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16 June, 16:09

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

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  1. 16 June, 19:26
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    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
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