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Please I need some help in Newtons Laws

  1. Oct 10, 2005 #1
    In the example i have A block (mass m1 = 22.1 kg) is on a surface inclined at = 33°. It is connected to a mass m2 = 11.2 kg by a massless cord passing over a pulley. The coefficients of friction are µk = 0.15 and µs = 0.25
    It asks me for the magnitud of the friction.

    After looking at this problem I solved for Fs= 45.4 & the Fk=27.24. So i said that Fk<Fs. Therefore there is no acceleration. I really dont know where to go from there to get the magnitud of this friction. That its not even moving the block.

    Please any sugestions??
     
  2. jcsd
  3. Oct 10, 2005 #2

    What no answers!!!:cry:
     
  4. Oct 10, 2005 #3

    Andrew Mason

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    You have to determine the other forces on the block along (parallel to) the surface. If the net force greater than the maximum static friction force, the block will move. In that case, the friction is kinetic. Otherwise, friction is static. If it is static, you do not use the coefficient of friction to determine the static friction force. Static friction will be whatever is required to balance the forces so that the block does not move.

    AM
     
  5. Oct 10, 2005 #4
    Ok so what i got is [tex] m_1gSin\theta-F_f-F_T=m_1a[/tex] Thats the only forces I should have.

    I have my [tex]F_{net}=85.46[/tex] and the [tex]F_f=29.49[/tex], So i think the block is moving. Then if its moving what I should do next??
     
    Last edited: Oct 10, 2005
  6. Oct 10, 2005 #5

    Andrew Mason

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    I get a net sideways force of:

    [tex]m_1gsin(33) - T = m_1gsin(33) - m_2g = 118 - 110 = 8 N.[/tex]

    Since this is less than the maximum force that static friction can provide, the block does not move. What is the magnitude of [itex]F_s[/itex]?

    AM
     
  7. Oct 11, 2005 #6
    I think if I use [tex]F_s=\mu_s*F_n[/tex] It will give me a magnitud would that be right.
     
  8. Oct 11, 2005 #7

    Andrew Mason

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    No. You do not use the co-efficient of static friction to find the actual static friction force. You use it to find the maximum possible static friction force. The static friction force provides just enough force to balance the other forces so the block does not move (zero acceleration).

    AM
     
  9. Oct 11, 2005 #8
    Im sorry about not understanding. I believe that the static friction if its maximum will be [tex]\mu_s*F_n[/tex] but how would I know for the amount of [tex]\mu_s[/tex]?
     
    Last edited: Oct 11, 2005
  10. Oct 11, 2005 #9

    Andrew Mason

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    The forces on the block sum to 0 since the block does not move.

    [tex]m_1gcos\theta - m_2g + F_s = 0[/tex]

    Therefore:

    [tex]F_s = - m_1gcos\theta + m_2g = 118 - 110 = 8 N.[/tex]

    AM
     
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