An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a positive magnitude of 6.7N; a second force has a magnitude of 4.6 N and points in the negative y direction. (1)"Find magnitude of the third force acting on the object.

Question

(2) Find the direction of the third force acting on the object in terms of pheta=-----------degrees from the + x direction.

Answer 1

constant velocity means no net force. So add the three forces, and equal zero.

6.7x -4.6y + z=0
let z have two components, in the x direction, and the y direction.

then zx=-6.7x
and zy=4.6y
but zx= ZcosTheta and zy=ZsinTheta
so ZcosTheta=-6.7
ZsinTheta=4.6
which makes (dividing the first equation into the second)
tanTheta=-.687
Theta=180-34.5 degrees where theta is counterclockwise from the x axis.
Z sinTheta=4.6, so you can solve for Z

Answer 2

1,2. F1+F2+F3 = M*a = M*0 = 0.
6.7 - 4.6i + F3 = 0.
F3 = -6.7 + 4.6i = 8.13 N.[34.5o] N. of W. = 145.5o CCW from +x-axis.