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cassytaylor92

  • one year ago

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)

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  1. phi
    • one year ago
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    Did you sketch a picture ?

  2. cassytaylor92
    • one year ago
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    No, I don't know how to begin it.

  3. phi
    • one year ago
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    |dw:1382104719667:dw|

  4. phi
    • one year ago
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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 ?

  5. cassytaylor92
    • one year ago
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    No, i can't @phi

  6. phi
    • one year ago
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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 ?

  7. phi
    • one year ago
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    you should get \[ 2 x^2 = 25^2 \\ x^2 = \frac{25^2}{2} \\ x= \sqrt{\frac{25^2}{2} }= \frac{25}{\sqrt{2}}= \frac{25 \sqrt{2}}{2} \]

  8. cassytaylor92
    • one year ago
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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

  9. cassytaylor92
    • one year ago
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    why is it 2x^2 = 25^2 ?

  10. phi
    • one year ago
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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. phi
    • one year ago
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    If the pythagoraean theorem does not ring any bells, you should learn it.

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

  13. phi
    • one year ago
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    now you have this picture |dw:1382106175853:dw|

  14. phi
    • one year ago
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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

  15. cassytaylor92
    • one year ago
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    Is the downward force: 25^2/2 and 50 N?

  16. phi
    • one year ago
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    25^2/2 means 25*25/2 do you mean \( 25 \sqrt{2}/2\) ?

  17. cassytaylor92
    • one year ago
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    No. What I mean is the 25 sqr(2)/2 and 50 N.

  18. phi
    • one year ago
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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"

  19. cassytaylor92
    • one year ago
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    i'm confused with the 25 sqr(2)

  20. phi
    • one year ago
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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

  21. cassytaylor92
    • one year ago
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    so that would be the answer for the downward force? @phi

  22. phi
    • one year ago
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    yes

  23. cassytaylor92
    • one year ago
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    thanks :) how about the force horizontally?

  24. phi
    • one year ago
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    look at the picture. How big is the vector that goes sideways ?

  25. cassytaylor92
    • one year ago
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    25 sqr(2)/2 ?

  26. phi
    • one year ago
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    yes. or as a decimal about 67.678 N

  27. cassytaylor92
    • one year ago
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    Thank you very much for helping! @phi ^_^

  28. phi
    • one year ago
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    yw

  29. phi
    • one year ago
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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)

  30. cassytaylor92
    • one year ago
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    its okay. thanks for reminding :) @phi

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