physics 1 problem (magnitude and angle of vector

Question

Forces of 50 N at 20 degrees, and 40 N at an angle of 250 degrees, measured counter-clockwise from the positive x-axis, act on an object.

a) What is the magnitude of the resultant force (in Newtons)?

b) What is the angle of the resultant force with respect to x-axis? (positive values only)

Answer 1

a) We have two different magnitude forces acting on an object at two different angles. To solve this problem, we are going to have to break down these forces into their respective x and y components, and then combine them to find the resultant force. We do this using trig functions. Note the angles are given with respect to the x axis, or horizontal axis. So the horizontal, x, component is found using cosine and the vertical component, y, is found using sine.

Force 1 = F₁ = 50 N force at 20 degrees

F_1x = 50*cos(20 deg) = 46.98 N

F_1y = 50*sin(20 deg) = 17.10 N

Force 2 = F₂ = 40 N force at 250 degrees

F_2x = 40*cos(250 deg) = -13.68 N

F_2y = 40*sin(250 deg) = -37.59 N

Next, we can combine our x components with each other to determine the resulting force in the x direction. And likewise with the y components.

F_Rx = F_1x + F_2x = 46.98 N + (-13.68 N) = 33.3 N

F_Ry = F_1y + F_2y = 17.10 N + (-37.59 N) = -20.49 N

From here, the magnitude of the resultant force is calculated using Pythagorean theorem: a² + b² = c². In terms of this problem, the equation is:

F_Rx² + F_Ry² = F_R² (or)

F_R = sqrt((F_Rx)² + (F_Ry)²)

F_R = sqrt((33.3 N)² + (-20.49 N)²)

F_R = 39.1 N

(b) To calculate the angle of the resultant force, we simply look at our right triangle from above, where F_Rx and F_Ry are the side lengths and F_R is the hypotenuse. Using trigonometry, tan = sin/cos = (opposite side) / (adjacent side). Therefore, our resultant angle is calculated as follows:

θ = arctan(F_Ry/F_Rx)

θ = arctan ((-20.49 N)/(33.3 N))

θ = -31.6 degrees, this implies 31.6 degrees clockwise. However, the problem wants us to provide a positive value. So we simply subtract from 360 degrees to arrive at the same point traveling counterclockwise.

θ = 328.4 degrees