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A triangle with a Force applied at one tip

  1. Jun 25, 2014 #1
    1. The problem statement, all variables and given/known data

    This is a general question about Statics. I was not able to find a specific question that includes this situation.

    I have a right triangle ABC with two (or three) members. Member AC is diagonal with a pin support (prevents translation) at C. Member AB is horizontal with a full support at B (prevents rotation and translation). B and C are against a wall (i.e. parallel to the Y-axis).

    I apply a Force F at point A (which is one tip of the triangle). This Force could be parallel to the Y-axis, but I'm interested in finding out about cases where it is not parallel to anything.

    I need to find the forces in all members of the triangle (there are only two members which experience force; there could be a member BC but it will be a zero-force member). I also need to find the reactions at B and C.


    2. Relevant equations

    ∑F(x) = 0
    ∑F(y) = 0
    ∑M = 0 at point N

    3. The attempt at a solution

    If we have a Force that is parallel to the Y-axis, M=F(d) where d is the length of AB
    Since the net moment at B is zero, we must have a counteracting moment provided by the member AC and/or by the full support at B (if this were a pin-type support instead, I *believe* but would like to confirm that the problem is simplified because the counteracting moment must be provided by AC, then applying M=F(d), though I'm not sure if that would work).

    In general, I would like to know how this force is distributed among the two involved members AB and AC.
     
  2. jcsd
  3. Jun 25, 2014 #2

    Simon Bridge

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    Lets make sure I've got this.
    Point C is on the ground and point B is in the corner between an upright wall and the ground.
    Force s applies at point A which is off the ground.
    The y axis points directly upwards.

    Lets say the length |AC| is 1 unit then |AB|=cosA and |BC|=sinA, where A is the angle AC makes to the horizontal.

    ... when analyzing forces that are not nicely aligned, you will have resolve them into components.

    In statics - the reaction forces along each member must add up to the applied force.
     
  4. Jun 25, 2014 #3
    More like you have a triangle that's completely suspended against a wall (no points on the ground), with B in the lower-right, C in the upper-right, and A aligned with B in the X-axis. Then the angle ABC is a right angle.

    Not like a traditional "Angle Bracket" like you would find at a hardware store, where you have a member against the wall and the upper-right angle is a right angle.
     
  5. Jun 25, 2014 #4

    Simon Bridge

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    Oh I see - a bit like a bracket for hanging plants on.
    That's a tad easier actually - you'd still be resolving the forces into components.
     
  6. Jun 25, 2014 #5
    Right. If you have a Force parallel to the Y-axis though, it only has 1 component. Then we need to figure out what effect that Y-Force will have on the X-aligned member AB and the slanted AC. We can say that AC is at angle (theta) to AB for simplicity. Then I'm not sure how to proceed with figuring out the distribution...
     
  7. Jun 25, 2014 #6

    Simon Bridge

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    It has only one component in the x-y axes, but it has two components wrt AC.
     
  8. Jun 26, 2014 #7
    Evidently the force in AC will have an X-component and a Y component, as will the reaction at C (the reaction at B will only have an X-component. I am largely concerned with identifying what these values might be, as a function of the moment, force, distance, angle etc. I am "tripped up" by the location of the applied force (at the tip A).
     
  9. Jun 26, 2014 #8
    Aha! A little bit of thought, some Google, and a practice problem have let me figure out that the moment at B due to both the reaction at B *and* the force at A must be considered. Thus, the moment due to F must be equal but opposite to the moment due to the reaction at B (i.e. M-sub-A minus M-sub-B = 0). Then we can figure out that since M=f(d), we can figure out the reaction as a function of dx, dy, and F:

    F(dx)-Fb(dy)=0
    (Fb being the reaction at B)
    Fb(dy)=F(dx)
    Fb=(dx/dy)F
     
  10. Jun 27, 2014 #9

    Simon Bridge

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    Oh well done.
    The other way of looking at these things is geometrically - the forces at point A will have an AC reaction and an AB reaction ... these have to sum, head-to-tail, with the applied force to get a total result of zero. The neat part is that this approach works for pretty much any geometry of bracket struts.

    This gives you a triangle where you know all the angles and the length of one side.

    When the applied force is straight down (say) it's a right triangle and you use trig.
    When the applied force is off the vertical, it isn't - but you know how to do trig for that case too.

    Note: you can do subscripts with the tag or you can use LaTeX.
     
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