Posted by **john** on Friday, October 24, 2008 at 6:29pm.

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.

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

- physics -
**bobpursley**, Friday, October 24, 2008 at 7:26pm
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

- physics -
**Henry**, Wednesday, December 23, 2015 at 6:04pm
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.

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