Question

Two of Robin Hood's men are pulling a sledge loaded with some gold along a path that runs due north to their hideout. One man pulls his rope with a force of \(62 \mathrm{N}\) at an angle of \(12^{\circ}\) east of north and the other pulls with the same force at an angle of \(12^{\circ}\) west of north. Assume the ropes are parallel to the ground. What is the sum of these two forces on the sledge?

Short answer

Answer: The sum of the two forces on the sledge is approximately \(26.06 \mathrm{N}\) north.

Step-by-step solution

Find the horizontal and vertical components of the forces

To find the horizontal and vertical components of each force, we can use the following equations: Horizontal component: \(F_x = F\cos(\theta)\) Vertical component: \(F_y = F\sin(\theta)\) For the first man pulling east: \(F_1 = 62 \mathrm{N}\) \(\theta_1 = 12^{\circ}\) east of north For the second man pulling west: \(F_2 = 62 \mathrm{N}\) \(\theta_2 = 12^{\circ}\) west of north Now, we can find the horizontal and vertical components of each force.

Calculate the components for the first man

For the first man pulling east, we have: \(F_{1x} = F_1\cos(\theta_1) = 62 \cos(12^{\circ}) \approx 60.22 \mathrm{N}\) \(F_{1y} = F_1\sin(\theta_1) = 62 \sin(12^{\circ}) \approx 13.03 \mathrm{N}\)

Calculate the components for the second man

For the second man pulling west, we have: \(F_{2x} = -F_2\cos(\theta_2) = -62 \cos(12^{\circ}) \approx -60.22 \mathrm{N}\) \(F_{2y} = F_2\sin(\theta_2) = 62 \sin(12^{\circ}) \approx 13.03 \mathrm{N}\) Notice that the horizontal components are opposite in direction, so they will cancel each other out.

Sum the components

Now, let's sum the horizontal and vertical components separately: \(F_{sum_x} = F_{1x} + F_{2x} = 60.22 + (-60.22) = 0 \mathrm{N}\) \(F_{sum_y} = F_{1y} + F_{2y} = 13.03 + 13.03 = 26.06 \mathrm{N}\) Since the horizontal components canceled each other out, the only force remaining is the vertical force.

Calculate the total force F_sum using Pythagorean theorem

Finally, we use the Pythagorean theorem to find the total force: \(F_{sum} = \sqrt{F_{sum_x}^2 + F_{sum_y}^2} = \sqrt{0^2 + 26.06^2} \approx 26.06 \mathrm{N}\) The sum of the two forces on the sledge is approximately \(26.06 \mathrm{N}\) north.