A crate 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