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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)

Physics
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  • phi
Did you sketch a picture ?
No, I don't know how to begin it.
  • phi
|dw:1382104719667:dw|

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Other answers:

  • phi
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 ?
No, i can't @phi
  • phi
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 ?
  • phi
you should get \[ 2 x^2 = 25^2 \\ x^2 = \frac{25^2}{2} \\ x= \sqrt{\frac{25^2}{2} }= \frac{25}{\sqrt{2}}= \frac{25 \sqrt{2}}{2} \]
25/sin 90 degrees * x/sin 45 degrees (25)(sin 45 degrees) = (x)(sin 90 degrees) i'm not really sure with this yet
why is it 2x^2 = 25^2 ?
  • phi
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 \[ x^2 + x^2 = 25^2 \\ 2x^2= 25^2\]
  • phi
If the pythagoraean theorem does not ring any bells, you should learn it.
I got the phythagorean theorem, thanks :) Can you still help with the next step on answering it?
  • phi
now you have this picture |dw:1382106175853:dw|
  • phi
can you answer the questions? the sideways force is only the one vector the downward force is the sum of two vectors
Is the downward force: 25^2/2 and 50 N?
  • phi
25^2/2 means 25*25/2 do you mean \( 25 \sqrt{2}/2\) ?
No. What I mean is the 25 sqr(2)/2 and 50 N.
  • phi
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"
i'm confused with the 25 sqr(2)
  • phi
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
so that would be the answer for the downward force? @phi
  • phi
yes
thanks :) how about the force horizontally?
  • phi
look at the picture. How big is the vector that goes sideways ?
25 sqr(2)/2 ?
  • phi
yes. or as a decimal about 67.678 N
Thank you very much for helping! @phi ^_^
  • phi
yw
  • phi
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)
its okay. thanks for reminding :) @phi

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