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March 24, 2017

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Consider four vectors ~F1, ~F2, ~F3, and ~F4, where their magnitudes are F1= 57 N, F2= 38 N, F3= 25 N, and F4= 67 N. Let θ1= 130◦, θ2=−130◦, θ3= 29◦, and θ4=−64◦, measured from the positive x axis
with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vector ~F, where ~F=~F1+~F2+~F3+~F4?
Answer in units of N.

  • Physics - ,

    x components of each

    57 cos 130 = -36.6
    38 cos -130= -24.4
    25 cos 29=
    67 cos -64=
    sum of x components = sumx =

    y components of each

    57 sin 130=
    38 sin -130=
    25 sin 29=
    67 sin -64=
    sum of y components = sumy =

    |F|^2 = sumx^2 + sumy^2

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