**Unformatted text preview: **2—49. If the magnitude of the resultant force acting on the
eyebolt is 600 N and its direction measured clockwise from
the positive x axis is 6 = 30°, determine the magnitude of
F1 and the angle <1). SOLUTION Rectangular Components: By referring to Figs. a and b, the x and y components of
F1, F2, F3, and FR can be written as (F1)x = F1 COS d) (171)}, = F1 sin (1) (F2): = 500 cos 60° = 250N (102)y = 500 sin 60° = 433.01 N
3 4 (F3)x = 450(g) = 270N (F3)y = 450<§> = 360N (FR)x = 600 cos 300 = 519.62N (my = 600 sin 30° = 300N Resultant Force: Summing the force components algebraically along the x and y axes,
i> 2(FR)x = EFx; 519.62 = F1 cos o + 250 — 270 F1 cos d) = 539.62 (1)
+TE(FR)y = 213; —300 = F1 sin¢ — 433.01 — 360 F1 sin ¢ = 493.01 (2) Solving Eqs. (1) and (2), yields
(1) = 42.4° F1 = 731 N Ans. ...

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