During a picnic, you and two of your friends decide to have a three-way tug- of-war, with three ropes in the middle tied into a knot. Roberta pulls to the west with \(420 \mathrm{~N}\) of force; Michael pulls to the south with \(610 \mathrm{~N}\). In what direction and with what magnitude of force should you pull to keep the knot from moving?

The forces of Roberta and Michael can be represented as vectors. Since Roberta is pulling to the west with 420 N of force, her vector can be represented as \((-420, 0) \mathrm{~N}\). Similarly, Michael is pulling to the south with 610 N of force, so his vector can be represented as \((0, -610) \mathrm{~N}\).

We can find the sum of the two force vectors by adding them component-wise: \((F_x, F_y) = (-420, 0) + (0, -610) = (-420, -610) \mathrm{~N}\)

Since the knot should not move, the third person has to apply an equal force in the opposite direction of the sum of Roberta's and Michael's forces. Thus, the third person's force vector is \((420, 610) \mathrm{~N}\). We can find the magnitude of this force vector using the Pythagorean theorem: \(F = \sqrt{F_x^2 + F_y^2} = \sqrt{420^2 + 610^2} \approx 738.73 \mathrm{~N}\) Now, we can find the direction of the force vector using the tangent function: \(\tan(\theta) = \frac{F_y}{F_x} = \frac{610}{420}\) \(\theta = \arctan\left(\frac{610}{420}\right) \approx 55.52^{\circ}\) Since the force vector is in the second quadrant, the actual angle from the positive x-axis is: \(180^\circ - 55.52^\circ = 124.48^\circ\) So the third person should pull with a force of \(738.73 \mathrm{~N}\) at an angle of \(124.48^\circ\) from the positive x-axis (east) to keep the knot from moving.