17,5 cm

    Is it correct?
  11. Jun 11, 2017 #10
    Looks like it
  12. Jun 11, 2017 #11
    Ok, thank you!
    But can I have one more question? How should I draw those forces at that picture? Should I draw Fs and Ft or only one of them? They should "touch" mass 2 or spring?
  13. Jun 11, 2017 #12
    You could explicitly show all of the forces (so that the spring looks like the green to the side):
    But I think that would be messy and overly complicated. I personally would only draw forces on objects with mass.
  14. Jun 11, 2017 #13
    Right, I don't think I have to do that much.
    To be 100% sure, it's ok?

    Attached Files:

  15. Jun 11, 2017 #14
    Looks fine to me
  16. Jun 11, 2017 #15
    Ok, thank you. :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted