Engineering Mechanics Statics - Intructor Solutions manual 2.44 - *244 The magnitude of the resultant force acting on the bracket is to be 400 N

# Engineering Mechanics Statics - Intructor Solutions manual 2.44

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This preview shows page 1 out of 1 page. Unformatted text preview: *2—44. The magnitude of the resultant force acting on the bracket is to be 400 N. Determine the magnitude of F1 if (I) = 30°. SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of F1, F 2, and F3 can be written as (F1)x = F1 cos 30° = 0.8660F1 (my = F1 sin 30° = 0.51?1 (F2)x = 650G) = 390N (F2)y = 650 (g) = 520N (F3)x = 500 cos 45° = 35355 N (F3)y = 500 sin 45° = 35355 N Resultant Force: Summing the force components algebraically along the x and y axes, we have i 2(FR),, = EFX; (FR); = 0.8660F1 — 390 + 353.55 = 0.8660F1 — 36.45 +TE(FR)y = 2F” (FR)y = 0.51?1 + 520 — 353.55 = 0.5F1 + 166.45 Since the magnitude of the resultant force is FR = 400 N, we can write FR = V (FR)x2 + (FR)y2 400 = \/(0.8660F1 — 36.45)2 + (0.51631 + 166.45)2 F12 + 103.321?1 — 130967.17 = 0 Ans. Solving, F1 = 314N or F1 = —417N Ans. The negative sign indicates that F1 = 417 N must act in the opposite sense to that shown in the ﬁgure. ...
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