Answer:
|F net| = 20.22 N
θ ≈ 19.8°
Step-by-step explanation:
F net = 15N i + 8cos(60°)N i + 8sin(60°)N j
= 15N i + 8×½N i + 8×√3/2N j
= 15N i + 4N i + 4√3N j
= 19N i + 4√3N j
|F net| = √(19²+(4√3)²) = √(361+48) = √409 ≈ 20.22N
tan(θ) = 4√3 ÷ 19 ≈ 0.36 → θ ≈ arctan(0.36) = 19.8°
Unfortunately, you inadvertently cut off the instructions for this problem when you photographed it. Could you try again, making certain to include the instructions?
Let me guess: perhaps the instructions are 1) determine the slope of the line passing through the given points, or 2) write the equation of the line passing through the given points.