Two forces with magnitudes 33 pound and 74 pounds act on an object at angles of 47° and 132° ,respectively, with the positive x axis. Find the magnitude of the resultant force.

Answer 1

First, break each force into its x and y components.

F₁ = 33 lb → F_1x = 33 lb (cos 47º) F_1y = 74 lb (sin 47º)

F₂ = 74 lb → F_2x = - 74 lb (cos 48º) F_2y = 74 lb (sin 48º)

(48º is the reference angle for 132º in the second quadrant. In the second quadrant, x is negative and y is positive so the x force is negative and the y force is positive.)

Then, find the sum of the x components and the sum of the y components.

ΣF_x = F_1x + F_2x = 33 lb (cos 47º) - 74 lb (cos 48º) = -27.01 lb

ΣF_y = F_1y + F_2y = 33 lb (sin 47º) + 74 lb (sin 48º) = 79.13 lb

Finally, find the magnitude of the resultant force by taking the square root of the x components squared plus the y components squared.

F_resultant = √[(ΣF_x)² + (ΣF_y)²] = √[(-27.01 lb)² + (79.13 lb)²] = 83.61 lb