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14 April, 00:39

Ellus

Three horizontal forces are pulling on a

ring, at rest. F1 is 12.0 N at a 0° angle, and

F2 is 18.0 N at a 90° direction. What is the

x-component of F3?

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Answers (1)
  1. 14 April, 01:18
    0
    Three forces are in equilibrium.

    Let us represent forces in vector form ie in the form of i and j.

    F₁ is acting along positive x axis (at a 0° angle)

    F₁ = 12 i

    F₂ is acting along positive y axis (at a 90° direction)

    F₂ = 18 j

    Suppose F₃ is represented by X i + Y j

    Total of three forces is zero.

    F₁ + F₂ + F₃ = 0

    12 i + 18 j + F₃ = 0

    F₃ = - 12i - 18 j

    So x - component of F₃ = - 12 i

    X - component of F₃ is equal to 12 N in magnitude and is acting in negative x direction ie at 180° direction.

    similarly y - component is 18 N acting in negative y direction.
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