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Homework Help: Pushing blocks - static friction, force, and accel

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data

    A block of mass 1.2 kg rests on top of another block of mass 1.8 kg, which rests on a frictionless surface. The coefficient of static friction between the blocks is mu_s = 0.3.

    (a) What is the maximum horizontal force that can be applied to the upper block so that the blocks accelerate together, without the upper block sliding on the lower one?

    (b) Suppose that the force is applied to the lower block instead; what is the maximum horizontal force that can be applied in this case so that the blocks will accelerate together, without any slipping between the two?

    2. Relevant equations



    3. The attempt at a solution

    Not sure how to start going about this one.

    I tried this but it didnt work out to be right and i think i am leaving something out.

    i found the normal force on the top block

    n=1.2(9.8) = 11.76

    then i found the fs

    0.3 = fs/11.76


    i knew that wasnt right cause i didnt include anything about the bottom block moving with the top block so i found the fs of this bottom block as well

    0.3 = fs/17.64

    fs = 5.292

    Not sure what to do now? i think if i add the two fs's together that will give me part (b) which would be 8.82 N for the force and i believe that is correct but im not sure how to get part (a)

    Any help would be appreciated.

    Thank you :)
  2. jcsd
  3. Feb 21, 2010 #2
    (Part a) Draw a FBD for the bottom block, and then determine the maximum acceleration for the bottom block w/o the top block slipping. This maximum acceleration will be the acceleration for the top block as well, since they are moving together (no slipping). Be careful when you draw the FBDs for the top and bottom block, there are action-reaction pairs, fs and N, that must be treated carefully.
    Last edited: Feb 21, 2010
  4. Feb 21, 2010 #3
    ok i drew my FBD for the bottom block but since i do not know the force that is pushing the blocks how can i find the acceleration?

    thank you for the help :)
  5. Feb 21, 2010 #4
    Hang on i may have found it.

    i found the fs for the bottom block


    fs=8.82 (which is the answer for part (b)

    8.82 = (1.2+1.8)(a)

    a = 2.94 m/s^2

    if this is right im not sure how to use this to find the force that can be applied to the top block with no slipping :(
    is that correct?
  6. Feb 21, 2010 #5
    Stick with (Part a) for now...

    When you draw the FBD for the bottom block, where does the fs force come from? What object is providing this force, and in which direction does it point?
  7. Feb 21, 2010 #6
    ok. so on my FBD the fs force comes from the normal force times by the coefficient of static friction -----> 0.3[(1.2+1.8)(9.8)] = 8.82

    this points in the opposite direction of the force being applied. so on my diagram i have the force we are trying to find pushing the blocks to the right so the fs force will be to the left.
  8. Feb 21, 2010 #7

    The fmaxs does not include the mass of the bottom block. Take a look at this picture, label the forces, and then see if it agrees with your FBDs:

    http://img16.imageshack.us/img16/4536/blocks3.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  9. Feb 21, 2010 #8
    yes you are right the fs is only determined by the top block cause its on a frictionless surface so there will be no friction on the bottom of the bottom block i forgot about that. yes that is what my FBD looks like.

    ok so the fs is then


    fs = 3.528

    is this correct?

    would i then use this to solve for acceleration?


    a= 1.176????
  10. Feb 21, 2010 #9
    Yes, this is fmaxs = mu_s m g

    here you're using (m+M). Why? Look at the FBD for the bottom block.
  11. Feb 21, 2010 #10
    i thought i had to use the total mass because both blocks are moving with a certain acceleration.


    3.528 = 1.8(a)

    a = 1.96 m/s^2

    then to solve for part (a)

    F = (3)(1.96)

    = 5.88 N and that is correct :)

    does that all look right?
  12. Feb 21, 2010 #11
    Yes, that's the answer, but note how you quickly wrote

    Fmax = (m+M)amax

    This equation comes from a FBD for the system (m+M), which is a FBD that we haven't drawn yet. See if you can get the same result using the FBD for the small block, which we have a picture for.
  13. Feb 21, 2010 #12
    so using the small FBD i would....

    F = (1.2)(1.96) = 2.352

    then i would add this force to the fs force correct to get the total force in the x direction that is allowed to be applied to the top block with no slipping.

    F_total = 2.352 + 3.528 = 5.88 N :)

    is that the right way to do it?

    hmmmmm now the right way to do part (b).....
  14. Feb 21, 2010 #13
    Sort of. You're confusing some ma's with F's, etc.

    Just use the FBD...

    The FBD for the top block says

    F-fs = max

    and therefore

    Fmax = fmaxs + mamaxx

    Just remember, for a problem like this, you can make a FBD for each mass separately, and then you can also make one for the system. The one for the system would not have fs and N, these are reaction-action pairs that cancel since they are "internal" to the system.

    Onward to (Part b)...
  15. Feb 21, 2010 #14
    Thank you so much for explaining that last part. that really makes sense and i think that is how my professor will want to see the work done.

    ok so part b

    the fs would be different so i found that

    fs = 1.8(9.8)(0.3) = 5.292

    now to find the Fmax i would do the same as in part a

    Fmax - fs = ma

    Fmax = (1.8)(1.96) + 5.292

    Fmax = 8.82 N

    Look ok?
  16. Feb 21, 2010 #15
    The answer is correct, 8.83N, but how you got it doesn't make sense to me.

    Draw new FBDs, for the top and bottom block. fs changes direction for part (b), and then write down the equations for Newton's 2nd postulate.
  17. Feb 21, 2010 #16
    i didnt even think about that. yes fs does change direction when pushing the bottom block

    im not sure how to come up with a newtons 2nd equation for this?

    would it be

    Fmax - fs(of the bottom block) = ma
  18. Feb 21, 2010 #17
    No, you have to be careful with m and M, they're not the same thing. Here's a picture to get you started again:

    http://img705.imageshack.us/img705/6676/blocks4.jpg [Broken]

    Just do what Newton said to do! :)
    Last edited by a moderator: May 4, 2017
  19. Feb 21, 2010 #18
    ok thats what i got for my FBD's

    newtons law is f=ma

    so you have the force we are trying to find pushing to the right and the fs of M is in the opposite direction.

    is the acceleration the same or is it different? is it maybe different because you are pushing on the bottom block now so the fs will be greater which allows the acceleration to be greater?
  20. Feb 21, 2010 #19
    i think i maybe got it but it may not be the right way to find it.....

    i found the new fs

    fs = 0.3(1.8*9.8)

    fs = 5.292


    5.292 = 1.8(a)

    a = 2.94

    Fmax = (m+M)(a)

    Fmax = 3(2.94)

    Fmax = 8.82 N

    correct or am i still thinking about it wrong?
  21. Feb 21, 2010 #20
    Newton says...

    for m:

    N - mg = 0
    fs = mam


    amaxm = fmaxs/m = us g

    for M:

    F - fs = MaM

    if aM = amaxm

    then, for (Part b)

    Fmax = us g (m+M)


    The result for (Part a) was, Fmax = us g (m+M) m/M


    I recommend that you do this problem over, from scratch, writing down three FBDs, and their attendant Fnet equations, for (1) m alone, (2) M alone, and (3) m+M together, just to make sure you understand action-reaction pairs.
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