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How to show force R^2=(Q-24)^2 + 7^2

  1. Dec 13, 2012 #1
    A particle is in equilibrium under the action of three horizontal forces P, Q and R. The magnitudes of P, Q and R are 25 N, Q N, and R N respectively, and the cosine of the angle between P and Q is -0.96.

    Show that R^2 = (Q - 24)^2 + 7^2.

    Hence find:

    (a) the least possible value of R,

    (b) the corresponding angle between Q and R.

    My working:
    R^2 = P^2 + Q^2 - 2(P)(Q) Cos (theta)
    R^2 = 25^2 + Q^2 - 2(25) (Q) (-0.96)
    R^2= 25^2 + Q^2 + 48 Q
    R^2 = Q^2 + 48Q + 25^2
    Now, I was trying completing the squares but... i don't know how to proceed. Please Help.
     
    Last edited: Dec 13, 2012
  2. jcsd
  3. Dec 13, 2012 #2
    If you have an equation in mathematics: ax^2 + bx +c = 0 Then how do we simplify to get the roots? do you remember the term that c = (b/2)^2 for it to be the same roots. (x+c_new)^2... Try doing this... if you get the new c put it into the equation but remember to subtract it again... Really trying hard not to give the answer...
    so ax^2 +bx +c_old = R^2 becomes ax^2 +bx +c_new -c_new +c_old = R^2... Try this and I'll help you further...
     
  4. Dec 14, 2012 #3

    ehild

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    The "minus" in red should be "+".

    ehild
     

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    Last edited: Dec 14, 2012
  5. Dec 15, 2012 #4
    Thanks. So now we are doing this.

    R^2= (Q)^2 + 2(Q)24 + (24)^2 - (24)^2 + 625
    R^2= (Q + 24)^2 -576 + 625
    R^2 = (Q + 24)^2 + 49
    R^2 = (Q + 24)^2 + 7^2
    it looks like we are almost there. But why is it (Q-24) in the required proof
    and not (Q + 24) as i have got in my proof.
    Am i making some mistake? please suggest. Thanks
     
  6. Dec 15, 2012 #5

    ehild

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    Read #3.

    ehild
     
  7. Dec 15, 2012 #6
    I looked up in 2 sites, "wikipedia" and "math is fun" for Cosine Law. There it is saying
    c^2=a^2 + b^2 -2ab cos(theta)

    Am i missing out on something? Please suggest...
     
  8. Dec 15, 2012 #7

    ehild

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    See the figure in #3. You applied the Law of Cosines to the triangle with sides P and Q and enclosed angle theta, but the third side of that triangle is not R.
    In the triangle with sides of length P, Q, R, the angle enclosed by P and Q is 180-theta.

    ehild
     
  9. Dec 20, 2012 #8
    But this is an equilibrium problem. I understand by this that the vector triangle will be closed. Which means that the heads of two arrows will not meet. But in the diagram you have given, the two arrows' heads are meeting. Is there an error? Please suggest!!!
     
  10. Dec 20, 2012 #9

    ehild

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    I don't quite follow you. I have shown the parallelogram method of addition. The sum of vectors R and Q is -P. R+Q +P=-P+P=0. You have a triangle (dotted) with sides P, Q, R and angle 180-θ, opposite to the side R.
    ehild
     

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    Last edited: Dec 20, 2012
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