Acrate is pulled to the right with a force of 85.7 n, to the left with a force of 125.9 n, upward with a force of 525.2 n, and downward with a force of 242.6 n.
1) what is the net external force in the x direction? 2) what is the net external force in the y direction?
3) what is the magnitude of the net external force on the crate?
4) what is the direction of the net external force on the crate (as an angle between −180◦ and 180◦, measured from the positive x axis with counterclockwise positive)?

1)  Rx = 40.2 N   (←)  

2) Ry = 282.6 N   (↑)  

3) R = 285.4449 N

4) ∅ = 98.096º  

Explanation:

Given

F₁ = 85.7 N  (→)

F₂ = 125.9 N  (←)

F₃ = 525.2 N   (↑)

F₄ = 242.6 N    (↓)

1)  ∑Fx = Rx  (→+)

⇒   Rx = F₁ - F₂ = 85.7 N - 125.9 N = -40.2 N

⇒   Rx = 40.2 N   (←)   is the net external force in the x direction

2) ∑Fy = Ry  (↑+)

⇒   Ry = F₃ - F₄ = 525.2 N - 242.6 N = 282.6 N

⇒   Ry = 282.6 N   (↑)   is the net external force in the y direction

3) In order to get the magnitude of the net external force on the crate we can use the formula

R = √(Rx² + Ry²)

⇒   R = √((-40.2 N)² + (282.6 N)²)

⇒   R = 285.4449 N

4) The direction of the net external force on the crate can be obtained as follows

∅ = tan⁻¹(Ry / Rx)

⇒  ∅ = tan⁻¹(282.6 N / -40.2 N)

⇒  ∅ = 98.096º   measured from the positive x axis with counterclockwise positive

answer: its c lololo

true tan speed = ang vel x radius