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ksaimouli

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.

  • one year ago
  • one year ago

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  1. henpen
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    |dw:1349552917635:dw|

    • one year ago
  2. ksaimouli
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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

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

    • one year ago
  4. henpen
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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 \[142 \cos(60)=71= \mu 425\]

    • one year ago
  5. henpen
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    So yes, you are correct.

    • one year ago
  6. ksaimouli
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    no but the answer is .235

    • one year ago
  7. henpen
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    |dw:1349553164246:dw|

    • one year ago
  8. ksaimouli
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    i know i am wrong somewhere

    • one year ago
  9. ksaimouli
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    answer should be .235

    • one year ago
  10. Decart
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    |dw:1349553244013:dw|

    • one year ago
  11. ksaimouli
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    ok how do u solve it

    • one year ago
  12. Decart
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    \[\sum_{F}^{?}=142\]

    • one year ago
  13. henpen
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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.

    • one year ago
  14. ksaimouli
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    i am confused

    • one year ago
  15. henpen
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    |dw:1349553460977:dw|

    • one year ago
  16. henpen
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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.

    • one year ago
  17. Decart
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    tan theta

    • one year ago
  18. ksaimouli
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    what is force of friction

    • one year ago
  19. henpen
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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.

    • one year ago
  20. ksaimouli
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    instead of that can ii say 142cos60?

    • one year ago
  21. ksaimouli
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    fx=71

    • one year ago
  22. ksaimouli
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    i got i got ohhhhhhhh it is simple guys

    • one year ago
  23. ksaimouli
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    |dw:1349554073776:dw|

    • one year ago
  24. ksaimouli
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    so friction is 71 by finding the Fn=302

    • one year ago
  25. ksaimouli
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    |dw:1349554155311:dw|

    • one year ago
  26. ksaimouli
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    thx u for helpinggggggggggggg me

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

    • one year ago
  28. ksaimouli
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    @henpen

    • one year ago
  29. henpen
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    http://gyazo.com/cef28f38aa4b15271c0397dce1f96da1

    • one year ago
  30. ksaimouli
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    i got u thx

    • one year ago
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