Two forces, 473 N at 11 degrees and 423 N at 28 degrees are applied to a car in

Question

Two forces, 473 N at 11 degrees and 423 N at 28 degrees are applied to a car in an effort to accelerate it.

a) What is the magnitude of the resultant of these two forces?
b) Find the direction of the resultant force (in relation to forward, with counter-clockwise considered positive). Answer in degrees from the positive x-axis, with counter-clockwise positive, within the limits of -180 to 180.

Note: I have tried calculating the individual x and y components for part a, then use the sq. root of x^2 + y^2, but I can't seem to get the correct answer (which is 844.772 N). I think for part b I need to use the tan^-1 (y/x), but I'm not sure. If needed, the mass of the car is 3234 kg. Any help is appreciated!!!

Explanation / Answer

The net force acting on the car along x-direction is           Fx = (473N)cos11 + (423 N)cos28                = 837.7964895 N The net force acting on the car along y-direction is         Fy = (473N)sin11 - (423 N)sin28              = -108.3338 N The magnitude of the net force is         F = [Fx^2 + Fy^2]             = [(837.7964895 N)^2 + ( -108.3338 N )^2 ]             =844.7716674 N Direction is             =tan-1[(-108.3338 N)/(837.7964895 N)]                = -7.307 Direction is             =tan-1[(-108.3338 N)/(837.7964895 N)]                = -7.307

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