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Blocks on a table problem

  1. Nov 19, 2005 #1
    Could anybody of you be so kind to help me a little bit with understanding what to do in this problem? Thanks in advance for your effort!

    A mass m2= 10 kg slides on a frictionless table. The coefficients of static and kinetic friction between m2 and m1 = 5 kg are us= 0.6 and uk= 0.4.

    [​IMG]

    a) What is the maximum acceleration of m1?

    Here I totally don't have a clue of the solution

    b) What is the maximum value of m3 if m1 moves with m2 without slipping?

    What I've done is the following:

    fs= us * Fn = 0.6 * 9.81 * (10+5)= 88.29 N= T= m3 * g
    m3= 88.29/9.81= 9 kg.

    (I've calculated Fn by thinking of m1 and m2 as one system, but is that correct?)


    c) If m3= 30 kg, find the acceleration of each body and the tension in the string.

    For m1:
    Fres= fk= uk * Fn = 0.4 * m1 * 9.81= 19.62 N
    Fres= m1 * a therefore a= 19.62/5= 3.924 m/s^2

    For m2:
    Fres= T- fk = Fzm3 - fk= m3 * g - uk * m1 * g= 30* 9.81 - 0.4 * 5 * 9.81= 274.68 N
    Fres= m2 * a therefore a= 274.68/ 10= 27.47 m/s^2

    For m3:
    Fres= Fz= m3 * g= 30 * 9.81= 294.3 N
    Fres= m3 * a therefore a= 294.3/30= 9.81 m/s^2

    Is this ok?!
     
  2. jcsd
  3. Nov 19, 2005 #2
    ^^^^^^ Up
    .......................
     
  4. Nov 19, 2005 #3

    daniel_i_l

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    Gold Member

    a) I think that also in this question they want to know the maximum acceleration so that it dosn't slide of the block, the solution is simular to b.
    b) In this case you want the friction between m1 and m2, so to calculate the normal you only use the mass of m1 - because the friction of m1 on m2 depends only on the mass of m1.
     
  5. Nov 19, 2005 #4
    Umm in the book there was an example like this:

    [​IMG]

    Here the only thing they said was that the coefficients of the two blocks are us=0.3 and uk=0.2. and again the maximum horizontal force F that can be applied to the 4 kg block if the 2 kg block is not to slip is asked. The answer according to the book is 17.7 N. So I puzzeled to get to know how they got it. It turns out F= fs= 0.3 * 9.81 * (4+2)= 17.7 N... I still don't know why.... What you said seems the most logical answer to me.... What do you think now?
     
  6. Nov 19, 2005 #5

    daniel_i_l

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    Gold Member

    Also in this question, the friction only has to do with the mass of the top block, but since before the slip the acceleration of the blocks are equal and you are pulling the bottom block, you need to find the force needed to accelerate both of the blocks, with the amount of acceleration needed to create the right amount of force on the top block in order to be equal to the friction. (I hope that that was clear)
     
  7. Nov 19, 2005 #6
    I think it was clear enough.... at least this is how I answered a & b:

    a) fsmax= us * Fn= us*m*g= 0..6 * 5 *9.81= 79.43 N
    F= m*a so F on m1= m1 * atot. Because F on m1 needs to equal fsmax 79,43= 5 * a and therefore a= 79,43/5= 15,886 m/s^2

    b) The dragging force is T= Fz3= m3 g.
    T= mtot * atot= 15 * 15.886= 238.29 N
    m3= 238.29/9.81=24.29 kg

    Is this correct? And what about c is that also correct?
     
  8. Nov 19, 2005 #7
    It is indeed a bit of a tricky problem...
     
  9. Nov 20, 2005 #8
    Lisa, what i meant to say is that in the second problem with the two blocks of 2 and 4 kg, the answer goes like this :

    Let us take 10 m/s² for g, ok ?

    Friction between the lower and upper block should be equal in magnitude (but opposite in direction) of the force (F) with which we pull.

    We have

    (4+2)a=F

    on the upper block we have

    2a = -F' and F' is friction between the two blocks

    F' = 0.3 * 10 * 2 = 6 N

    2a = 6, thus a = 3

    (4+2)3 = F

    thus F = 18 N

    This is my opinion

    marlon
     
    Last edited: Nov 20, 2005
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