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Force acting along a line

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine the components of F that act along rod AC and perpendicular to it.
    Given: B is located 3 m along the rod from end C.

    The problem is that the coordinates of B aren't given so i'm having some trouble in finding the vector components of F along BD. If someone can give me a hint in finding the coordinates of B.
     

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  2. jcsd
  3. Oct 21, 2013 #2

    Simon Bridge

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    Hint: The coordinates of A and C are given.
     
  4. Oct 21, 2013 #3
    Still no clue. Is this related to Pythagoras theorem?
     
  5. Oct 21, 2013 #4

    gneill

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    Sure, that'll work. Drop a perpendicular from B to the xy plane. Also draw line OC if it helps you to visualize the triangles.
     
  6. Oct 21, 2013 #5
    The only thing you don't know is the z coordinate of point B. But you do know that a vector drawn from point D to point B will be perpendicular to a vector drawn from point A to point C. So you know that the cross product of these two vectors must be zero.

    Chet
     
  7. Oct 21, 2013 #6

    Simon Bridge

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    But you know the coordinates of A and C.

    The rod stretches from A to C.

    What are the coordinates of A? What are the coordinates of C?
    What is the equation of the line representing the rod?

    You are also told:
    What are the coordinates of the point 3m along the rod from C?
     
  8. Oct 21, 2013 #7
    A(0,0,4) C(-3,4,0)
    rAC=(-3-0)i+(4-0)j+(0-4)k
     
  9. Oct 21, 2013 #8

    Simon Bridge

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    Do you not know how to find the coordinates of a point 3m back along that line?
    How far along the line is that?

    Compare with the length of the line: |rAC|=?

    Perhaps you'll find it easier to think of in terms of vectors... the vector equation of the line starting at A and going along the beam is:

    ##\vec{r}_{beam} = \vec{r}\!_A + \lambda\vec{r}\!_{AC}##

    i.e. ##(x,y,z) = (0,0,4)+\lambda (-3,4,-4)##

    ... notice that as λ goes from 0 to 1, rbeam=(x,y,z) goes from point A to point C.

    ... what does λ have to be to get you the point 3m back from C?
     
  10. Oct 22, 2013 #9
    Sorry but the equation you just wrote is out of the scope of the chapter.
    What does lambda stand for in that equation. Also isn't rbeam=rAC?
     
  11. Oct 22, 2013 #10

    Simon Bridge

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    Oh - you don't know about vector equations ... drat!

    ##\vec{r}_{beam}## is a vector pointing from the origin to a point on the beam.
    ##\vec{r}\!_{AC}## is a vector pointing along the beam. ##|\vec{r}\!_{AC}|=|AC|##

    lambda is a parameter ... ##0<\lambda<1##
    when ##\lambda=0## the equation just gives ##(x,y,z)=(0,0,4)## ... which is point A.
    when ##\lambda=1## the equation just gives ##(x,y,z)=(-3,4,-4)## ... which is point C.
    when ##\lambda## is in-between, then ##(x,y,z)## is a point on the beam in between A and C.

    It will get you your answer very directly.

    If you use pythagoras and maybe some trig - then you want to construct the triangle OAC - sketch it out. (point O being the origin.)

    Point B is 3m up the hypotenuse... you should be able to find it's z-coordinate from that using similar triangles.

    You get the x-y coordinates from where B projects onto the OC side.

    You'll see what happens if you sketch the side and overhead views of the beam, and mark out where B is.

    There's some geometry tricks to notice:
    See on the diagram where the 3m sight-line touches the y-axis?
    Call that point E.
    Then triangle OCE is a 3-4-5 triangle.... so the distance |OC| must be...

    So the hypotenuse of OAC must be equal to...


    Aside:
    You know you could always draw a scale diagram and measure?
     
    Last edited: Oct 22, 2013
  12. Oct 22, 2013 #11
    Let's do it one step at a time. Here is the triangle.
    Untitled.jpg

    I'm not sure if B is the midpoint of AC. What do you say?
    Also, I didn't get how the z coordinate of B is 3.
    Also, AC=5.65
     
  13. Oct 22, 2013 #12

    gneill

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    Harken back to your geometry classes; invoke similar triangles:

    https://www.physicsforums.com/attachment.php?attachmentid=63213&stc=1&d=1382439539
    $$\frac{Be}{BC} = \frac{AO}{AC}$$
    The other coordinates of B can be found by recognizing other similar triangle opportunities in the setup.
     
  14. Oct 22, 2013 #13
    Where is point E on the triangle?
     
  15. Oct 22, 2013 #14

    gneill

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    I added point e to your diagram when I dropped the vertical line from B to the xy plane (where the base of the triangle sits).
     
  16. Oct 22, 2013 #15
    Sorry, i can't view that image. The attachment is broken.
     
  17. Oct 22, 2013 #16

    gneill

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    Really? It shows up fine for me in post #12. The link looks fine.

    In words then: Take your drawing and drop a vertical line from B to the base of the triangle. Call the point where it meets the base point e. BeC then forms a triangle similar to AOC. Use ratios to find the length of Be.
     
  18. Oct 22, 2013 #17
    $$\frac{Be}{BC} = \frac{AO}{AC}$$
    $$\frac{Be}{3} = \frac{4}{5.65}$$
    Be = 2.12 = zB
    eC also equal to 2.12 which is equal to yB
    BC = (-3-2.12)i + (4-2.12)j + (0-zB)k
    BC = -5.12i + 2.12j + zBk
    BC2 = (-5.12)2 + (2.12)2 + zB2
     

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  19. Oct 22, 2013 #18
    zB 2 is being negative when i subtract.
     
  20. Oct 22, 2013 #19

    gneill

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    I'm not sure what it is you're trying to find with these calculations. I thought you were going to find the coordinates of the point B.

    Even so, it looks to me like the value you're plugging in for the length AC is not correct. How did you obtain it?
     
  21. Oct 22, 2013 #20
    Using Pythagoras theorem i calculated the length of AC which is 5.65.
    I made a mistake early on, it should be;
    BC = (-3-x)i + (4-2.12)j + (0-2.12)k
    By the above equation im trying to get the x coordinate of B.
     
    Last edited: Oct 22, 2013
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