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Force equations: Force on ground by angled supports

  1. Sep 28, 2016 #1
    1. The problem statement, all variables and given/known data
    14513699_1123621714339842_1207432723_o.jpg

    P=60kN
    angle alpha = 45 degrees
    angle beta = 20 degrees

    The task is to calculate the force on the base A and C, and to draw vectors that represent those forces.

    2. Relevant equations


    3. The attempt at a solution
    I did similar examples at the class, but when I had to do this one on the test, I just could not get it done.

    If I could get the balanced equations down correctly, I would manage to do the rest. (balanced force equations on x and y axis)

    Sorry if there are wrong terms used, Im not a native speaker. Ill try to explain if there is anything unclear.

    Thanks!
     
  2. jcsd
  3. Sep 28, 2016 #2

    RUber

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    What have you tried so far?

    I'm no physicist, but I would think that some there might exist some r, such that r*(sin 20 + sin 45 ) = 60. This might help you balance out the y direction.
     
  4. Sep 28, 2016 #3
    Well the R should be right above the P in that case, which cancels the P itself. But that didnt help me out. Or im drawing something incorectly
     
  5. Sep 28, 2016 #4

    RUber

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    I was thinking that r would be a scalar...rsin(20) is the y component of the vector acting at point C, r sin 45 would be the y component of the vector acting at point A.
     
  6. Sep 28, 2016 #5

    RUber

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    Or perhaps that is too simple. Let's take another look at this...
    You have two force vectors W and V, each one has magnitude |W| and |V|, respectively.
    You want to find those magnitudes such that the following equations are satisfied:
    ## |W| \sin 45 + |V| \sin 20 = 60##
    ##|W| \cos 45 = |V| \cos 20##
    Does that seem more reasonable?
     
  7. Sep 28, 2016 #6
    1475086289183.jpg
    I am a bit confused.

    Is this equation for the projection of all forces on y axis wrong?
     
  8. Sep 28, 2016 #7
    I also dont get why on the second equation both magnitudes have to be equal? Id say they wont be equal, since the angles are different
     
  9. Sep 28, 2016 #8
    If F and G are the compression forces in rods AB and BC, what are the components of these forces in the x- and y directions?
     
  10. Sep 28, 2016 #9
    In the picture F and G is Fw and Fv. Is that right? In that case components are. P/sin45 and P/sin20 for the y axis. And P*cos45 and P*cos20 for the x axis. Atleast if I got it right.
     
  11. Sep 28, 2016 #10
    You didn't get it right.

    Step 1 - Forget about P.

    Step 2 - From the trigonometry, what are the components of F and G in terms of F, G, and the angles.
     
  12. Sep 28, 2016 #11
    1475088641021.jpg

    Like this?
     
  13. Sep 28, 2016 #12
    Yes. Very nice.

    The signs on the G components should be - in the x direction and + in the y direction. Now, using the upper junction point as your free body, do a force balance in the x direction and a force balance in the y direction on the junction point. What do you get?
     
  14. Sep 28, 2016 #13
    I hope I understood correctly what you asked due to the language barier.

    But I guess this is what you meant:
    1475089658584.jpg
     
  15. Sep 28, 2016 #14
    Excellent. Now solve for F and G.

    Also, now compare with Ruber's post #5.
     
  16. Sep 28, 2016 #15
    Is this correct?
    1475090812690.jpg
     
  17. Sep 28, 2016 #16
    Your methodology is correct. I didn't check your algebra. If you want to do that, just substitute your results into the two force balance equations and see if the equations are satisfied.
     
  18. Sep 28, 2016 #17
    Great! Thanks! To both of you.
    So my mistake was that I tried to make components of F and G from P? While I actually had to make them from the F and G itself.
     
  19. Sep 29, 2016 #18
    Im sorry. Could you please tell me where Im wrong again? I am using the same methodology, but I have a feeling its incorrect, cause I cant find the second force momentum equation.

    The A, Ax, Ay, C , Cy and Cx is added by me. Everything else was given.
     
  20. Sep 29, 2016 #19
    Why am I getting a negative Cy, when it should be positive?
     
  21. Sep 29, 2016 #20

    haruspex

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    I do not understand the connection between this post and the rest of this thread. What are Ax etc.?
     
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