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Help with statics question please.

  1. Sep 29, 2007 #1
    does anyone know how to do this question?
    [​IMG]
    i've tried to work it out but i just can't seem to get it.

    i've tried the using the usuall trigonometry stuff but i can't seem to appply it to this question.

    thanks
    scott.
     
  2. jcsd
  3. Sep 29, 2007 #2

    Doc Al

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    Show what you've tried.
     
  4. Sep 29, 2007 #3
    i've just tried to see if i could work out the angle using SOH CAH TOA.

    i thought i could do: theta= cos(to the minus1) 700/400 but i get a mathmatical error on my calculator, anything else i can think of seems like a dead end.

    I'm not really looking for the answer, just some help on how to do the problem.
     
    Last edited: Sep 29, 2007
  5. Sep 29, 2007 #4

    Doc Al

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    What you have to do is add those two forces and set the sum equal to 1000 N. Then you can solve for the angle.

    There are several ways to add vectors. One way is to add them graphically and use some triangle geometry. (The law of cosines.) Another way is to find their components, add those, then find the resultant. Either way should work.
     
  6. Sep 29, 2007 #5
    ok, i don't know why i can't work this out, i've not had any problems with any of the questions before it.

    i was going to add them the tail to tail method but that won't work because i don't know the angles.

    if you're not busy could you show me how you would do it with different values?
    or show me the first step on how to solve it, once i get started on it i'll probably see how to do it.
     
  7. Sep 29, 2007 #6

    Doc Al

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    I assume you mean "head to tail" method. This will work just fine. You'll get a triangle for which you can use the law of cosines to find the resultant (which would be one side of that triangle). You don't have to know the angle, since that's what you're trying to find.

    This might help: Law of cosines method for adding two vectors

    You can also just add the components and find the resultant that way. The magnitudes of the horizontal and vertical components of the 400 N force would be [itex]{400}\cos\theta[/itex] and [itex]{400}\sin\theta[/itex]
     
  8. Sep 29, 2007 #7
    i don't know if it is correct(the answer for this particular question is missing in my book) but i got 18.19 degrees for the angle.

    does that sound about right?

    i made A the angle for theta and did this:

    COSA= (700squared + 1000squared - 400squared)/ (2*700*1000)
    COSA= 0.95
    A= COS to the minus 1(0.95) = 18.19

    thanks for the help.
     
    Last edited: Sep 29, 2007
  9. Sep 29, 2007 #8

    Doc Al

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    No, that's not correct. You can check by finding the components of the resultant: You'll find that the horizontal component alone will be greater than 1000 N.

    Where did you get that equation?
     
  10. Sep 29, 2007 #9
  11. Sep 29, 2007 #10

    Doc Al

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    D'oh! I didn't recognize it--my bad.

    You want the angle opposite the 1000 N side, not opposite the 400 N side. But you're on the right track with the law of cosines. Be sure to draw yourself a picture with the sides and angles labeled.
     
  12. Sep 29, 2007 #11
    i ended up with the angle=90degrees although i have a feeling it is wrong.

    i've been at this one question all day, i think i'm just going to leave it for another time, i'm getting very frustrated.
     
  13. Sep 29, 2007 #12

    Doc Al

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    Just use the same formula, but chose the 1000 N side as the side opposite the angle. You won't get 90 degrees.
     
  14. Sep 30, 2007 #13
    ok i think i've got it, i got 128.68 degrees for the angle.

    what does it mean by " for this condition, what will be the angle between R and the horizontal P?"

    should there be a vertical line going from the 700N arrow to form a 90 degree angle?

    Does the R stand for resultant and the P stand for the vertical line going from the arrow of the 700N force?

    so i would have the 128.68 degrees i worked out and the 90degrees from the 700N arrow and i've just to work out the other angle?
     
  15. Sep 30, 2007 #14

    Doc Al

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    Almost. Did you draw yourself a diagram of the vector addition? It would look like: A horizontal arrow (length = 700 N) pointing left plus a 400 N arrow at some angle (diagonally up and to the left) adding up to the resultant (at some other angle up and to the left) of length 1000 N.

    That's the triangle that you are working with. What you found--correctly!--is the angle within that triangle opposite to the 1000 N resultant. Now use that to find the angle that they call theta.

    I assume that they want the angle the resultant makes with the horizontal. That's the angle in your triangle that's opposite to the side 700 N. (No idea what "P" means.)
     
  16. Sep 30, 2007 #15

    Doc Al

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    a diagram

    Here's a diagram of the situation.

    Note that you found the angle labeled [itex]\alpha[/itex], but you need the angle labeled [itex]\theta[/itex].
     

    Attached Files:

  17. Sep 30, 2007 #16
    yeah, ok , i see what i've done now, i'm getting 51.32degrees for the angle theta.

    i guess i'll be able to work out the angle across from the 700N now too.

    thanks for the help.
     
  18. Sep 30, 2007 #17

    Doc Al

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    Good. That's the angle I think they want.
     
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