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jadefalcon Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. Taufique Group Title
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    |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.

    • 2 years ago
  2. hosein Group Title
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    angle with normal or horizontal? if horizontal:\[-8\cos62+90=ma \]&\[N+8\sin62=mg\]

    • 2 years ago
  3. jadefalcon Group Title
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    yeh if finding the xcoordinate use cos from what i know

    • 2 years ago
  4. hosein Group Title
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    ordinarily angle with horizontal|dw:1332050614025:dw|

    • 2 years ago
  5. jadefalcon Group Title
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    i get 1.75m/s^2 but the answer is 2.9 m/s^2

    • 2 years ago
  6. Taufique Group Title
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    @hosein here Normal force will be equal to weight of the body.hence they will be cancel to each other.

    • 2 years ago
  7. jadefalcon Group Title
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    i will try use pythagoras with vector notation

    • 2 years ago
  8. hosein Group Title
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    one the force has angle so it has component throgh normal

    • 2 years ago
  9. hosein Group Title
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    my ans is same to you @jadefalcon are you sure about answer?

    • 2 years ago
  10. jadefalcon Group Title
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    straight from the book...you don't consider gravity in this problem

    • 2 years ago
  11. hosein Group Title
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    why we use gravity !we have'nt got friction here so we don't use it

    • 2 years ago
  12. Taufique Group Title
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    @hosein,yes ,you can also do here but it is not necessary here

    • 2 years ago
  13. jadefalcon Group Title
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    ah k i was just looking at your diagram...

    • 2 years ago
  14. hosein Group Title
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    so the answer is..........?

    • 2 years ago
  15. jadefalcon Group Title
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    2.9 m/s^2 and working is???

    • 2 years ago
  16. jadefalcon Group Title
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    im still fiddling with angles to come up with an answer

    • 2 years ago
  17. jadefalcon Group Title
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    ohh i see now

    • 2 years ago
  18. jadefalcon Group Title
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    i think

    • 2 years ago
  19. hosein Group Title
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    i don't understand what are you saying

    • 2 years ago
  20. jadefalcon Group Title
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    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

    • 2 years ago
  21. Taufique Group Title
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    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)

    • 2 years ago
  22. jadefalcon Group Title
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    ahhh yes he is right...but you forgot to divide by 3 at the end

    • 2 years ago
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