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 force on the cart?
The net force on the cart has to be calculated by adding the two forces that are acting on the cart. As force is a vector quantity, we need to perform vector addition.
The first force of 200 N pulls the cart towards the south.
The second force of 150 N, pulls the cart in a direction 25 degrees west of south. The component of the force towards the south is 150*cos 25 = 135.94 N. The component towards the west is 150*sin 25 = 63.39 N.
The total force towards the south is 335.94 N and the force towards the west is 63.39 N.
This gives the magnitude of the net force as sqrt(335.94^2 + 63.39^2) = 341.86 N acting in the direction arc tan(63.39/335.94) = 10.68 degrees west of south.
The net force of 341.86 N acts in a direction 10.68 degrees west of south.
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
Fx1 = 0
Fy1 = -200N
Fx2 = -150sin25º
Fx2 = -63.39N
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