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Resultant Force and its Components

  1. Jun 27, 2016 #1
    1. The problem statement, all variables and given/known data
    The Resultant of the system of forces shown is R = 425 j lbs. Determine the possible values of P and Θ.
    2. Relevant equations
    R = F1 +F2+P

    3. The attempt at a solution
    I ended up with two equations, one in the "i" direction and one in the "j" direction. I think you have to use the quadratic formula because I know that you will get two values for P and two values for Θ. Thanks!
     

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  2. jcsd
  3. Jun 27, 2016 #2

    BiGyElLoWhAt

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    So what's the problem? What specifically?
     
  4. Jun 27, 2016 #3
    The Problem is to find P and Θ. You will need to refer to the figure I attached on the original post. I get two equation but then I cant get past that point.
     
  5. Jun 27, 2016 #4

    BiGyElLoWhAt

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    2 equations, 2 unknowns? Solve for one of the variables and use substitution.
     
    Last edited: Jun 27, 2016
  6. Jun 27, 2016 #5

    BiGyElLoWhAt

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    Example:
    ##a = x + y##
    ##b = 2x - 3y##
    Solve for x in the first equation:
    ##x= a- y##
    Sub it into the second equation for x
    ##b = 2(a-y) - 3y##
    Then solve for y.
    Once you have found y, plug it into one of the equations and solve for x.
     
  7. Jun 27, 2016 #6

    SteamKing

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    Or substitution.
     
  8. Jun 27, 2016 #7

    BiGyElLoWhAt

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    :sorry:... fixed
     
  9. Jun 27, 2016 #8
    2 Equation and 2 unknowns should result in an answer for P and an answer for Θ. The solution has two answers for P and two answers for Θ.
     
  10. Jun 27, 2016 #9
    Can someone please just solve this problem?
     
  11. Jun 27, 2016 #10

    BiGyElLoWhAt

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    That's against policy, sorry. I'm not seeing multiple solutions. It's kind of a mess, but I only see one solution for theta. Use the trig identity sin(theta) = cos(theta +90) or vice versa cos(theta) = sin(theta - 90), then factor, and solve.
     
  12. Jun 27, 2016 #11
    Yeah that's what I have done and only got one solution. Unfortunately the solution manual says that P=545.46 and 53.89, Θ=33.30° and -87.44°
     
  13. Jun 27, 2016 #12

    SteamKing

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    If you take the original force diagram and decompose the vectors shown into components, you should obtain two equations in P and θ. You know the x-component of the resultant is equal to 0, while the y-component must equal 425 lbs. Further algebraic manipulation yields and equation in sin θ and another in cos θ. By squaring and adding these two equations together, one can use the Pythagorean relation to eliminate θ, leaving a quadratic in P, which of course gives two values as solutions.

    By substituting these values of P back into one of the earlier equations involving θ, you can obtain a pair of angles which are also solutions, but only one angle will work with a given solution to P, while the other angle works with the other solution to P. This is why there are four answers given.

    Now, instead of dancing around this problem, if you would like to share your work towards solution, we may be able to see where you went wrong in your original attempt.
     
  14. Jun 28, 2016 #13
    This is what I have gotten so far.
     

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  15. Jun 28, 2016 #14

    SteamKing

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    This looks good.

    Re-write the two final equations so that sin θ and cos θ are on one side with P and the rest on the other side of the equals.

    Square the LHSides and eliminate θ using the Pythagorean relation, sin2 θ + cos2 θ = 1.

    After that, it becomes a matter of collecting the remaining terms and writing a quadratic equation in P.
     
  16. Jun 28, 2016 #15
    So you are saying I can square both sides and then set the RHSides equal to one? and then solve for P using quadratic equation?
     
  17. Jun 28, 2016 #16
    I did that and got some extremely small numbers for P
     
  18. Jun 28, 2016 #17

    SteamKing

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    Then you made some mistakes somewhere in you calculations.
    Not necessarily. You have one equation in sin θ and another equation in cos θ. By squaring these two equations and adding them together, you wind up with a third equation where you have something like A (sin2 θ + cos2 θ) = RHS with P terms. sin2 θ + cos2 θ = 1 by the Pythagorean relation, and the terms involving θ drop out.
     
  19. Jun 28, 2016 #18

    SammyS

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    From your work:
    upload_2016-6-28_18-37-56.png

    Isolate the terms with sin(θ) and cos(θ) on one side of each equation. You can leave each with a coefficient of 275 .

    You may want leave .866 as ##\displaystyle\ \frac{\sqrt3} {2}\ .##

    Be careful in squaring the other side. It will have two terms. Remember: ##\ (a-b)^2=a^2-2ab+b^2\ .##
     
  20. Jun 28, 2016 #19

    SammyS

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    Another method of solution you may want to consider can be done using a graphing calculator, such as a TI-83 or TI-84 .

    Solve each equation for P.
    The first equation gives P as some function of θ, the other equation gives P as some other function of θ.

    With the TI calculators, P of the first equation will be treated as Y1 , and P of the second equation will be treated as Y2. In both cases use X in place of θ .

    I suppose you should leave the calculator in DEGREE mode.
    Use the trig ZOOM to set the X scale. You may need to fiddle with the Y scale. It looks like the Y range should be about ±750 .

    Once the graphs look good, use the intersect feature of the calculator.
     
  21. Jun 28, 2016 #20
    I got here and then I add the RHSide together correct? and set that equal to 1? Because when I do that you start getting extremely small numbers...
     

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