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Math Help - force question - urgent help please

  1. #1
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    Question force question - urgent help please

    Four forces Fsub1 , Fsub2, Fsub3, Fsub4 act on a particle in 2-space. The particle is in static equilibrium.
    Fsub1 has magnitude 75N and points due east.
    Fsub2 has magnitude 50N and points 30 degress north of east.
    Fsub3 has magnitude 40N and points 60 degrees west of north.

    Find the magnitude and direction of Fsub4.

    Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    Four forces Fsub1 , Fsub2, Fsub3, Fsub4 act on a particle in 2-space. The particle is in static equilibrium.
    Fsub1 has magnitude 75N and points due east.
    Fsub2 has magnitude 50N and points 30 degress north of east.
    Fsub3 has magnitude 40N and points 60 degrees west of north.

    Find the magnitude and direction of Fsub4.

    Thank you very much.
    Express in components
    (I understand that north of east means 30 degree above the east line)
    (I understand that west of north means 60 degree to the left of north line).

    Force 1)75 i
    Force 2)50*cos(30)i+50*sin(30)j
    Force 3)-40*cos(60)i+40*sin(60)
    Force 4)F_xi+F_yj
    Thus, sum of components,
    75+50*cos(30)-40*cos(60)+F_x=0
    50*sin(30)+40*sin(60)+F_y=0
    You can solve.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Jenny20 View Post
    Four forces Fsub1 , Fsub2, Fsub3, Fsub4 act on a particle in 2-space. The particle is in static equilibrium.
    Fsub1 has magnitude 75N and points due east.
    Fsub2 has magnitude 50N and points 30 degress north of east.
    Fsub3 has magnitude 40N and points 60 degrees west of north.

    Find the magnitude and direction of Fsub4.

    Thank you very much.
    As F_1+F_2+F_3+F_4=0

    F_4=-( F_1+F_2+F_3 )

    Resolve the three forces you have into components in the N and E directions.
    and sum the components of the three vectors, then F_4 is minus this sum.

    You know how then to find the magnitude and direction of a vector in
    component form.

    RonL
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  4. #4
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    I did not work out this question right in my homework. Can anyone show me the complete work of this question. Thank you very much!
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Jenny20 View Post
    I did not work out this question right in my homework. Can anyone show me the complete work of this question. Thank you very much!
    First do a diagram. See attachment.

    We have:

    f1=(75,0)
    f2=(50cos(30),50sin(30))
    f3=(-40sin(60),40cos(60))

    so:

    f4=-(75+50cos(30)-40sin(60), 50sin(30)+40cos(60))~=(83.66,45)

    RonL
    Attached Thumbnails Attached Thumbnails force question - urgent help please-gash.jpg  
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  6. #6
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    Quote Originally Posted by Jenny20 View Post
    Four forces Fsub1 , Fsub2, Fsub3, Fsub4 act on a particle in 2-space. The particle is in static equilibrium.
    Fsub1 has magnitude 75N and points due east.
    Fsub2 has magnitude 50N and points 30 degress north of east.
    Fsub3 has magnitude 40N and points 60 degrees west of north.

    Find the magnitude and direction of Fsub4.

    Thank you very much.
    Here is one way.

    In a system in static equilibrium, the summation of forces is zero.
    So F4, which is unknown, is the exact opposite of the Resultant of the 3 given forces.

    You know how to get the Resultant?

    Let the East-West axis be the x-axis.
    And the North-South axis be the y-axis.

    Rx = 75 +50cos(30deg) -40sin(60deg) = 83.66 newtons
    Ry = 0 +50sin(30deg) +40cos(60deg) = 45 newtons

    R = sqrt[(83.66)^2 +(45)^2] = 94.9947, or, 95 newtons
    Angle from x-axis = arctan(45 /83.66) = 28.275 degrees north of East.

    Therefore, F4 = 95 newtons, and points 28.275 degrees south of West. ----answer
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