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(Newtons second law) One of the two problems I really need to solve this weekend!

  1. Nov 20, 2005 #1
    Could anybody of you verify my answers here please... it's really important for me to know if I'm doing it right!

    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?

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

    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



    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/s2

    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/s2

    For m3:
    Fres= Fz= m3 * g= 30 * 9.81= 294.3 N
    Fres= m3 * a therefore a= 294.3/30= 9.81 m/s2

     
    Last edited: Nov 20, 2005
  2. jcsd
  3. Nov 20, 2005 #2

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    a) fsmax should have been 29.43 rather than 79.43. otherwise correct.

    b) When m1 is accelerating, so also is m2, at the same acceleration. The tension from the string attached to m3 must provide that acceleration to both m1 and m2, not just m1. Also m3 will have the same acceleration as m1 (they are attached) What are the forces on m3 giving this (same) acceleration ?

    for part c) remember that m1 and m3 are attached to each other and so will have the same accln. (one horizontal, the other vertical)
     
  4. Nov 20, 2005 #3
    Thank you very very much for your help!

    Ahwww how stupid could I be :D I accidentally wrote the two in 29.43 as a seven..... Now the acceleration of m1= 5.886 m/s2 (and also of m1 and m3 as you said, because the whole system is one if m1 doesn't move off the block).


    Concerning b) I've done the following:
    Fres= mtot * a= Fz on m3 therefore
    (15+m3)5.886= m3 g
    88.29 + 5.886 m3= 9.81 m3
    88.29= 9.81 m3 - 5.886 m3= 3.924 m3

    m3= 88.29/3.924= 22.5 kg.

    This is the right way you ment right?

    c) So now the answer would be:

    - For m3:
    Fres= Fz= m3 * g= 30 * 9.81= 294.3 N
    Fres= m3 * a therefore a= 294.3/30= 9.81 m/s2

    -For m2:

    Fres on m3= Fres on m2 therefore m3 * g= m2* a
    294.3= 10* a
    a= 294.3/10= 29.43 m/s2

    -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/s2

    Is this still correct?
     
  5. Nov 20, 2005 #4

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    Part a) is correct, now for parts b) and c)

    According to your image attachment, m3 is not touching anything, so will have no horizontal acceleration. Nor has it any vertical movement, so the tension in the string will be T = m3.g.
    This tension accelerates only m1 and m2. m3 isn't affected.

    For part c) when m3 is 30 kg, then m1 will be moving on top of m2. Both m1 and m2 will not only be moving, but also moving wrt each other. You should note that the movement of m3 is the difference in movement (i.e. accelerations) of m1 and m2. Since m1 and m3 are attached, then they will have the same acceleration (one horizontal,one vertical)
     
  6. Nov 20, 2005 #5
    Aha thank you! Now my answers are:

    b) The dragging force on m1 and m2 is T= Fz3= m3 g.
    T= mtot * atot= (m1+m2)*atot= 15 * 5.886= 88.29 N
    m3= 88.29/9.81= 9 kg

    c) - 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 and m3:
    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

    Is this the right way?
     
  7. Nov 20, 2005 #6

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    b) 9kg is correct, yes :smile:

    c)The kinetic friction force isn't what accelerates m1. It's the difference between T, the string tension, and Fk that accelerates m1

    T - Fk = m1.a1

    What force is pulling m2 forward ?
    This gives the acceleration of m2, call it a2.
    Now that you know a1 and a2 are, what must be the acceleration of m3, call it a3, say.
    What are the forces acting on m3 that produce this (vertical) acceleration ?

    You should now have a set of simultaneous eqns that you can solve.

    Edit: T does NOT = Fzm3 (=g.m3) That is only the case if m3 is static (or moving with constant velocity)

    Edit2: you should note that a1, the acceleration of m1, is its acceleration wrt the ground, and not wrt the block, m2.
     
    Last edited: Nov 20, 2005
  8. Nov 20, 2005 #7
    Ok so what I've done till now is:

    Fres on M1= T- Fk= m1 * a1= T- 19.62= 5 * a1
    Fres on M2= T= m2*a2= 15 *a2
    Fres on M3= Fz3 -T= 30 * 9.81 -T= 30 * a3
    294.3 -T= 30 * a3

    I've tried substituting the formula of M2 into the other equations, but I don't seem to get any further :S
     
    Last edited: Nov 20, 2005
  9. Nov 20, 2005 #8
    Ah wait a sec... I think I've got it :D

    So we had:
    -Fres on M1= T- Fk= m1 * a1= T- 19.62= 5 * a1
    -Fres on M2= T= m2*a2= 15 *a2
    -Fres on M3= Fz3-T= 30 * 9.81 -T = 30 * a3
    294.3 -T = 30 * a3

    First I subsituted the M2 formula (T= m2*a2= 15 *a2) into the one of M3:
    294.3- 15 a2= 30 * a3

    a2= a3 as they are attached to the same rope, therefore:

    294.3- 15 a3= 30 * a3
    294.3= 45 a3
    a3= 294.3/45= 6.54 m/s^2 = a2

    Ok so now we know a2 as well, therefore T is to be calculated from the formula of M2:

    T= 15 a2= 15 * 6.54= 98.1 N

    The last thing to do is calculating the a1:

    T- 19.62= 5 a1
    98.1-19.62= 5 a1
    78.48= 5 a1
    a1= 78.48/5= 15.696 m/s^2

    Is this ok now? :)
     
  10. Nov 20, 2005 #9

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    correct.

