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Forces and acceleration - two masses instead of one

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    A red box and a blue box sit on a horizontal, frictionless surface. When horizontal force F is applied to the red box, its acceleration has magnitude a = 5.09 m/s2.

    a) If force F is applied to the blue box, its acceleration has magnitude a = 1.24 m/s2. Find mred/mblue, the ratio of the mass of the red box to the blue box.

    b) Now, the two boxes are glued together. If horizontal force F is now applied to the combination, find a-combo, the magnitude of the acceleration of the combination of boxes.


    2. Relevant equations
    F = ma


    3. The attempt at a solution
    I already correctly answered question a, which came to be a ratio of .2436. The second part, however, always confuses me because there has to be some sort of manipulation. Intuitively, when we put together the two boxes, we are combining their masses, making the total acceleration even smaller than what they were individually. This is true because we are applying the same amount of force to both of the objects. To get the numerical answer for the combined acceleration, I have no idea. I already know that adding the two accelerations together (two vectors in the same direction) and then multiplying it by the ratio (6.33 m/s^2 * .2436) gives me an incorrect answer, so that is probably a poor approach. I'm just stuck, like the rest of the problems I've had. lol it's quite frustrating.
     
  2. jcsd
  3. Feb 20, 2012 #2

    tiny-tim

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    hi iJamJL! :smile:
    yes, that's right :smile:

    so imagine there's a green box, whose mass equals the sum of the masses of the red and blue boxes …

    what then? :wink:
     
  4. Feb 20, 2012 #3
    Hi tiny-tim,

    I have to admit, I'm still stuck at where I am. I'm imagining this scenario:

    We have three hockey pucks that are colored blue, red, and green (to your suggestion :approve:). The red one travels at 5.09 m/s^2 when I hit it with a certain force. When I hit the blue one, it travels at 1.24 m/s^2. If both of these pucks were to be hit at the exact same time, the blue one would be traveling .2436 times slower than the red one, leaving a gap that is increasing. Now if we introduce the third puck, and hit them all at the same time, the green one is even slower than the blue one because it is the heaviest.

    I understand that the balance has to remain. If we have a certain force, that force will not change (in the formula F = ma). When mass increases, acceleration must decrease to keep F the same. I just don't know how to apply numbers to this scenario.
     
  5. Feb 20, 2012 #4

    tiny-tim

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    hi iJamJL! :smile:

    yes, everything you say is correct …

    i don't understand why you're not getting this :confused:
    in other words: mass is inversely proportionate to acceleration

    so if the masses were say 2 3 and 5,

    the the accelerations would be in the ratios 1/2 1/3 1/5​
     
  6. Feb 20, 2012 #5
    Sorry, I get mind blocks with these types of problems. I don't know why I don't understand it either :cry:


    LOL I'm going to take two wild shots at what you're saying because my mind is struggling.

    1) Since the mass is inversely proportionate to acceleration, I can say that the accelerations should then be 1/1.24 + 1/5.09, leaving me with a total of about 1.0 m/s^2?

    2) The ratio is actually .2436, so that means it should be .2436/1.24 + .2436/5.09, leaving me with a total of .2443 m/s^2.
     
  7. Feb 20, 2012 #6

    tiny-tim

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    no, the accelerations are 1.24 and 5.09

    the masses should be in the ratios 1/1.24 and 1/5.09
     
  8. Feb 20, 2012 #7
    I must be stupid, lol. Let me see if I can type this out.

    If we start from the beginning, then we have the formula F = ma.

    For red, F = m(red) * a(red)
    F = m(red) * 5.09

    Blue: F = m(blue) * a(blue)
    F = m(blue) * 1.24

    We combine them, calling it green:
    F = m(green) * a(green)
    F = [m(red)/1.24 + m(blue)/5.09] * a

    lol I'm lost here. If we put the masses in ratios like that, then our formula F = ma becomes F = m(decrease)*a(increase). I don't see where the .2436 ratio applies.

    Sorry, tiny-tim. I don't want to have you work it out for me, but I'm just not seeing the solution.
     
  9. Feb 20, 2012 #8

    tiny-tim

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    So m(red) = F/5.09

    m(blue) = F/1.24
     
  10. Feb 20, 2012 #9
    :( I don't know what to do next..
     
  11. Feb 20, 2012 #10
    I suppose if we add those two together, m(red) + m(blue) = F/5.09 + F/1.24, and then it'll be 5.1F/5.09, and then we divide it to make 1.0019=a
     
  12. Feb 20, 2012 #11
    I think I might've come to another solution, but I do not know how to complete it. Let's take this try:

    Red, F = m(red)*a(red)
    Blue, F = [m(red)/.2436] * [a(red)*.2436]

    If I add the masses, then m(red) + m(red)/.2436 = [1.2436*m(red)]/.2436 = 5.1149

    Therefore, the acceleration is 1/5.1149 * 5.09 = .9951
     
  13. Feb 21, 2012 #12

    tiny-tim

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    hi iJamJL! :smile:

    (just got up :zzz:)

    i don't follow your last post at all :confused:

    this one is nearly correct :smile: (although difficult to follow at the end) …
    you'll find these questions easier if you write them out properly, one step at a time

    F = 5.09 ma, so ma = (1/5.09)F

    F = 1.24 mb, so mb = (1/1.24)F

    so ma&b = ma + mb =(1/5.09 + 1/1.24)F = 1.003 F

    so the acceleration of a&b = F/ma&b = … ? :wink:
     
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