Engineering Mechanics Statics - Intructor Solutions manual 2.45 - 245 If the resultant force acting on the bracket is to be directed along the positive

# Engineering Mechanics Statics - Intructor Solutions manual 2.45

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This preview shows page 1 out of 1 page. Unformatted text preview: 2—45. If the resultant force acting on the bracket is to be directed along the positive u axis, and the magnitude of F1 is required to be minimum, determine the magnitudes of the resultant force and F1. 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¢ (F1)y=Flsin¢ (F2)x = 650G) = 390N (F2), = 650 (g) = 520N (F3), = 500 cos 45° = 353.55 N (F3), = 500 sin 45° = 353.55 N (FR)x = FR cos 45° = 0.7071FR (FR)y = FR sin 45° = 0.7071FR Resultant Force: Summing the force components algebraically along the x and y axes, we have i, 2(FR)X = 213,; 0.7071FR = F1 cos ¢ — 390 + 353.55 (1) +12%)y = 213,; 0.7071FR = F1 sin ¢ + 520 — 353.55 (2) Eliminating F R from Eqs. (1) and (2), yields 202.89 F = — 3 1 cos¢ — sin¢ ( ) The first derivative of Eq. (3) is dF1 sin (1) + cos (1) = — 4 dd) (cos d) — sin ¢)2 ( ) The second derivative of Eq. (3) is W 2 ' + 2 1 = (smqﬁ cos qb) + 1 (5) d¢2 (cos (I) — sin (1))3 cos d) — sin ¢ For F1 to be minimum, —1 = 0. Thus, from Eq. (4) 01¢ sin (15 + cos ()5 = 0 tan (b = —1 ¢ = —45° Substituting 45 = —45° into Eq. (5), yields d2F1 2 = 0.7071 &gt; 0 d¢ This shows that 45 = —45° indeed produces minimum F1.Thus, from Eq. (3) (F3)? 202.89 C“) F1 = 143.47 N = 143 N Ans. = cos (—45°) — sin (—45°) Substituting ()5 = —45° and F1 = 143.47 N into either Eq. (1) or Eq. (2), yields FR = 919 N Ans. ...
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