Seth M. answered • 10/12/19

BS in Physics and Engineering, Taught High School and College Students

The detailed way to approach this is to break each force into its orthogonal (x and y) components, sum these, then re-combine them to arrive at the resultant force vector. There are short-cut ways of doing this, but for the sake of providing a clear example, I'll be a bit tedious.

It doesn't really matter which direction and rotation we use for our angles as long *as we are consistent throughout the entire problem*.

Find the sum of the forces in the x-direction:

F1x = 23.2N x cos(14.1) = 22,5N

F2x = 35.4N x cos(139) = -26.7N

F3x = 77.7N x cos(289) = 25.3N

Fx = 21.1N

Find the sum of the forces in the y-direction:

F1_{y} = 23.2N x sin(14.1) = 5.65N

F2_{y} = 35.4N x sin(139) = 23.2N

F3_{y} = 77.7N x sin(289) = -73.5N

F_{y} = -44.7N

The magnitude of the force is found by taking the square root of the sum of the squares of our orthogonal vector forces:

F = √(21.1^{2} + (-44.7)^{2}) = √(445 + 2000) = √2445 = 49.4N

The angle of movement can be found via a number of methods. An easy one is to us arc-tangent:

TANθ = O/A = Fy / Fx = -44.7 / 21.1 = -2.12

θ = TAN^{-1}(-2.12) = -64.7^{o} or 295^{o}

So, the magnitude of the sum of the forces is 49.4N and the block will accelerate in the 295^{o} direction.

Does that make sense?