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

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

Step by step solution

01

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.
02

Calculate the components for the first man

For the first man pulling east, we have: \(F_{1x} = F_1\cos(\theta_1) = 62 \cos(12^{\circ}) \approx 60.22 \mathrm{N}\) \(F_{1y} = F_1\sin(\theta_1) = 62 \sin(12^{\circ}) \approx 13.03 \mathrm{N}\)
03

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.
04

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.
05

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.

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