**Unformatted text preview: ***2—48. If F1 = 600 N and (I) = 30°, determine the magnitude of the y
resultant force acting on the eyebolt and its direction
measured clockwise from the positive x axis. SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of each
force can be written as (F1), = 600 cos 30° = 519.62N (F1), = 600 sin 30° = 300N (F2), = 500 cos 60° = 250N (F2), = 500 sin 60° = 433.01N (F3)x = 450(2) = 270N (F3)y = 450%) = 360N Resultant Force: Summing the force components algebraically along the x and y axes,
i) 2(FR)X = EFX ; (FR)x = 519.62 + 250 — 270 = 499.62 N —> +TE(FR)y = EFy; (FR), = 300 — 433.01 — 360 = —493.01 N = 493.01 N l The magnitude of the resultant force FR is FR = \/(FR)§ + (FR), = \/499.622 + 493.012 = 701.91 N = 702N Ans. The direction angle 0 of FR, Fig. b, measured clockwise from the x axis, is (FR)y 493 01
= ‘1 = ‘1 —‘ = 44. ° A .
0 tan |: tan 499.62 6 ns (FR)x FE) =4?3-o/~ J
(b) ...

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