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Yet another numerical on friction

  1. Mar 10, 2013 #1
    1. The problem statement, all variables and given/known data

    See the image: http://postimage.org/image/o19oxb5pj/ [Broken]

    You can click on the image to enlarge it.


    2. Relevant equations

    f<=fssN
    fkkN


    3. The attempt at a solution

    Ok, so displacement covered by big block + l= displacement covered by small block, I think.

    As whole system accelerates due right, static friction acts on big block on right , while the one by big block on small block acts on left. Am I correct up till ? Any hints will do. I am stuck.

    Please help !!

    Thanks in advance... :smile:
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 10, 2013 #2

    PeterO

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    The blocks are ititially both travelling to the right [set in motion at speed v].

    I believe friction will cause both of them to slow down, so won't both accelerations be towards the left?
     
    Last edited by a moderator: May 6, 2017
  4. Mar 10, 2013 #3
    First of all, thank you PeterO...

    But the question asks the time taken by small block to separate from big block. If the blocks were both accelerated due left, what would be the use of length "l" as given in the question ?

    More hints please.....

    Edit: Yes, static friction, or better kinetic in motion on big block will be towards left by ground, friction by big block on small bock will be towards right. Am I correct ?
     
    Last edited: Mar 10, 2013
  5. Mar 10, 2013 #4

    PeterO

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    As I see it, both blocks are slowing down [accelerating to the left], it is just that the small block is slowing down at a smaller rate of acceleration.

    When the large block has slowed from v to v/2, the small block might have only slowed to 3v/4, but will none the less have slowed down. Because the smaller block is slowing at a smaller rate [of acceleration], it will be - at all times except t=0 - travelling faster than the larger block, so it is only a matter of time before it falls off the right hand end. I think it is that time you seek.
     
    Last edited: Mar 10, 2013
  6. Mar 11, 2013 #5
    Ok, so kinetic friction will be towards left, in both the cases (i.e. friction by ground on big block and friction by big block on small block), correct ? Then using free body diagrams, I can calculate acceleration of both the blocks due left. The acceleration of the two blocks will be different.

    Then shall I calculate relative acceleration of the small block with respect to big block which will be a vector towards right ? And then I can apply kinematic one dimension equation. Correct ?
     
  7. Mar 11, 2013 #6

    PeterO

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    Sounds like a plan ! :-)
     
  8. Mar 11, 2013 #7
    Thanks!!
    Ok, so I am on it !!

    FBD of small block :

    Forces acting are :

    1. Normal reaction R by big block upwards.
    2. Force mg downwards.
    3. Kinetic friction acting on it due left : µR/2

    So owing to vertical equilibrium of small block :

    mg=R
    and,

    µR/2=ma1
    µmg/2=ma1
    a1=µg/2 ...(i)

    Now, FBD for big block :

    Forces acting are:

    1. Normal reaction R1 upwards by floor.
    2. Force Mg downwards.
    3. Reaction force R of small block on it downwards (Newton's third law)
    4. Kinetic friction by floor on it due left : µR1
    5. Reaction force of friction on big block by small block due right : µR/2

    Thus, vertical equilibrium :

    R1=R+Mg

    as R=mg, so

    R1=g(m+M)

    and,

    µR1- µR/2 = Ma2
    µ[g(M+m)] - µmg/2 = Ma2

    From here :

    a2= µg(m+2M)/2M

    Ok, so now I have both the acceleration of big block and small block, Now what can I do next ?
     
  9. Mar 11, 2013 #8

    PeterO

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    In your previous post your plan was:

    Then shall I calculate relative acceleration of the small block with respect to big block which will be a vector towards right ?

    SO that would probably be the way to go.
     
  10. Mar 11, 2013 #9
    Ok, so I get relative acceleration of small block with respect to big block as:

    aS/G = a1-a2= µg/2- µg(m+2M)/2M = -µg(m+M)/2M

    Oh, I get a minus sign. Fine that means the relative acceleration of small block with respect bigger one is due right. Now I see, why will small one separate from bigger one.

    But I am not sure, how to apply equations of motions ? What about length of big block ? Will not small block travels its own displacement relative to big block plus the length "l" of big block itself, before smaller one breaks off from bigger one?
     
  11. Mar 11, 2013 #10

    ehild

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    Yes, that is also a good approach the problem. Write the displacements of both blocks as vot+a/2 t2, and take the difference between them. It should be L.

    ehild
     
  12. Mar 11, 2013 #11
    Hello ehild,

    PeterO advised me to calculate relative acceleration of small block with respect to big block, and that I calculated as:

    aS/G = a1-a2= µg/2- µg(m+2M)/2M = -µg(m+M)/2M

    Thus |aS/G|=µg(m+M)/2M

    Now what to do ? The approach which you gave does not involve the relative acceleration of smaller one with respect to bigger one. I want to continue with the method in which I found relative acceleration.

    With this method,

    |aS/G|t2/2=L

    With this, I get,

    t=√[(4ML)/{µg(m+M)}]

    And this is the correct answer !! Thanks PeterO and ehild !! :smile:

    Oh , then why they give initial velocity "v", if it was already to cancel in the end ?
     
    Last edited: Mar 11, 2013
  13. Mar 11, 2013 #12

    ehild

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    Why not if it confuses the student??? :tongue2:

    (They just wanted to indicate that initially the blocks were moving.)

    You would have got the same result from the difference of the displacements.

    ehild
     
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