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Simple Force in Equilibrium

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi everyone, I've been stuck on this relatively very simple problem for quite a while now and even after solving I'm still getting two answers while the book only lists one.

    Question: If block D weighs 300 lb and block B weighs 275 lb, determine the required weight of block C and the angle (theta) for equilibrium.

    So I did the usual since it's in equilibrium then Fx = 0 and Fy = 0.

    So, Fc(cos30) - 275(cos[itex]\theta[/itex]) = 0
    and Fy(sin30) + 275(sin[itex]\theta[/itex]) - 300 = 0

    Then after finding cos[itex]\theta[/itex] in terms of Fc and substituting, you get this equation: 158.7cos[itex]\theta[/itex] + 275sin[itex]\theta[/itex] = 300

    The answer listed is: [itex]\theta[/itex] = 40.9 and Fc = 240lb

    Can anyone please tell me how to solve the equation above? I tried using the trig identity (square of cos plus square of sin = 1) but I got two answers, one was similar to the book ([itex]\theta[/itex] = 40.9, and one was [itex]\theta[/itex] 79). Thing is, when I use the second value I got (79) in the equations above, it works for all.

    I'd appreciate some clarification on this problem! Thanks.

    d9053b6a-35de-4064-a670-de15385a16a6.jpe
     
  2. jcsd
  3. Sep 25, 2011 #2

    ehild

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    Both angles are correct.

    ehild
     
  4. Sep 25, 2011 #3
    first realize that (275lbs)/tantheta= C/tan30 since the weights are all in the y direction right?

    where C= 275lbs(tan30)/tantheta and theta=inverse tan ((275lbs)(tan30)/C)

    and so, you must find c first.

    c would be found using forces in the y direction such that sum of all forces is zero since equalibrium occurs.

    so, 300lbs + 275lbs + D=0N

    D=575N

    what do you say to this?
     
  5. Sep 25, 2011 #4
    sorry C=575N
     
  6. Sep 25, 2011 #5
    Rayquesto, I'm sorry I don't understand exactly what you're trying to do or how you're doing and why we need to find C first? Also, wouldn't the force in the y direction need to take into account the angle the string is making with the horizontal?

    ehild, so then there are also two possible weights as well, correct?
     
  7. Sep 25, 2011 #6

    ehild

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    Yes, of course. There is less weight for the greater angle.

    ehild
     
  8. Sep 25, 2011 #7
    Yes the two answers for the angle are correct and the less weight mentioned by ehild is 59.9lb.
     
  9. Sep 25, 2011 #8
    yes, but ok, well Im confused too then, because wouldnt the weights already be acting in the y direction? I mean, to me, your calculuation would be correct if the forces were affected such that it can be treated as a hypotenuse of the force, but It looks like the weights can only be thought of as a Y leg.
     
  10. Sep 26, 2011 #9
    I cannot follow what Rayquesto is saying. Can he be more specific?
    The weights are acting in the Y-direction (i.e. vertically).But here we are examining the equilibrium at point A and the equations in the first post are correct.
     
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