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