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Resolution of Forces

  1. Sep 25, 2005 #1
    I can't figure out how to do this one:

    "Mr. Porter has attached a sign that has a weight of 495N to a wall outside his home. Determine: a) The magnitude of the tension in the chain; b) The thrust force exerted by the rod, if the angle is 35 degrees"

    I'm not seeing how the answers are a) 863N and b)706.9N (my teacher gave us the answers)
     
  2. jcsd
  3. Sep 25, 2005 #2

    Päällikkö

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    Draw a diagram.
    Weight is the y-component.
     
  4. Sep 25, 2005 #3
    Can you show us what you've done so far on getting the answers?

    From the answers, the rod must be at 35 degrees to the wall, rather than the horizontal. It's really only two simple calculations, just draw a diagram of forces, all there is to it.
     
  5. Sep 25, 2005 #4
    What I'm not understanding is how the tension is 863N? If the signing is hanging straight down, shouldn't is be 495N [495sin(90)]?
     
    Last edited: Sep 25, 2005
  6. Sep 25, 2005 #5

    Päällikkö

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    It isn't hanging straight down.
     
  7. Sep 25, 2005 #6
    Where do you see that?
     
  8. Sep 25, 2005 #7

    Päällikkö

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    There's the sign, a chain, a rod and 35 degrees. I'd say the rod's placed horizontally.
     
  9. Sep 25, 2005 #8
    This is all I can get.

    Physics is not exactly my best subject :redface:
     

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    Last edited: Sep 25, 2005
  10. Sep 25, 2005 #9

    Päällikkö

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    Last edited: Sep 25, 2005
  11. Sep 25, 2005 #10
    That makes a lot more sense
    Thanks
     
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