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Spring, Amplitude and Acceleration

  1. Nov 28, 2005 #1
    Ok, I'm stuck again.....

    A block of mass M1=0.490 kg sits on a block with a mass of M2=4.50 kg that rests on a frictionless surface and is connected to a spring. The spring has a spring constant of k=16.6 N/m. The block is displaced and undergoes Simple Harmonic Motion. What is the largest amplitude (in meters) the block can have for the smaller block to remain at rest, relative to the larger block? The coefficient of friction between the two blocks is m=0.300


    Ok so I tried like 5 different ways to work the problem and I know at least one of the ways I tried was wrong. I was working with F=ma. I was using a=g (acceleration=gravity) and so F=mg, and also mg=kx. I found my force (44.15 N) and multiplied it by my friction coefficient (0.300) and got 13.245 N so then I divide by my spring constant and get 0.79789 m. Which is wrong. My question is "What am I doing wrong and how do I fix this problem? :confused:

    Thanks abunch.
     
  2. jcsd
  3. Nov 28, 2005 #2

    Doc Al

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    Hint: What's the maximum force that the bottom block can exert on the top block? What's the corresponding maximum acceleration?
     
  4. Nov 28, 2005 #3
    Is the max. force exerted on the second block by the first block (0.49 kg)(9.81 m/s^2)? Which would make the max. acceleration (9.81 m/s^2) right?
     
  5. Nov 28, 2005 #4

    Doc Al

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    No. What I should have asked is what is the maximum horizontal force that the bottom block can exert on the top. Hint: The only horizontal force between the two blocks is the friction force.
     
  6. Nov 28, 2005 #5
    But friction force is the coefficient of friction multiplied by the normal force, and I can't figure out the normal force because I don't know the acceleration. I am very confused.
     
  7. Nov 28, 2005 #6

    Doc Al

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    One block sits on top of the other. What's the normal force between them? (Note: There's no acceleration in the vertical direction, which is the only direction that could affect the normal force.)
     
  8. Nov 28, 2005 #7
    I am really sorry. I know that I am probably missing the most simple solution, and I am truly not trying to get you to do the problem for me, but I honestly don't know what to do. I don't have either force or acceleration and need them. I am confused and on top of it have been told that this problem is basically as hard as it gets in the course I am taking. Maybe that should be a good thing, but right now it is very daunting. Thank you so much for your patience with me!
     
  9. Nov 29, 2005 #8

    Doc Al

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    We'll lick this problem, step by step. But try to answer my questions as I ask them.

    The first thing to realize is that the only horizontal force acting on the top block is the friction from the bottom block. So let's find the maximum value of that friction. To do that, we need the normal force between the blocks. So, answer my question: One block sits on top of the other. What's the normal force between them? (Hint: This one's easy.)
     
  10. Nov 29, 2005 #9
    0.3(M1)(g)=0.3(0.49 kg)(9.81 m/s^2)
     
  11. Nov 29, 2005 #10

    Doc Al

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    The normal force is just the weight of the top mass: [itex]N = mg[/itex]. You went one step further and found the maximum static friction force that the bottom mass can exert on the top mass: [itex]f_{max} = \mu N = \mu mg[/itex]. Good!

    Now, knowing the maximum horizontal force that can be applied to the top mass, use Newton's 2nd law to find the maximum acceleration of the top mass. (Hint: Don't be in a rush to plug in numbers. Work with symbols as much as possible; only plug in numbers at the last step.)
     
  12. Nov 29, 2005 #11
    so (a=F/m), correct?
     
  13. Nov 29, 2005 #12

    Doc Al

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    Yes. So what does that maximum acceleration equal? (In symbols, not numbers.)
     
  14. Nov 29, 2005 #13
    a=(M1)(g)(mew)/(M1) so a=(g)(mew)
     
  15. Nov 29, 2005 #14

    Doc Al

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    Right. Now, since the top and bottom block must move together, this is also the maximum acceleration of both blocks. Once again, use Newton's 2nd law to find the force needed to produce this acceleration on both blocks together.

    Since the spring is what exerts this force, use Hooke's law (the spring force law) to find the maximum displacement that corresponds to this maximum force.
     
  16. Nov 29, 2005 #15
    a=(M1+M2)(g)(mew)/(M1)? I confused myself again........


    ps-i will be back at noon est.
     
  17. Nov 29, 2005 #16

    Doc Al

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    You have the acceleration ([itex]a = \mu g[/itex]) and the mass of both blocks ([itex]M_1 + M_2[/itex]). Now combine them (using Newton's 2nd law) to find the force.
     
  18. Nov 29, 2005 #17
    ohhhhhhh.......so F=ma so F= (mew)(g)(M1+M2) right?
     
  19. Nov 29, 2005 #18

    Doc Al

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    So far, so good. Now what spring displacement would produce that force?
     
  20. Nov 29, 2005 #19
    d=F/k correct? so that would mean d=(M1+M2)(g)(mew)/(k) right?
     
  21. Nov 29, 2005 #20

    Doc Al

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    Right. That's all there is to it, if you understand what you've done. As long as the amplitude of the motion remains less than d, the friction will be enough to keep the top block stuck to the bottom block.
     
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