A point is pulled by two ropes...

Question

A point is pulled by two ropes: one to the left and up 34° with a magnitude of 5, the other to the right and up 16° with a magnitude of 7. What is the resultant force on the point?

(A) 28.98∠9.4°

(B) 5.38∠61.5°

(C) 0.85∠1.0°

(D) 6∠25.0°

(E) none of these

Answer 1

Adding vectors requires you to break each vector into its x-component and y-component, then to add the two x-components together to get the x-component of the resultant and then repeat for y.

If F₁ = left and up by 34° then F_1x = 5cos(34) = 4.145 however, since this is LEFT, we use a negative making F_1x = -4.145. F_1y = 5sin(34) = 2.796

If F₂ = right and up 16° the F_2x = 7cos(16) = 6.729 and F_2y = 7sin(16) = 1.929

So adding the two x-components we get: -4.145 + 6.729 = 2.584 and adding the two y-components we get 2.796 + 1.929 = 4.725

Putting those back together, we can use the Pythagorean Theorem where the magnitude of the resultant force (F_R) is calculated as F_R = √(2.584² + 4.725²) = √29 = 5.386

The direction is calculated using inverse tangent where θ = tan^-1(4.725/2.584) = tan^-1(1.829) = 61.33° so I'd say answer B looks pretty close