a asked in Science & MathematicsMathematics · 6 months ago

Can anyone help me with this asap thanks?

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3 Answers

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  • 6 months ago

    The sum of forces=

    8[cos(30*)i+sin(30*)j]+

    6[cos(-60*)i+sin(-60*)j]+

    4[cos(90*+15*)i+sin(90*+15*)j]

    =

    8.893i+2.668j

    =>

    [magnitude=

    sqr(8.893^2+2.668^2)~9.285,

    angle=tan^-1(2.668/8.893)~16.7*]

    =

    9.285[cos(16.7*)i+sin(16.7*)j]

    =

    9.3 n in the direction of 16*1'59"

    from the x-axis.

  • 6 months ago

    you want the vector sum of the three vectors shown?

    F1x = 6 cos 60

    F1y = –6 sin 60

    F2x = 8 cos 30

    F2y = 8 sin 30

    F3x = –4 sin 15

    F3y = 4 cos 15

    total is

    Fx = F1x + F2x + F3x

    Fy = F1y + F2y + F3y

    F = √(Fx² + Fy²)

    θ = arctan (Fy/Fx)

    surely you can do the arithmetic...

    edit:

    I'll fill in some of the numbers

    F1x = 6 cos 60 = 3

    F1y = –6 sin 60 = –6(√3/2) = –3√3 = –5.20

    F2x = 8 cos 30 = 8√3/2 = 4√3 = 6.93

    F2y = 8 sin 30 = 4

    F3x = –4 sin 15 = –4((√6–√2)/4) = –√6 + √2 = –1.035

    F3y = 4 cos 15 = 4 ((√6+√2)/4) = √6 + √2 = 3.86

    Fx = 3 + 6.93 – 1.04 = 8.89

    Fy = –5.20 + 4 + 3.86 = 2.66

    F = √(8.89² + 2.66²) = √(79.0321 + 7.0756) = 9.28

    θ = arctan (Fy/Fx) = arctan (2.66/8.89) = 16.7º

    but check the math.

  • Anonymous
    6 months ago

    I believe it's about 8.4N at 28°

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