    The only force acting on m2 is the kinetic friction between m1 and m2. You should replace T by Fk,

    Fk = m2.a2

    btw, m2 = 10 kg, rather than 15 kg

    correct.

    m2 is moving with an acceleration of a2, m1 is moving with an acceleration of m1. So m1 is moving with an acceleration of a1 - a2, wrt the block, m1. But m1 and m3 are attached, so the acceleration of m3 is the same as the (relative the block) acceleration of m1. i.e a3 = a1 - a2.

    Can you now substitute and eliminate ?
     
  11. Nov 20, 2005 #10
    But there's one more thing : the first question was: what is the maximum acceleration of m1 the answer was 5.886 m/s^2 and that was ok. Now we calculated an acceleration of 15.696 m/s^s that is way above the answer of question a... so that can't be right, right?
     
  12. Nov 20, 2005 #11
    Why is the only force acting on m2 Fk?! It is attached to a rope and therefore there must be a T somewhere :S correct?
     
  13. Nov 20, 2005 #12
    Perhaps you accidently mixed up M1 and M2?

    EDIT: No on a second thought I don't think so...
     
  14. Nov 20, 2005 #13

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    That max accln of m1 is for when static friction is applied. The static friction can only be so much and the resulting accln is also only so much.

    When m1 is moving, only kinetic fricton (less than static friction) opposes motion. And the accelerating force now comes from a 30 kg weight, when in the static case, it was only 9 kg.
    m1 will move much faster now.

    About the only force on m2 being friction.
    m2 isn't attached by a rope to anything. The rope/string is attached between m1 and m3. If the surface of m2 were frictionless, then m1 would slide along the top of it and m2 wouldn't move. It is only the friction between m1 and m2 that moves m2.

    I know that the pulley applies a force on m2, but m2 applies the same force straight back on the pulley and the net result is no action. These are internal forces/actions and can be disregarded.
     
  15. Nov 20, 2005 #14
    Hey but the rope is attached between m2 and m3 as you can see on the picture :S? M3 falls down, pulling m2 to the right that is attached to a string that is attached to m3....
     
  16. Nov 20, 2005 #15
    I believe the force on m1 (no rope attached only friction with m2)= Fk
    The force on m2 (rope attached, friction with m1)= T- fk
    The force on m3 (rope attached, gravity)= Fz3- T
     
    Last edited: Nov 20, 2005
  17. Nov 20, 2005 #16

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    Oh good grief!

    I sketched that image onto my own notes (the wrong way) and I've been working with that all along.

    My apologies, Lisa.
    Please forget all my previous posts. I'll check through them all again.

    Once again, sorry about that
     
  18. Nov 20, 2005 #17
    No problem at all Fermat! I appreciate your help very very much! You're also only a human being and things like that can happen to anyone of us!
     
  19. Nov 20, 2005 #18

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    ok, I've gone over the question again and gotten results very similar (or the same ?) to your own - the ones I thought were wrong :redface:

    a)
    Fs = us*m1 = 0.6*5*9.81 = 29.43 N
    Fs = m1*a
    a = Fs/m1 = 29.43/5 = 5.886
    a = 5.886 m/s²
    ===========

    b)
    (m1 + m2)a = T
    15 * 5.886 + T
    T = 88.29 N
    =========

    m3*g - T = m3*a
    m3(g - a) = T
    m3 = 88.29/(9.81 - 5.886)
    m3 = 22.5 kg
    ==========

    c)
    Fk = uk*m1 = 0.4*5*9.81 = 19.62 N
    a1 = Fk/m1 = 19.62/5
    a1 = 3.924 m/s²
    ============

    (T - Fk) = m2*a2
    m3*g - T = m3*a3 (and a3 = a2)

    adding the eqns,

    m3*g - Fk = a2(m2 + m3)
    a2 = (m3*g - Fk)/(m2+m3)
    a2 = (30*9.81 - 19.62)/(10 + 30)
    a2 = 274.68/40
    a2 = 6.874 m/s²
    ============

    T = m3(gt - a3) = 30(9.81 - 6.867)
    T = 30*2.943
    T = 88.29 N
    =========
     
  20. Nov 20, 2005 #19
    Wow thank you! Yes you're right the answers you've obtained are similar to the ones I found along the way... thanks again!
     
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