Engineering Mechanics Statics - Intructor Solutions manual 2.51 - 25L Determine the magnitude and direction measured counterclockwise from the positive

Engineering Mechanics Statics - Intructor Solutions manual 2.51

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Unformatted text preview: 2—5L Determine the magnitude and direction measured counterclockwise from the positive x axis of the resultant force of the three forces acting on the ring A. Take F1 = 500Nand6 = 20°. SOLUTION Scalar Notation: Summing the force components algebraically, we have 4 i) FRI = EFX; FRI = 500 sin 20° + 400 cos 30° — 600(5) = 37.42N —> +TFRy = my; FR, = 500 cos 20° + 400 sin 30° + 600(%) = 1029.8 N T The magnitude of the resultant force F R is FR = VFfix + Ffiy = \/37.422 + 1029.82 = 1030.5 N = 1.03 kN Ans. The direction angle 0 measured counterclockwise from positive x axis is FR 1029.8 : _l y = _1 — = o 0 tan FRX tan ( 37.42 > 87.9 Ans. ...
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  • Spring '16
  • DR.Khan
  • Statics, Force, resultant force, kN Ans, positive x axis, Scalar Notation, tan FRX tan

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