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ksaimouli Group Title

A block rests on a horizontal surface and weighs 425 N. A force is applied to the block and has a magnitude of 142 N. The force is directed upward at an angle relative to the horizontal. The block begins to move horizontally when the angle is 60 degrees. Determine the static coefficient.

  • 2 years ago
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  1. henpen Group Title
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    |dw:1349552917635:dw|

    • 2 years ago
  2. ksaimouli Group Title
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    Correct me if I am wrong but here goes: Yea I get the x component to be 71N as well. The point at where the block is about to move is where the applied force is equal to the friction force. Friction = static coefficient * normal force I get 71 = static coefficient * 425 static coefficient = 71/425 = 0.167

    • 2 years ago
  3. henpen Group Title
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    The normal force, N\[N=425\] The force along the horizontal, Fx, is\[F_x=142\cos(\theta)\]

    • 2 years ago
  4. henpen Group Title
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    http://en.wikipedia.org/wiki/Friction#Dry_friction When it is just about to move, the force of friction equals the normal force times the coefficient of friction. So \[142 \cos(60)=71= \mu 425\]

    • 2 years ago
  5. henpen Group Title
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    So yes, you are correct.

    • 2 years ago
  6. ksaimouli Group Title
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    no but the answer is .235

    • 2 years ago
  7. henpen Group Title
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    |dw:1349553164246:dw|

    • 2 years ago
  8. ksaimouli Group Title
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    i know i am wrong somewhere

    • 2 years ago
  9. ksaimouli Group Title
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    answer should be .235

    • 2 years ago
  10. Decart Group Title
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    |dw:1349553244013:dw|

    • 2 years ago
  11. ksaimouli Group Title
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    ok how do u solve it

    • 2 years ago
  12. Decart Group Title
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    \[\sum_{F}^{?}=142\]

    • 2 years ago
  13. henpen Group Title
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    @Decart is correct- I did not take into account F_y, the fact that the force's y-component weakens the normal force. @ksaimouli , just subtract 142sin60 from the normal force.

    • 2 years ago
  14. ksaimouli Group Title
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    i am confused

    • 2 years ago
  15. henpen Group Title
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    |dw:1349553460977:dw|

    • 2 years ago
  16. henpen Group Title
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    Okay- you push the block at an angle. This does 2 things- push the block ALONG (which is resisted by friction) and UP (opposite to gravity, so this reduces the block's force on the ground, as the force of the block DOWN is decreased, so it reduces the normal force the ground exerts on the block). It's as if you have a man pushing horizontally with force Fx and a cable with tension Fy pulling up on the block.

    • 2 years ago
  17. Decart Group Title
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    tan theta

    • 2 years ago
  18. ksaimouli Group Title
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    what is force of friction

    • 2 years ago
  19. henpen Group Title
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    The normal force changes, so the friction changes. @Decart , I'm not sure what you're doing, but Fy is defintely using sine This gives the right answer by the way.

    • 2 years ago
  20. ksaimouli Group Title
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    instead of that can ii say 142cos60?

    • 2 years ago
  21. ksaimouli Group Title
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    fx=71

    • 2 years ago
  22. ksaimouli Group Title
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    i got i got ohhhhhhhh it is simple guys

    • 2 years ago
  23. ksaimouli Group Title
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    |dw:1349554073776:dw|

    • 2 years ago
  24. ksaimouli Group Title
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    so friction is 71 by finding the Fn=302

    • 2 years ago
  25. ksaimouli Group Title
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    |dw:1349554155311:dw|

    • 2 years ago
  26. ksaimouli Group Title
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    thx u for helpinggggggggggggg me

    • 2 years ago
  27. ksaimouli Group Title
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    i have only one question how did u get the equation Fn=mg-fy

    • 2 years ago
  28. ksaimouli Group Title
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    @henpen

    • 2 years ago
  29. henpen Group Title
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    http://gyazo.com/cef28f38aa4b15271c0397dce1f96da1

    • 2 years ago
  30. ksaimouli Group Title
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    i got u thx

    • 2 years ago
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