# 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)

• 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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