A 100 N force acts along the x axis and a 150 N force acts at an angle of 40 degrees above the x axis.?

A 100 N force acts along the x axis and a 150 N force acts at an angle of 40 degrees above the x axis.

a) What is the resultant force vector?

b) What is the magnitude of the resultant force

c) What is direction of the resultant(hint think arctan)

3 Answers

  • The first step is to determine the x and y components of the 150 N force.

    x = 150 * cos 40 ≈ 115 N

    y = 150 * sin 40 ≈ 96.4 N

    Total x = 100 + 150 * cos 40 ≈ 215 N

    F = √(Fx^2 + Fy^2)

    F = √[(100 + 150 * cos 40)^2 + (150 * sin 40)^2]

    This is approximately 236 N.

    Tan θ = y ÷ x

    Tan θ = (150 * sin 40) ÷ (100 + 150 * cos 40)

    The angle is approximately 24.2˚ above horizontal .

    The resultant force vector is approximately 236 N at angle of 24.2˚ above horizontal. I hope this is helpful for you.

  • You didn’t tell whether the 150N is sorta helping the 100N or sorta opposing.

    RealPro assumed the 150N acts so that its x axis component points the same direction as the 100N. If that is true, go with that advice. If it was the other way around, you should have told us which.

  • Break up each force into its x- and y-components and then add them.

    Fx = (100 + 150cos40°)i

    Fy = (150sin40°)j

    You can get magnitude using pythagorean theorem.

    F^2 = (Fx)^2 + (Fy)^2

    Direction = arctan( 150sin(40°) / [100 + 150cos(40°)] ) = 24.2° above x-axis

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