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Dynamics Question

  1. Mar 2, 2005 #1
    If friction is negligible, and masses of 100g and 50g are pushed across with an applied force of 1N of the greater mass, how do I find the magnitude of the force exerted by each of the two masses on eachother?
     
  2. jcsd
  3. Mar 2, 2005 #2
    What do you mean by "applied force of 1N of the greater mass"? That's kinda confusing. I might be able to help you.
     
  4. Mar 2, 2005 #3
    oops sorry...

    I meant applied 1 N of force on the greater mass, that being the 100g mass.
     
  5. Mar 2, 2005 #4
    OOOHHH!!

    Ok, so let me see if i've got this...your 100g mass is moving and the 50 g mass is stationary...right?
     
  6. Mar 2, 2005 #5
    Both masses are in contact with eachother when they are being moved. Like the 50g mass is on top of the 100g one.
     
  7. Mar 2, 2005 #6
    Alright, that makes this a bit easier. What do you have for work so far if anything? Because you can count both of them as one mass since there in contact. That elminates one variable.
     
  8. Mar 2, 2005 #7
    I've only found that acceleration is 6.7m/s^2
     
  9. Mar 2, 2005 #8
    Ok...I have notes on a similar problem. Let me see if I can point you in the right direction.
     
  10. Mar 2, 2005 #9
    Couldn't you use the Fnet=ma formula?

    Input the total mass as m and use the 6.7 m/s^2 as your a.
     
  11. Mar 2, 2005 #10
    That's what I thought but it didn't work. I think that 6.7 is the horizontal acceleration, thats why.
     
  12. Mar 2, 2005 #11
    Hmm, this is interesting. It would've been eaiser had they asked you for the coefficient of kinetic friction. lol.

    Well, there is no vertical acceleration so that's out. Are you looking for WORK or FORCE?
     
  13. Mar 2, 2005 #12
    Do you have any other information?
     
  14. Mar 2, 2005 #13
    I'm looking for the magnitude of the force exerted by each of the masses on the other. I've mentioned all the info thats given.
     
  15. Mar 2, 2005 #14
    Grr, unfortunately I've never seen a problem like this. All I can think of is newton's thrid law: the forces of two objects acting on one another or of one object acting on another object, the contact forces will be equal and opposite.

    That's as far as my thinking takes me right now. Sorry love.
     
  16. Mar 2, 2005 #15

    learningphysics

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    Draw the freebody diagram of the top mass. Draw all the forces being exerted on the top mass. There is weight.... what else is there?

    Then use F=ma in both the horizontal and vertical directions to solve for the unknowns. You know a=0 in the vertical direction, and a=6.7 in the horizontal direction.
     
  17. Mar 2, 2005 #16

    chroot

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    harhar,

    Can you please state the problem exactly as it was given to you, all in one post?

    - Warren
     
  18. Mar 2, 2005 #17
    How did you come up with the acceleration of 6.7 m/s^2?
     
  19. Mar 2, 2005 #18
    Ok the EXACT question is the same concept but w/ different numbers cuz I just tried to use easy numbers to deal with..

    2 masses in contact w/ eachother are 0.113kg and 0.139kg. No friction. The applied horiz. force of magnitude 5.38x10^-2 N is exerted on the heavier mass.

    Determine:
    a)magnitude of accel. of the 2 mass system.
    b)magnitude of the force exerted by each of the 2 masses on the other

    I've already gotten a) which is 0.213m/s^2
     
  20. Mar 2, 2005 #19

    chroot

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    I assume there's a picture, also, showing one mass above the other. You need to tell us about those parts of the question when posting your question here, or we won't be able to make sense of it.

    The answer to b) is simply the weight of the mass on top. The top mass presses its weight down on the bottom mass, and the bottom mass in turn pushes back up, supporting it.

    - Warren
     
  21. Mar 2, 2005 #20

    learningphysics

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    Are the masses on top of each other, or just side by side with one pushing the other along? Is that mentioned in the question? That will affect the answer to b). If it is one on top of each other, have a look at my previous post.
     
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