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Homework Help: A problem about friction with three blocks stacked together

  1. Aug 19, 2016 #1
    1. The problem statement, all variables and given/known data
    I think I am lacking some concept. I can't solve this one:
    There are three blocks of masses 2kg, 3kg and 7kg lying on a frictionless table with the 7kg block on bottom 3kg in the middle and 2kg on the top. A horizontal force of 10N is applied on the bottom block. The coefficient of friction between the top block and the middle block is 0.2 and that between the middle block and bottom block is0.3. Find the accelerations of all the three blocks. Take g=10m/s^2. Please help.

    2. Relevant equations
    Newton's 2nd and 3rd laws and max static friction=uN where N is normal force and u is coefficient of friction.

    3. The attempt at a solution
    The max value of static friction between the bottom block and the middle block =
    =0.3*5*10 =15N
    Since force on the bottom block is only 10N , so the bottom block and the middle one cannot have relative motion, so they must move together. Let their common acceleration be 'a' and the friction between them be F1, then for the bottom block,
    10-F1=7a, ................(1)
    Now, for the middle block the forces, the acceleration of this block must also be 'a' because it is moving together with the bottom one. The forces on the middle block are 'F1'(forward) and F2 ( the frictional force from the top block in the backward direction). So, for the middle block,
    F1-F2=3a, ................(2)
    and, for the top block, Let it's acceleration be 'b' . Now, the only forward force on the top block is F2 which is the friction between it and the middle one. So, for the top block,
    F2= 2b ...................(3)
    I can't figure out what to do further. I don't know where I am supposed to use the coefficient of friction between the top and the middle block. If I don't know the value of F1, then how am I supposed to know if force on the middle block is greater than or less than the limiting static friction between the top and the middle block (which is 4N). So, how can I know if there will be relative motion between the top and the middle block or if the top one will also move with the same acceleration 'a'?
  2. jcsd
  3. Aug 19, 2016 #2


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    Much better!

    You now focus on the interface between the lower blocks. Fine, but: why ?
    (Just one question: suppose no sliding between blocks takes place. What would be the common acceleration ? -- then you can check if max frictions are exceeded).

    Remember the advice: "make a free body diagram of each of the blocks" ?
  4. Aug 19, 2016 #3
    I have already made free body diagrams. That's how I got those three equations. Now, you're saying to find the common acceleration of all three if they moved together and then see if F1 and F2 are exceeded. Thank you, I did that and found that all three had the same acceleration.
    BUT is there a more straightforward method to solve this type of question except this trial and error method?
  5. Aug 20, 2016 #4


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    With different parameters, this problem could be trickier. E.g. if the force at the bottom is 16N.
    A more systematic approach is to imagine the force at the bottom increasing until something slips. Given some set of masses and coefficients, which interface would slip first? Which second...?
  6. Aug 20, 2016 #5


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    With the remainder of your exercise (At least, from your preceding thread I understand there's more to do) you have a chance to weigh the intuitive approach against the more systematic approach. It's not all that complicated, but the non-linear behaviour of friction makes that this category of problems needs branching and that's always difficult to catch in general expressions. In fact you already applied that by noticing 10 N < 15 N, thereby excluding a branch. Similarly, my question was inspired by recognizing (10 N / mtotal ) < (4 N / mC), thereby excluding the sliding of the upper block.

    This exercise is fairly compact. In real life branch-and-bound problems, e.g. when designing a chemical process plant, the work to calculate through all branches is often impossible, so choosing promising branches to evaluate (if necessary by intuition, experience, etc.) is very important.​
  7. Aug 20, 2016 #6
    I think even with 16 N force on bottom the situation would have been the same. If we calculate the common acceleration of the three in this case, it'll be
    so frictional force on the bottom block, F1= 16-7a =16-7*5/3=4.33N which does not exceed 15N so bottom one will move with middle one.
    and F2=2a =2*5/3=10/3N = 3.33N which does not exceed 4N, so middle one will move with top one.
    So, the three will move with common acceleration even with 16N force.
    However, we can definitely think of parameters which could make this problem trickier.
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