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22 November, 23:23

A 20-kg object is subjected to three forces which produce an acceleration a = - 8 m. s^-2 i + 6.0 m. s^-2 j on the object. Two of the forces are:F1 = 3.0 N i + 16.0 N jF2 = - 12.0 N i + 8.0 N jFind the third force.

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  1. 22 November, 23:30
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    F₃ = - 151 N i + 96 N j

    Explanation:

    Newton's second law:

    ∑F = m*a Formula (1)

    ∑F : algebraic sum of the forces in Newton (N)

    m : mass s (kg)

    a : acceleration (m/s²)

    Forces acting on the object

    F₁ = 3.0 N i + 16.0 N j

    F₂ = - 12.0 N i + 8.0 N j

    F₃ = F₃x N i + F₃ y N j

    x component of the net force on the object

    Fx=F₁x+F₂x+F₃ x

    Fx = 3.0 N-12.0 N + F₃x

    Fx = F₃x - 9 N

    y component of the net force on the object

    Fy=F₁y+F₂y+F₃ y

    Fy = 16.0 N + 8.0 N + F₃y

    Fy = F₃y + 24 N

    Newton's second law to the object:

    a = - 8 m/s² i + 6.0 m/s² j

    ∑Fx = m*ax m=20 kg, ax = - 8 m/s²

    F₃x - 9 = 20 * (-8)

    F₃x = - 160+9

    F₃x = - 151 N

    ∑Fy = m*ay m=20 kg, ay = 6 m/s²

    F₃y + 24 = 20 * (6)

    F₃y = 120 - 24

    F₃y = 96 N

    F₃ = - 151 N i + 96 N j
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