Two forces pull on ropes attached to a cart: 1 with a force of 200 N South, and the other with a force of 150 N at 25º W of S. What was the net...

To sole the problem we

let F1
= 200N, South

F2 = 150N, 25º West of
South

R the Net Force

R² = Fx²
+ Fy²

where the x-axis along the East-West and the y-axis
along the Meridian line

Assuming that component force
towards the N & E as positive and component force towards S & W as
negative, then

For F1:

Fx1 =
0

and

Fy1 =
-200N

For F2:

Fx2 =
-150sin25º

Fx2 =
-63.39N

and

Fy2 =
-150cos25º

Fy2 = -135.95N

Fx =
Fx1 + Fx2

Fx = -63.39 N

Fy =
Fy1 + Fy2

Fy = -200 +
(-135.95)

Fy = -335.95 N

R² =
(-63.38)² + (-335.95)²

R =
341.87N

The direction (θ) of the net force can be computed
by

tan θ = |Fx/Fy|

tan θ =
|-63.39/-335.95|

Note (-) Fx means the direction is West
and (-) Fy means the direction is South, then

θ =
Arctan|-63.38/-335.95|

θ = 10º30'17.5"
South-West

Therefore the net force is 341.87N
at 10º30'17.5" South-West