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Statics problem(chapter 2 of the book Vector Mechanics for Engineers)

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Three forces are applied to a bracket. The directions of the two
    30-lb forces can vary, but the angle between these forces is always 50. Determine
    the range of values of  for which the magnitude of the resultant of
    the forces applied to the bracket is less than 120 lb.

    [​IMG]

    2. Relevant equation:

    Fnet,x=o
    Fnet,y=0

    3. The attempt at a solution

    i tried to get this problem solved by using different ways, but none of these have given me any answers.
    @= the angle between 30-lb forces
    i tried this one :

    Fnet,x= 100 cos(30)-30 cos(@)-30cos(@+50)=0
    Fnet,y=100sin(30)-30sin(@)-30sin(@+50)=0

    but, this does not seem right for me because i am not going to have range of the angle and i did not use 120 ,and alos the TI-89 did not give me any thing.

    however, i used another way,

    120> -100 cos(150)+30cos(150)+30cos($)+30cos($+50)

    i tried to solve by Ti-89 but also nothing happned.

    $= the angle between the x-postive axis to the first 30-lb force.






    ------------


    my problem is i do not know how to deel with (the which the magnitude of the resultant of
    the forces applied to the bracket is less than 120 lb.) .....


    can anyone helps me with ASAP
     
  2. jcsd
  3. Sep 14, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi Moayad! Welcome to PF! :smile:

    (have a theta: θ :smile:)
    It's Fnetx = 100 cos(30)+30 cos(@)+30cos(@+50)=0
    Hint: F2 = Fx2 + Fy2 :smile:

    (oh … and it might be easier to combine the two 30lbs forces into one force at θ + 25º first :wink:)
     
  4. Sep 14, 2008 #3
    thanks man ...



    can i ask more questions

    is not Fnet,x = -100 cos(30)+30 cos(@)+30cos(@+50)=0
    what about Fnet,y?

    is it goin to be = 100sin(30)+30sin(@)+30sin(@+50)?


    Thank you for help


    Moayad was here
     
  5. Sep 14, 2008 #4

    tiny-tim

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    (what happened to that θ i gave you?)

    if x is to the right, it's Fnet,x = -100 cos(30)-30 cos(@)-30cos(@+50)

    if x is to the left, it's Fnet,x = 100 cos(30)+30 cos(@)+30cos(@+50)

    (where did your "=0" come from? :confused:)

    Your original Fnet,y was correct.
     
  6. Sep 14, 2008 #5
    from the graph


    is not the x to the right ?



    sorry i did not use what you gave me, i' am gonna use from now




    thank you again
     
  7. Sep 14, 2008 #6
    here again

    i did this :

    i used the x to the right, so

    Fx=-100cos(30)-30cos(25-θ) , ( i combined the two 30lbs forces into one as you told me )
    Fy=100sin(30)-30sin(25+θ)

    F^2 = Fx^2 + Fy^2

    l let F=120 ........(is it right?)

    so,

    Fx= sqrt((120)^2+Fy^2)

    i pulged to to(Fx=-100cos(30)-30cos(25-θ) )

    it became :
    sqrt((120)^2+Fy^2)=-100cos(30)-30cos(25-θ)

    and i pulged (Fy=100sin(30)-30sin(25+θ))

    so,

    my final equation is :


    sqrt((120)^2+(100sin(30)-30sin(25+θ))^2)=-100cos(30)-30cos(25-θ)


    then , i put it to the Ti-89 using the solve comand. i solved to the θ

    but, it did not gave me a number or range of values
     
  8. Sep 14, 2008 #7

    tiny-tim

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    Hi Moayad! :smile:

    i agree x is usually to the right …

    but it's your decision …

    the maths is only there to help you :rolleyes:

    you're in control! :smile:

    you can make x to the left if you want to, and in this case it's much easier, because it avoids all those minus signs! :wink:
     
  9. Sep 14, 2008 #8

    tiny-tim

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    Hi Moayad! :smile:

    hmm …

    i] you can't just write 30cos(25-θ) … the magnitude of two 30lb forces combined won't still be 30 (or even 60), will it? :wink:

    ii] don't be so complicated :cry:

    just square Fx and Fy, add them, and use cos²θ + sin²θ = 1 :smile:
     
  10. Sep 14, 2008 #9
    Hi again ...


    i usually make stupid mistakes ... sorry heh



    i added them so what i got is 45.69 lb




    i am gonna work this out .. and i'll be back



    thank you sooooo much
     
  11. Sep 14, 2008 #10

    tiny-tim

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    how did you get that?

    you didn't use 50º, did you?
     
  12. Sep 14, 2008 #11
    i used

    30/sin(25)=R/sin(40)

    R= the new force
     
  13. Sep 14, 2008 #12

    tiny-tim

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    wherever did 40º come from? :confused:

    resolve in the direction of R: R = 30cos(25º) + 30cos(-25º) = … ? :smile:
     
  14. Sep 15, 2008 #13
    i got it ..... thank you for you help


    i used the three equation

    1- F=sqrt((Fx)^2+(Fy)^2)

    F=120


    2-Fx = -100 cos(30)-30 cos(θ)-30cos(θ+50)

    3-Fy=100sin(30)-30sin(θ)-30sin(θ+50)


    i pluged equation 2,3 to 1
    and solve for θ


    i got 27.39>θ>223
     
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