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ksaimouli

  • 2 years ago

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.

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

  2. ksaimouli
    • 2 years ago
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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

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

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

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

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

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

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

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

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

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

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

  13. henpen
    • 2 years ago
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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.

  14. ksaimouli
    • 2 years ago
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    i am confused

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

  16. henpen
    • 2 years ago
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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.

  17. Decart
    • 2 years ago
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    tan theta

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

  19. henpen
    • 2 years ago
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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.

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

  21. ksaimouli
    • 2 years ago
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    fx=71

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

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

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

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

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

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

  28. ksaimouli
    • 2 years ago
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    @henpen

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

  30. ksaimouli
    • 2 years ago
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    i got u thx

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