Engineering Mechanics Statics - Intructor Solutions manual 2.52 - *252 Determine the magnitude of force F so that the resultant FR of the three forces

# Engineering Mechanics Statics - Intructor Solutions manual 2.52

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This preview shows page 1 out of 1 page. Unformatted text preview: *2—52. Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible. What is the minimum magnitude of FR? 5kN SOLUTION Scalar Notation: Summing the force components algebraically, we have \$FRX = EFX; FRX = 5 — Fsin30° = 5 — 0.50F—&gt; +TFRy = My; FRy = Fcos 30° — 4 = 0.8660F — 4 T The magnitude of the resultant force FR is FR = W l = \/ (5 — 0.501?)2 + (0.8660F — 4)2 = m a) 1 na ..... r F2 = F2 — 11.93F + 41 ’ I :3 “ 9 I —X. dFR f 2F—=2F—11. 2 “ RdF 93 () I 22----..&quot; ‘n F d2FR+ﬂXdFR _1 (3) 1‘sz dF dF _ dFR F = 0. From Eq. (2) In order to obtain the minimum resultant force FR, ZFﬂ-ZF—1193—O RdF _ ‘ _ F = 5.964 kN = 5.96 kN Ans. Substituting F = 5.964 kN into Eq. (1), we have FR = V59642 — 11.93(5.964) + 41 = 2.330 kN = 2.33 kN Ans. dF Substituting F R = 2.330 kN with 51—; = 0 into Eq. (3), we have dZF [(2,330) de + o] = 1 dZFR dF2 = 0.429 &gt; 0 Hence, F = 5.96 kN is indeed producing a minimum resultant force. ...
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• Spring '16
• DR.Khan
• Statics, Force, resultant force FR, minimum resultant force

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