If the magnitude of the resultant force acting on the eyebolt is 570 N and its direction measured clockwise from the positive x axis is θ = 33 ∘, determine the magnitude of

F1 = 1210.65 N

Q = 65.7081 degrees

Explanation:

Sum of Forces in x - direction:

F1 * cos (Q) + F2*sin (30) - F3 * (3/5) = Fres*cos (theta)

F1 * cos (Q) + (500) * (0.5) - 450 * (3/5) = 570*cos (33)

F1*cos (Q) = 498.0422237 N ... Eq1

Sum of Forces in y - direction:

F1 * sin (Q) - F2*sin (60) - F3 * (4/5) = Fres*sin (theta)

F1 * sin (Q) - 500*sin (60) - 450 * (4/5) = 570*sin (33)

F1 * sin (Q) = 1103.45692 N ... Eq 2

Divide Eq 2 by Eq 1

tan (Q) = 2.21558916

Q = arctan (2.21558916) = 65.7081 degrees

F1 = 1210.65 N