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Mechanics prb

  1. Mar 21, 2004 #1
    hey firstly, im not so certain this is the correct place for this thread, sorry!feel free to move!!
    basically, if you look at the file attached, im trying to find friction up the slope, but i cannot do it, and i cant see how you will do it, so if anyone could shine some light, i would be most apreciative!!!
    thanks in advance!!
    si
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2004 #2

    NateTG

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    Well, for starters you've got the wrong magnitude for the normal force.
     
  4. Mar 21, 2004 #3
    You know the body doesn't move in the Y axis so [tex]\Sigma F_y = 0[/tex].

    [tex]\Sigma F_y = N - mg\cos \alpha = 0[/tex]

    So now that you have the N force, do you know how to find the friction force?
     
  5. Mar 21, 2004 #4

    uart

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    You know that is really a horrible description of the problem at hand. Has anyone else ever noticed the high correletion between someones inabilty to do a given problem and to accurately describe what it is that they are even trying to do.


    Anyway, if you're asking what I think you're asking then the answer is,
    [tex] F = mg \sin(\theta)[/tex],
    Assuming that the block is not accelerating.

    You may be concerned that the solution does not mention or require the coefficient of friction ([tex]\mu[/tex]). That is actually embedded in the assumption that the block is not accelerating. In particular, if [tex]\mu > \tan(\theta)[/tex] then the block will remain at rest and the value of the frictional force will be less then the maximum available frictional force and just sufficient to keep the block at rest, namely [tex]mg \sin(\theta)[/tex]
     
  6. Mar 21, 2004 #5
    thankyou for your constructive replies,
    NateTG Well, for starters you've got the wrong magnitude for the normal force. ok fair enough say its wrong, but i dont see you correcting it?

    bout my description, ok im trying to find the frictional force working against the block which is MOVING and accelerating down the plane, i was getting at the fact that couldnt do it, and didnt think it was possible, if it is accelerating, is it possible?
    cheers
    si
     
  7. Mar 21, 2004 #6
    Did you even read my post? Do you know how the kinetic friction force is defined? Do you know how to find it if you have the magnitude of the normal force?
     
  8. Mar 21, 2004 #7

    NateTG

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    It's not my homework is it?

    If you have [tex]\mu_k[/tex]
    [tex]F_{friction}=\mu_k N[/tex]

    If you know all the other forces, and the acceleration you can use
    [tex]F_{net}=ma[/tex]
     
    Last edited: Mar 21, 2004
  9. Mar 21, 2004 #8
    ur hwk? no.....its not even mine? i found this prb in the math book an thought it was impossible, u dont know 'mu'so i aint so sure on what to do, findin the coeficient of friction would help, or even having it would help!! so its not possible without either of those two?
    cheers
    si
     
  10. Mar 21, 2004 #9

    uart

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    Ok, that's the vital piece of information that I was referring to when I compained about poor problem description. The cases of moving versus static are very different.

    In the moving case the motion will be resisted by the full available frictional force of,
    [tex]F = \mu m g \cos(\theta)[/tex].


    In the static case the frictional force will never be greater than that which is required to oppose the other forces and keep the block static. That is,
    [tex]F = m g \sin(\theta)[/tex].

    Two very different cases as you can see. The first case (moving) is possible for any value of the incline [tex]\theta[/tex] as you can alway give the block some initial velocity.

    The second case (static) is only applicable if [tex] \mu > \tan(\theta)[/tex], otherwise motion will be spontanious.
     
  11. Mar 22, 2004 #10

    NateTG

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    Actually there's a [tex]\mu_k[/tex] for kinetic friction and a [tex]\mu_s[/tex] for static friction. Typically [tex]\mu_s>\mu_k[/tex].

    If you know the rate of acceleration of the block, then you can still figure out what [tex]\mu_k[/tex] is.
     
  12. Mar 22, 2004 #11
    thanks a lot mate
    cheers
    si
     
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