Question

An object of mass m = 1.00 kg is observed to have an acceleration of a(arrow above) with a magnitude of 17.6 m/s^2 in a direction θ = 57.5° east of north. The figure below shows a view of the object from above. The force F(arrow above)^2 acting on the object has a magnitude of 9.46 N and is directed north. Determine the magnitude and direction of the force F(arrow above)1 acting on the object.

Answer 1

The object's acceleration is from the total force acting on it. Ftotal = F1 + F2. You can figure out Ftotal (it's m times the given acceleration vector, a), and they give you F2. Use vector addition to find F1 = Ftotal - F2.

Answer 2

I am considering East to be the +x axis, and expressed the angle of Ftotal in those terms, because that's how trigonometry works.

Ftotal = F1 + F2
(17.6N @32.5 degrees) = F1 + (9.46N @90degrees)
convert to rectangular coordinates to do the addition:
F1 = (14.84, 9.46) - (0, 9.46)
F1 = (14.84, 0)
So F1 = 14.84N @0 degrees (directed east).