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Slipping off a car seat

  1. Feb 25, 2015 #1

    Yam

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    1. The problem statement, all variables and given/known data
    A car travelling at 20 m/s stops in a distance of 60 m. Assume that the deceleration is constant. The coefficients of static and kinetic friction between a passenger and the seat are Us and Uk respectively. You may assume that the passenger is not in contact with anything else. Which of the following statements below is true?

    a) The passenger will slip off the seat if he is light and remain on the seat if he is heavy.
    b) The passenger will not slip off the seat regardless of the values of Us and Uk .
    c) The passenger will slip off the seat regardless of the values of Us and Uk .
    d) The passenger will not slip off the seat if Us> 0.4
    e) The passenger will slip off the seat if Uk<0.3

    2. Relevant equations

    Kinematics : v^2 = u^2 +2as
    Newton's second Law: F=ma
    Force of Friction: Us*Normal Force

    3. The attempt at a solution
    a) The passenger will slip off the seat if he is light and remain on the seat if he is heavy.
    The statement is too ambiguous.
    b) The passenger will not slip off the seat regardless of the values of Us and Uk .
    As long as F>UsN, the passenger will slip
    c) The passenger will slip off the seat regardless of the values of Us and Uk .
    As long as F>UsN, the passenger will slip
    d) The passenger will not slip off the seat if Us> 0.4
    e) The passenger will slip off the seat if Uk<0.3

    I will attempt to find out the minimum Us that is required so that the passenger will not slip.

    0 = 20^2 + (2)(a)(s)
    a = 10/3
    (m*g*Us)=ma
    Us = 10/3g =0.34, which doesnt fit into any option above.

    May i know if i made any errors in concept or calculation


     
  2. jcsd
  3. Feb 25, 2015 #2

    haruspex

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    a) can be answered.
    For all of them, you need to spot how to eliminate mass from the equation. Let the mass of the person be m. What is the normal force? What (given the speed and stopping distance) is the decelerating force needed?
     
  4. Feb 25, 2015 #3

    Yam

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    0 = 20^2 + (2)(a)(s)
    a = 10/3

    frictional force = decelerating force
    (m*g*Us)=ma <=== the mass is eliminated here from the equation
    The normal force is mg, from the individual.

    Us = 10/3g =0.34.

    Is that right??
     
  5. Feb 25, 2015 #4

    haruspex

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    Sorry, I should have noticed you'd posted all that. I only looked at the answers you were proposing in red. For each question, you only need to answer true or false. Can you correct your answer to a)?
    The 0.34 doesn't need to match anything exactly. What do you think the 0.34 represents?
     
  6. Feb 25, 2015 #5

    Yam

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    I would say that 0.34 represents the minimum coefficient of static friction that the seat must have.

    I see, so the best answer would be option A.
     
  7. Feb 25, 2015 #6

    SammyS

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    That's the correct acceleration, except for the sign, but that doesn't matter here.

    The value for μS looks good.

    That is covered by one of the choices.
     
  8. Feb 25, 2015 #7

    haruspex

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    Option A? No, there are five questions and you must answer true or false to each.
     
  9. Feb 26, 2015 #8

    Yam

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    a) The passenger will slip off the seat if he is light and remain on the seat if he is heavy.
    False, a heavy passenger can still slip off with a low Us

    b) The passenger will not slip off the seat regardless of the values of Us and Uk .
    False, Us<0.34 will cause the passenger to slip

    c) The passenger will slip off the seat regardless of the values of Us and Uk .
    False, for the passenger to slip

    d) The passenger will not slip off the seat if Us> 0.4
    True, Calculated Us>0.34

    e) The passenger will slip off the seat if Uk<0.3
    Uk is not invloved.

    Is this right?
     
  10. Feb 26, 2015 #9

    haruspex

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    Pretty much, but your reason for (a) could be refined. As your equation tells you, the mass of the passenger has no relevance.
     
  11. Feb 26, 2015 #10

    Yam

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    Oyes, i understand now, thank you for your help!
     
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