Law of Cosine|Please Help!

Question

Two tugboats are pulling a disabled speedboat into port with forces of 1240 lb. and 1480 lb. The angle between these forces is 28.2°. Calculate the magnitude of the equilibrant force.

Answer 1

I like to provide a detailed explanation and an approach to solving similar problems. First, we provide a vector representation of the forces acting on the disabled speedboat. We use the following notations: F₁ and F₂ denote the pulling forces of the two tugboats, and F_R denotes the equilibrant force. We use vector manipulation to form a parallelogram where the angle between the forces and their magnitude are as shown in the figure. Using the property of a parallelogram, we find all the angles shown in the figure. This representation allows us to apply the law of cosines to compute the equilibrant force using the equation:

Using the values of F₁ and F₂, F_R = 2638.70 lb.

F_R = sqrt(F₁²+F₂² - 2F₁F₂cos(θ)), θ=180-28.2 = 151.2

Answer 2

Here is a sketch that might help with the work that Yefim S showed:

Answer 3

R = (1240² + 1480² - 2·1240·1480cos(180° - 28.2°))^1/2 = 2638.7 lb