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alfers101 Group Title

Physics Problem: A boy pulls a 30.0kg box on a horizontal surface with a force that makes an angle 30degrees the horizontal. If the coefficient of static friction and kinetic friction between the box and the surface are 0.45 and 0.35 respectively, find: a.) the force that will start the box to move b.) the acceleration of the box of the boy maintains the same force that moves the box

  • 2 years ago
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  1. alfers101 Group Title
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    what is the exact formula for letter a and b ??

    • 2 years ago
  2. badreferences Group Title
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    The maximum value for the opposing force given by the coefficient of static friction is the product of that and the box's normal force.

    • 2 years ago
  3. badreferences Group Title
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    This means the boy's force must at least equal said value. (Sorry, I can't write formulas well since I'm not good with LaTeX.)

    • 2 years ago
  4. badreferences Group Title
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    Let me rephrase. The boy's x-component force imparted on the box must equal said value.

    • 2 years ago
  5. alfers101 Group Title
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    aw. this is the only problem left here in my problem set. i cant answer it because i forgot the formula

    • 2 years ago
  6. badreferences Group Title
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    JamesJ help me with your knowledge of typing equations on this site, lol.

    • 2 years ago
  7. JamesJ Group Title
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    First step draw a diagram. In that diagram, you can see that the x-component of the force F on the box is \[ F_x = F \cos(30) = \frac{\sqrt{3}}{2}F \] Now, the friction on the box at exactly the force level that makes it move is \[ F_f = -\mu_{static}N = -\mu_{static}mg \] This force is - negative, because it acts in the opposite direction to the force pulling the box F_x - and N = mg, because this is the normal reaction force to the weight of the box

    • 2 years ago
  8. alfers101 Group Title
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    ok, then ?

    • 2 years ago
  9. JamesJ Group Title
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    When the box starts to move, the force F is just sufficiently large that \[ F_x + F_f \] is infinitesimally larger than 0. Or for practical purposes equal to zero. Hence set this quantity equal to zero. You know every variable in that equation except F. Solve for that variable.

    • 2 years ago
  10. alfers101 Group Title
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    keep going. i think i might get it

    • 2 years ago
  11. JamesJ Group Title
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    That will give you the answer to part a.

    • 2 years ago
  12. JamesJ Group Title
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    Actually, one twist on this. You need to modify N. N is not exactly equal to mg. It would be if there was no other vertical force on the box. But there is.

    • 2 years ago
  13. badreferences Group Title
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    The y component of the boy's pull.

    • 2 years ago
  14. JamesJ Group Title
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    \[ N_{modified} = N_{unmodified} - F\sin(30) \]

    • 2 years ago
  15. badreferences Group Title
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    Yes, that.

    • 2 years ago
  16. JamesJ Group Title
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    Hence \[ F_f = -\mu_{static}N_{modified} = -\mu_{static}(mg - F\sin(30)) \] Now solve for F.

    • 2 years ago
  17. alfers101 Group Title
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    ok thank you. i get it now.

    • 2 years ago
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