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cassytaylor92 Group Title

HELP ME PLEASE ! (reposted) (Resolving Vector) A woman pushes a 50-N lawn mower with a force of 25 N. If the handle of the lawn mower is 45 degrees above the horizontal, how much downward force is the lawn mower exerting on the ground? How much force causes the motion of the lawn mower horizontally?(neglect friction)

  • 11 months ago
  • 11 months ago

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  1. phi Group Title
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    Did you sketch a picture ?

    • 11 months ago
  2. cassytaylor92 Group Title
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    No, I don't know how to begin it.

    • 11 months ago
  3. phi Group Title
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    |dw:1382104719667:dw|

    • 11 months ago
  4. phi Group Title
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    do you see the 25N force along the handle can be broken into sideways and up/down ? do you see you have a 45-45-90 triangle. You know the hypotenuse. can you find the length of the (equal) legs ?

    • 11 months ago
  5. cassytaylor92 Group Title
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    No, i can't @phi

    • 11 months ago
  6. phi Group Title
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    to do this problem you need to know some physics and some math Let's start with the math |dw:1382105210794:dw| can you solve for x ?

    • 11 months ago
  7. phi Group Title
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    you should get \[ 2 x^2 = 25^2 \\ x^2 = \frac{25^2}{2} \\ x= \sqrt{\frac{25^2}{2} }= \frac{25}{\sqrt{2}}= \frac{25 \sqrt{2}}{2} \]

    • 11 months ago
  8. cassytaylor92 Group Title
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    25/sin 90 degrees * x/sin 45 degrees (25)(sin 45 degrees) = (x)(sin 90 degrees) i'm not really sure with this yet

    • 11 months ago
  9. cassytaylor92 Group Title
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    why is it 2x^2 = 25^2 ?

    • 11 months ago
  10. phi Group Title
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    First, using the Law of Sines works. sin 45 = sqr(2)/2 and sin 90 = 1 you get x= 25 sqr(2)/2 which is good. I used geometry. you have a right triangle which is also isosceles, so its 2 sides are equal. You can use pythagorean theorem a^2 + b^2 = c^2, where a and b are the legs and c is the hypotenuse \[ x^2 + x^2 = 25^2 \\ 2x^2= 25^2\]

    • 11 months ago
  11. phi Group Title
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    If the pythagoraean theorem does not ring any bells, you should learn it.

    • 11 months ago
  12. cassytaylor92 Group Title
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    I got the phythagorean theorem, thanks :) Can you still help with the next step on answering it?

    • 11 months ago
  13. phi Group Title
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    now you have this picture |dw:1382106175853:dw|

    • 11 months ago
  14. phi Group Title
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    can you answer the questions? the sideways force is only the one vector the downward force is the sum of two vectors

    • 11 months ago
  15. cassytaylor92 Group Title
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    Is the downward force: 25^2/2 and 50 N?

    • 11 months ago
  16. phi Group Title
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    25^2/2 means 25*25/2 do you mean \( 25 \sqrt{2}/2\) ?

    • 11 months ago
  17. cassytaylor92 Group Title
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    No. What I mean is the 25 sqr(2)/2 and 50 N.

    • 11 months ago
  18. phi Group Title
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    then yes, you add those two values... the vectors are pointing in the same direction so you can just add their lengths to get the total length of the "resultant vector"

    • 11 months ago
  19. cassytaylor92 Group Title
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    i'm confused with the 25 sqr(2)

    • 11 months ago
  20. phi Group Title
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    first, it is \[ \frac{25\sqrt{2}}{2} \] you can add 50 to it like this: \[ 50+ \frac{25\sqrt{2}}{2} \] that is an exact answer. you can use a calculator and find a decimal number, but you will have to round the decimal number (because in theory the answer has an infinite number of digits on the right side of the decimal point) see http://www.wolframalpha.com/input/?i=+100+digits+of+50+%2B+25*sqrt%282%29%2F2

    • 11 months ago
  21. cassytaylor92 Group Title
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    so that would be the answer for the downward force? @phi

    • 11 months ago
  22. phi Group Title
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    yes

    • 11 months ago
  23. cassytaylor92 Group Title
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    thanks :) how about the force horizontally?

    • 11 months ago
  24. phi Group Title
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    look at the picture. How big is the vector that goes sideways ?

    • 11 months ago
  25. cassytaylor92 Group Title
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    25 sqr(2)/2 ?

    • 11 months ago
  26. phi Group Title
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    yes. or as a decimal about 67.678 N

    • 11 months ago
  27. cassytaylor92 Group Title
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    Thank you very much for helping! @phi ^_^

    • 11 months ago
  28. phi Group Title
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    yw

    • 11 months ago
  29. phi Group Title
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    oops I just noticed I posted the wrong decimal value for the horizontal force 25* sqrt(2)/2 is about 17.678 N (NOT 67.678 N which is the downward force)

    • 11 months ago
  30. cassytaylor92 Group Title
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    its okay. thanks for reminding :) @phi

    • 11 months ago
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