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Finding an angle in 3D

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the angle θ between the 300-lb force and the line OC.

    I have attached an image of the question


    2. Relevant equations

    P°Q = PQcos(θ)



    3. The attempt at a solution

    I tried finding the vectors AB and OA as well as their magnitudes first
    O is the origin
    A = (9,0,0)
    B = (0, 16, 6)
    C =(9, 16, 0)

    AB = <0-9, 16-0, 6-0>
    AB = <-9, 16, 6>

    |AB| = sqrt((-92) + 162 + 62)
    |AB| = 19.31

    OA = <9-0, 0-0, 0-0>
    OA = <9, 0, 0>

    |OA| = 9

    AB°OA = |AB||OA| cos(θ)

    <-9, 16, 6>°<9, 0, 0> = (19.31)(9)cos(θ)

    arccos(-81/173.79)

    θ = 117.78

    It says this answer is not correct and I realize that I haven't included the 300-lb force, but how would I take that force into account?
     

    Attached Files:

  2. jcsd
  3. Sep 22, 2012 #2

    Simon Bridge

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    Looks like you've found the angle of AB with line OA ... but it asks for the angle with OC.
     
  4. Sep 22, 2012 #3
    ...Excellent observation...it's been a long day.

    So I ran the calculations again, this time with OC. and I got an angle of 34.046 which is still wrong. :/
     
  5. Sep 22, 2012 #4
    I ran the calculations, for the angle with OC using the given info and got a larger angle than yours, double check your calculations.
     
  6. Sep 22, 2012 #5
    So I checked my calculations and still got 34.056°. My calculations are as follows:

    O is the origin
    A = (9,0,0)
    B = (0, 16, 6)
    C =(9, 16, 0)

    AB = <0-9, 16-0, 6-0>
    AB = <-9, 16, 6>

    |AB| = sqrt((-92) + 162 + 62)
    |AB| = 19.31

    OC = <9-9, 16-0, 0-0>
    OC = <0, 16, 0>

    |OC| = 16

    AB°OC = |AB||OC| cos(θ)

    <-9, 16, 6>°<0, 16, 0> = (19.31)(16)cos(θ)

    θ = arccos(256/308.96)

    θ = 34.046
     
  7. Sep 22, 2012 #6
    Shouldn't OC = <9,16,0> ?
     
  8. Sep 23, 2012 #7
    Yes, you are correct. I think I've been at this too long :/.

    My calculations are as follows

    A = (9, 0, 00
    B = (0, 16, 6)
    C = (9, 16, 0)

    AB = <0-9, 16-0, 6-0>
    AB = <-9, 16, 6>

    |AB| = sqrt((-9)^2 + 16^2 + 6^2)
    |AB| = 19.31 kN

    OC = <9-0, 16-0, 0-0>
    OC = < 9, 16, 0>

    |OC| = sqrt(9^2 + 16^2)
    |OC| = 18.358 kN

    AB°OC = |AB||OC|cos(θ)

    <-9, 16, 6>°<9, 16, 0> = (19.31)(18.358) cos(θ)

    θ = arccos(175/354.47)
    θ = 60.42°

    This answer is correct. Thanks for all your help.
     
  9. Sep 23, 2012 #8

    Simon Bridge

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    There's been a bunch of people struggling through these and it is usually just down to attention to detail. It gets everybody like that - you'll get used to it.
     
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