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Acceleration & tension in a cable

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-10-11_21-28-49.png

    2. Relevant equations

    F=ma

    3. The attempt at a solution
    Does the total Force F=50N divide evenly between m1+m2 and m3? Meaning the tension between m1/m2 & m3 is 25N given m1+m2=m3. Also, is the acceleration between m1/m2 & m3 the same as the acceleration at m3?

    a=50/.098=510.2

    a=25/.049=510.2
     
  2. jcsd
  3. Oct 11, 2015 #2
    Have you tried drawing free body diagrams, or do you feel like you have advanced beyond the point where you need to draw free body diagrams?

    Chet
     
  4. Oct 11, 2015 #3
    Not quite yet. And yes I have. Thought the answer might be pretty straight forward though? No?
     
  5. Oct 11, 2015 #4
    Apparently not. Please show us your force balances on the three masses.

    If the length of the cable doesn't change, how do the displacements, velocities, and accelerations of the three masses compare?

    chet
     
  6. Oct 11, 2015 #5
    upload_2015-10-11_23-8-9.png

    Guess I'm not sure about displacement and velocity.

    Robb
     

    Attached Files:

  7. Oct 11, 2015 #6
    I don't see any force balance equations on any of the masses. Let T be the tension on the cable, and let f be the contact force between masses 1 and 2. Let's see some force balance equations (in the horizontal direction). One equation for each mass.

    If the cable doesn't get any longer, how would it be possible for the velocities and accelerations of the three masses to be anything but the same? They're "joined at the hip."

    Chet
     
  8. Oct 11, 2015 #7
    What do the 0.098 and 0.049 represent?
     
  9. Oct 11, 2015 #8
    You say that 3 kg = 0.029 N,
    2 kg = 0.020 N,
    5 kg = 0.049 N.

    This doesn't make sense to me. What are you doing to come up with this?
     
  10. Oct 11, 2015 #9
    upload_2015-10-11_23-53-24.png


    Sorry for the bad conversion! Anyway, we have learned zero about contact force (engineering physics 1) so not sure about that. .098 is m1+m2+m3 and .049 is m1 + m2.
     
  11. Oct 11, 2015 #10
    I want a separate free body diagram for each of the masses, and a separate force balance equation for each of the masses. Didn't they teach you to do this in your course?

    Chet
     
  12. Oct 11, 2015 #11
    upload_2015-10-12_0-37-2.png

    Is this more like it?
     
  13. Oct 12, 2015 #12
    No. That's only slightly like what I had in mind. Here is what I really had in mind:

    Capture.PNG
    Please tell me if it makes sense. If not, please pose questions. Also, now please solve these equations for a, T, and f.

    Chet
     
  14. Oct 12, 2015 #13
    You've learned about the normal force? That's a contact force. The table pushes up on the block, the block pushes down on the table.

    Block 1 pushes rightward on Block 2, Block 2 pushes leftward on Block 1. Those are contact forces.

    Huhh? ##3+2+5=10##.

    By the way, multiplying a mass in kilograms by 9.8 m/s² to get a force in newtons is not a conversion. Also, it's not good to get in the habit of writing things like 3 kg = 29.4 N.
     
  15. Oct 12, 2015 #14
    Thanks, Chet. Yes it makes sense. I'm not sure how to find T, though. Does this make sense for m3? I tried a couple of different ideas for solving for a and T. Obviously very different results.


    upload_2015-10-12_11-27-44.png

    I'm not believing T is .02N. This is like the example we have from class except we weren't finding T.

    upload_2015-10-12_11-28-22.png
     
  16. Oct 12, 2015 #15
    None of this makes any sense to me. We are not dealing with the force balances in the y direction at all. We are only looking at the force balances in the x direction. So I have no idea why you are using angles and y-components.

    The solution to this problem involves taking the three linear algebraic equations in post #12 and solving for the three unknowns, a, T, and f. This is a 9th grade algebra problem. The easiest thing to do is to add the three equations together; this eliminates T and f immediately, so that you are left with an equation for a. What is the algebraic solution to this equation for a (in terms of m1, m2, m3, and F)?

    Chet
     
  17. Oct 12, 2015 #16
    Sorry, Chet. Believe it or not I'm a pretty good math student I've just never had physics before and I feel like I have to figure a lot of this out on my own. Anyway, given that T is eliminated does that suggest that there is no tension on the connecting cable?
     
  18. Oct 12, 2015 #17
    Yes, but you will to need understand things in more depth if you want to be able to predict what will happen when you have a less simple relationship between m1, m2, and m3.

    Many times in math classes the focus is on answer-making. But in a physics class that's just one layer. The other layer is sense-making.

    In this case, Blocks 1 and 2 are being treated like a single object, making the contact forces they exert on each other internal forces.
     
  19. Oct 12, 2015 #18
    If you are good at math, you should be able to add my 3 equation together. What do you get? No, the.tension is not zero
     
  20. Oct 12, 2015 #19
    I think I understand that. Basically they null each other mathematically. So if the acceleration is constant then the acceleration at m1/m2= acceleration at m3?
     
  21. Oct 12, 2015 #20
    adding them and solving for a I get a=F/(m1+m2+m3)= 50/98= .51m/s

    m1a=T-f
    m2a=f

    T=a(m1+m2)=(.51)(49)=25N
     
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