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Static Friction

  1. Jan 8, 2016 #1
    1. The problem statement, all variables and given/known data
    If you have 2 boxes over each other in a car that is slowing down . The mass of each one of them is 10kg and Us = 0.4 for all surfaces. What is the maximum deceleration that doesnt make the boxes fall?
    And will it make different if you change the masses of the objects?
    3. The attempt at a solution
    I calculated the deceleration it is equal to -4m/s2
    By doing the following
    200*0.4 = 20 * a

    And, no it doesnt make difference. I have a feeling that this question is so weird...
     
  2. jcsd
  3. Jan 8, 2016 #2
    when you say fall, you mean slide or topple? I'm guessing slide.
    if so i think
    that what you have to do is evaluate for both boxes as a hole, evaluate for the top box, and remember that the two box system, the bottom box will feel the inertia from the top box
    i might be wrong though

    also you can think of the system as beeing at rest and then starting to accelerate, and in the end reverse the sign, if it helps... also drawing free body diagrams helps a ton
     
  4. Jan 8, 2016 #3

    haruspex

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    It knows nothing of the inertia of the top box. It will feel the normal and frictional forces.
    However, you are right that you can break the problem into two parts as
    1. If the top box slides on the lower one then....
    2. Else we can consider the two boxes as a unit, combining their inertias.
    As Biker found, the mass is irrelevant. Consequently the answer will be the same for the top box in isolation as it is for the two as a unit.
     
  5. Jan 8, 2016 #4
    what i meant is that as the bottom box is accelerating, the top box (because of friction) will impede the movement. isnt this correct?

    again this might be wrong, this is based on a exercice i've done a while ago which was similar, but the blocks were not inside a car, this might be what makes it wrong...
     
  6. Jan 8, 2016 #5
    Try drawing some free body diagrams, and you will see what haruspex is saying.
     
  7. Jan 8, 2016 #6

    haruspex

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    Yes, that's ok. I objected to your saying it felt the inertia of the top box. What it feels is the frictional force where they meet. It may seem like a subtle distinction, but confusing proximate causes with distal (ultimate) causes is a common source of error in multibody problems. One of the benefits of drawing an FBD per rigid body is that it forces you to focus on the forces directly experienced by the body.
     
  8. Jan 8, 2016 #7
    I understand bad phrasing on my part, and now that i've drawn everything out and solved it, i arrive at the same conclusion you did... Anyway i was confusing this with that other exercise i did, shame i can't get access to it, i really wanted to check it out now
     
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