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Two blocks - one on a table and one hanging - acceleration with friction

  1. Apr 19, 2008 #1
    [SOLVED] Two blocks - one on a table and one hanging - acceleration with friction

    1. The problem statement, all variables and given/known data



    h ttp://bildr.no/view/186531

    that's a link to an image of the problem

    Block A (mass 2.25kg) rests on a table top, it is connected by a horizontal cord passing over a light , frictionless pulley to a hanging block B (mass 1.30kg). The coefficient of kinetic friction between block A and the table is 0.450

    After the blocks are released from rest, find the speed of each block after moving 3 cm.


    2. Relevant equations



    3. The attempt at a solution

    I've calculated the normal force for both blocks

    Block A = M1g = 2.25*9.81 = 22.07N
    Block B = M2g = 1.30*9.82 = 12.7N

    I'm a bit unsure on how to calculate the friction force fk for this one, if it had been just a single block sitting on a horizontal then it would've been the friction coefficient * mg



    My plan was to use the formula : V^2 = V0^2 + 2a (x-x0)
    to find the velocity after 3cm

    The blocks will have the same acceleration and the same velocity, but i don't know how to calculate either due to the frictional coefficient.

    If the friction wasn't involved, i could've used a=(M2g)/(m1 + m2 )

    could someone please point me in the right direction ?
     
  2. jcsd
  3. Apr 19, 2008 #2

    Hootenanny

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    You're on the right lines, but there are two blocks so you need to take into account both their masses.

    As for what to do next, I'd make use of Newton's Second Law.
     
  4. Apr 19, 2008 #3
    Cheers for the reply

    That's what i thought, so i tried going for friction force k = (friction coefficient * m1) + (friction coefficient * m2) = 15.64N

    Is that correct ?

    In that case, i know that the force holding it back is 15.64N , i know the total pull is normal force A + Normal force B is = 34,82N

    Am i going in the right direction ?

    I don't really know what to do next with newtons second law

    F=ma

    According to the solution in the back of the book, the velocity is supposed to be 0.218m/s
     
    Last edited: Apr 19, 2008
  5. Apr 19, 2008 #4

    Hootenanny

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    Hang on sorry, scratch my previous post. I thought the two blocks were on top of each other, which is not the case. Since only block A experiences friction, you only need to calculate the frictional force for this block. For this bit you can ignore the second block and treat block A in isolation. Do you follow?
     
  6. Apr 19, 2008 #5
    Ah, in that case, i'll end up with a frictional force fk that is:

    fk=frictional coefficient * m1g = 9.93

    I still have no idea what to do next though :/

    The magnitude of F is the sum of both blocks, right ?

    So basicly F = (frictional coefficient * M1g) + (m2g)

    F = 22.68N

    Correct ?
     
  7. Apr 19, 2008 #6

    Hootenanny

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    Correct :approve:
    Not quite, but you have the right idea. Now can you start by writing down the sum of the forces acting on block A? Don't try and put numbers in, just use words or symbols.
     
  8. Apr 19, 2008 #7
    The sum of forces working on Block A is:

    the normal force n, which is the same as mg
    fk = the frictional force that tries to hold it back
    then you have the force from block B which is trying to pull block A
     
  9. Apr 19, 2008 #8

    Hootenanny

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    Correct, since Block A doesn't move in the vertical direction we can ignore the normal force. So now we have,

    [tex]\sum F_A = T - \mu mg[/tex]

    Where T is the tension in the string. Do you agree? Can you do the same for the second block?
     
  10. Apr 19, 2008 #9
    I see

    I'm on my way out, so i'll try when i get back in

    Just one more thing, i tried doing this and i got the right acceleration, but i don't really know if i'm allowed to do it

    I know the frictional force fk that is trying to hold the blocks back

    I know the force that block B is trying to pull with

    So i tried force block b = mg = 12.753N

    Total force = force block b - frictional force = 12,753 - 9.93 = 2.823

    F=ma
    F=2.823
    m=block a + block b = 2.25 + 1.30 = 3.55

    a = F/m = 2.823 / 3.55 = 0.792

    When i then use V^2 = V0^2 + 2a (x-x0)

    I get V= 0.2179

    Which is the correct answer
     
  11. Apr 19, 2008 #10

    Hootenanny

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    I'm afraid your solution isn't correct. It's just shear luck that your answer came out close to the actual answer of 0.21833...m/s
     
  12. Apr 20, 2008 #11
    Hmm, i'm not sure what the second block will be

    [tex]\sum F_B= N + (-mg)[/tex] ?
     
    Last edited: Apr 20, 2008
  13. Apr 20, 2008 #12

    Hootenanny

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    Careful, there is no normal reaction force acting on block B.

    Which two forces are acting on block B? In what direction to the act in relation to the motion of B (i.e. in the same direction of motion or in the opposite direction)?
     
  14. Apr 20, 2008 #13
    Hmm, the two forces acting on B is it's mass and g. They are acting downwards, which is in the same direction of the motion of B

    Right ?
     
  15. Apr 20, 2008 #14

    Hootenanny

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    No, mass isn't a force and neither is g (g is acceleration due to gravity, which is not a force). It's weight is one force, but what is the other one? What is block B attached to? Look at my expression in post #8.

    Drawing a free body diagram of block B may help.
     
    Last edited: Apr 20, 2008
  16. Apr 20, 2008 #15
    Hmm

    Well, it's weight is trying to pull block B down, while Block A and the friction is basicly holding back.
     
  17. Apr 20, 2008 #16

    Hootenanny

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    Correct :approve:
    Correct, and this resistance to motion is transferred through the tension is the rope.

    Can you now write an expression for the sum of the forces on the second block, as I did for the first?
     
  18. Apr 20, 2008 #17
    The tension in the rope connecting the blocks is \mu * m(a)g

    So the sum of all forces working on block B is it's weight trying to pull it down, minus the tension of the rope ?
     
  19. Apr 20, 2008 #18

    Hootenanny

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    No it isn't, if it was, your previous method would be correct. The tension has to account for the frictional force, the weight of the block B and the acceleration of the two blocks.
    This however, is correct. Therefore,

    [tex]\sum F_B= m_Bg - T[/tex]

    Do you follow? Can you now apply Newton's second law to both these net forces?
     
  20. Apr 20, 2008 #19
    Hmm, i don't really know what to do next

    It would be more of a guess than actual knowledge :/
     
  21. Apr 20, 2008 #20

    Hootenanny

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    A guess is better than nothing :biggrin:! Newton's second law states that,

    [tex]\sum \vec{F}=m\vec{a}[/tex]

    So, can you make an educated guess?
     
    Last edited: Apr 20, 2008
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