anonymous
  • anonymous
Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62° north of west. What is the magnitude of the body's acceleration?
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1332049690832:dw| the net force on the body =(9-8sin62)i^+(8cos62)j^ Apply vector method to find the magnitude of the net force. after it find the acceleration: acceleration=(magnitude of the net force)/mass of the body.
anonymous
  • anonymous
angle with normal or horizontal? if horizontal:\[-8\cos62+90=ma \]&\[N+8\sin62=mg\]
anonymous
  • anonymous
yeh if finding the xcoordinate use cos from what i know

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anonymous
  • anonymous
ordinarily angle with horizontal|dw:1332050614025:dw|
anonymous
  • anonymous
i get 1.75m/s^2 but the answer is 2.9 m/s^2
anonymous
  • anonymous
@hosein here Normal force will be equal to weight of the body.hence they will be cancel to each other.
anonymous
  • anonymous
i will try use pythagoras with vector notation
anonymous
  • anonymous
one the force has angle so it has component throgh normal
anonymous
  • anonymous
my ans is same to you @jadefalcon are you sure about answer?
anonymous
  • anonymous
straight from the book...you don't consider gravity in this problem
anonymous
  • anonymous
why we use gravity !we have'nt got friction here so we don't use it
anonymous
  • anonymous
@hosein,yes ,you can also do here but it is not necessary here
anonymous
  • anonymous
ah k i was just looking at your diagram...
anonymous
  • anonymous
so the answer is..........?
anonymous
  • anonymous
2.9 m/s^2 and working is???
anonymous
  • anonymous
im still fiddling with angles to come up with an answer
anonymous
  • anonymous
ohh i see now
anonymous
  • anonymous
i think
anonymous
  • anonymous
i don't understand what are you saying
anonymous
  • anonymous
so 180deg mins 62degs =118degs so (9-8cos-118)/3=4.25 so (9-8sin118)/3=1.94 squareroot(4.25^2 + 1.94^2) = wrong answer arrrgh
anonymous
  • anonymous
oh sorry the diagram is wrong , see true diagram here.|dw:1332051564007:dw| then net force =(9-8cos62)i^+(8sin62)j^ then follow same way for finding the acceleration. acceleration must be 2.9m/sec^2(about)
anonymous
  • anonymous
ahhh yes he is right...but you forgot to divide by 3 at the end

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