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Finding the Magnitude and Direction of Force one with the Resultant Given

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine the magnitude and direction theata of F1 so that the resultant force is directed vertically upward and has a magnitude of 800N.

    2. Relevant equations
    Question 2-37.gif


    3. The attempt at a solution

    I have difficulty of where to start the problem? do i sum up the forces of the x axis with unknowns in it and forces of the y axis with the unknown forces...where should i start....and how do i start solving the problem????

    Please someone help me with this....

    Thanks,

    Nacha
     
  2. jcsd
  3. Jul 6, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi Nacha ! Welcome to PF! :smile:

    Yes … split the force F1 into Fx and Fy

    then do x and y separately …

    finally combine Fx and Fy to give you F1 and theta. :smile:
     
  4. Jul 6, 2008 #3
    Re: Welcome to PF!

    Hi,

    Thanks for helping me, but one more question, my Fx and Fy for F1 is unkown as shown in the diagram that i have posted earlier, so how do i go about solving for the magnitude of F1 and the direction of F1. Sorry for the trouble....

    thanks
    Nacha
     
  5. Jul 7, 2008 #4

    tiny-tim

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    Hi,

    Have you got Fx and Fy yet?

    You can get them from the diagram.

    Then use Pythagoras, and tangents. :smile:
     
  6. Aug 3, 2010 #5
    can anyone explain how to solve this problem in complete details? i don't understand the explanations. i have the same problem too. :confused:
     
  7. Aug 3, 2010 #6

    tiny-tim

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    welcome to pf!

    hi sandra! welcome to pf! :wink:

    you need to get all the y-components adding to 800, and all the x-components adding to zero

    that will give you two equations involving two unknowns (F1 and θ), which should enable you to find the unknowns

    what do you get? :smile:
     
  8. Aug 3, 2010 #7
    owhh.. how about to solve this question:

    determine the magnitude and direction of P so that the resultant of P and the 900N force is a vertical force of 2700N directed downward.

    is it using the same method also? it's almost the same right?
     
  9. Aug 3, 2010 #8

    tiny-tim

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    right! :smile:
     
  10. Aug 3, 2010 #9
    thanks so much tiny-tim! i'm doing my engineering mechanics tutorial problems right now.. if i have any problem again, i'll ask your help later.. ;)
     
  11. Aug 3, 2010 #10
    err... do you online often? then, i'll know when can i ask for your help..
     
  12. Aug 3, 2010 #11

    tiny-tim

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    sandra, just post a new thread any time :smile:

    somewhere in the world, there'll be someone awake and logged in ready to help, whatever time it is :wink:
     
  13. Aug 3, 2010 #12
    tiny-tim

    okie! will do.. :approve: :smile: :wink:
     
  14. Aug 14, 2010 #13
    i got a similar problem but i dont have a resultant force :\ not sure on how to solve ive found theata using a parallelogram
    theata is the one between the X axis and F1
     

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  15. Aug 14, 2010 #14

    tiny-tim

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    Welcome to PF!

    Hi Zlash! Welcome to PF! :smile:

    (have a theta: θ :wink:)

    Sorry, I dont understand the question …

    you have three forces (100N 350N and R), and also a line called X at 30º …

    if X is equivalent to F1 in Nacha's question, then changing the magnitude of X will give different directions for the resultant R (but a unique magnitude for each direction) …

    if neither the magnitude nor the direction of R is given, I don't see how you can answer the question :confused:
     
  16. Aug 14, 2010 #15
    Re: Welcome to PF!

    nah X' is a vector that cuts through.
    in the quiz im doing it doesnt say anything about a resultant but im assuming we have to use the one given in the text book. so the resultant measured from the positive X' -the line- is 600N
    i found θ as 60. θ being between F1 and the X axis( not x' )
    im not sure how i can work through forces based on different axis kinda messes me up =/
    im trying it now
     
    Last edited: Aug 14, 2010
  17. Oct 13, 2010 #16
    hi all
    ∑F x= 400cos30 – 600 cos36.87 + F1 sinØ =0
    ∑F y= 400sin30 – 600sin36.87 + F1 cosØ =0
    R=√(〖fx〗^2+〖fy〗^2 )
    800=√((〖298.35+f1sinØ)〗^2+〖(560+f1 cosØ)〗^2 )

    As we see ; we have two equation and two unknown so we can get the magnitude and direction of f1
     
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