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Please help with static friction question

  1. Mar 18, 2009 #1
    A 2.0kg box rests on a plank that is inclined at a angle of 65 degrees above the horizontal. The upper end of the box is attached to a spring with a force constant of 360 N/m. If the coefficient of the static friction between the box and the plank is 0.22, what is the maximum amount the spring can be stretched and the box remain at rest?

    Can someone show me how to complete this problem. The answer is 6.6cm however I don't know to get there. thanks
     
  2. jcsd
  3. Mar 18, 2009 #2

    LowlyPion

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    Welcome to PF.

    How about you start the problem?

    What forces will the spring need to balance?
     
  4. Mar 18, 2009 #3
    I know that I have to find the required force and divide that by 295 N/m--I know I said 360 but it is 295 in order to find the maximum amount how would the 2.0kg box factor in.
     
  5. Mar 18, 2009 #4

    LowlyPion

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    The box has a weight.
    The weight has friction.

    At maximum distance you will rely on both to get the maximum elongation of the spring.

    That of course will balance against the Hookes Law F = k*x for the spring.
     
  6. Mar 18, 2009 #5
    still confused I'M thining I will get the answer if I do the following F/295

    is that right and to find force would I do F=295/2.0?

    Please fill in the blanks if you can
     
  7. Mar 18, 2009 #6

    LowlyPion

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    You will ultimately determine the distance that way, but you need to determine what F is.

    So what is the weight of gravity down the incline?
    And what is the additional force of friction that you will use to get it to stretch to max?
     
  8. Mar 18, 2009 #7
    That is the problem I don't know how to find Force. I am assuming the additional force is .19 I put .22 in the problem. I was looking at something else. How do I find force so I can solve.
     
  9. Mar 18, 2009 #8

    LowlyPion

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    Before you go assuming forces and distances and guessing just work it out directly.

    What is the weight of the box?

    Weight is force.

    How much of that weight is directed down the incline? You know the angle.
     
  10. Mar 18, 2009 #9
    okay so the weight is 2.0kg so that's force right and the upper end is directed down the incline. and right the angle is 65 degrees still lost
     
  11. Mar 18, 2009 #10

    LowlyPion

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    2 kg is the mass

    Weight is m*g.
     
  12. Mar 18, 2009 #11
    okay so force would be (9.81)(2.0)? Is that correct? and if it is if I divide that by 295 I do not get 6.6
     
  13. Mar 18, 2009 #12

    LowlyPion

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    That's the weight all right.

    But ... it's on an incline isn't it?

    So not the whole weight will pull against the spring.
     
  14. Mar 18, 2009 #13
    so would it be (cos 65)(2.0)? since its on the horizontal X axis?
     
  15. Mar 18, 2009 #14

    LowlyPion

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  16. Mar 18, 2009 #15

    LowlyPion

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    Actually if you look at the link I provided you will see that it is Sin65 that you want. But again you want sin 65 of 2*9.8 the weight.
     
  17. Mar 18, 2009 #16
    I was looking for how to solve the equation You are confusing me. Thanks for your help anyways
     
  18. Mar 18, 2009 #17

    LowlyPion

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    Before you can solve an equation you have to be able to write it.

    Just giving you the equation isn't going to help you learn the material.

    Besides that would be against the spirit of the general forum guidelines.
     
  19. Mar 18, 2009 #18
    no it wouldn't people do it all the time within the physics forum. I have LEARNED so much by individuals writing the steps out including when my professor writes the equation for our class. It's okay if you choose to be difficult I will just ask someone else who is willing to help and so I can learn by example instead of toying with me. Again thanks. We just uphold to different styles of learning, and I can respect ur's and in the "spirit of general forum" Please respect mine and others who learn the same way.
     
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