HW34+ - PROBLEM 10.74 In Problem 10.40 determine whether...

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Unformatted text preview: PROBLEM 10.74 In Problem 10.40, determine whether each of the positions of equilibrium is stable, unstable, of neutral. (See the hint for Problem 10.73.) SOLUTION Equilibrium For 01‘ Solving numerically. 01” Stability At 6 = 78.7°: At 0 z 3218": At 19 = 379.1°: Potential Energy V : émz — stine d—V = K19 — P100519 110 ‘H: = K + PlsinB d0“ 11: = 0: 0050 = £6 1119 PI P =1260lb, 1 =10 111., andK = 1501b‘f‘t/rad c050: 1501b~fifrad 0 (12601b)(g a] c0549 = 2 7 9 = 1.37333 rad, 5.652 rad, and 6.616 rad 0 = 78.7°, 323.8", 379.1° 2 ‘1 V = (1501b-ft/rad) + (126011))[911 sin78.7° 0119‘ 12 = 1179.6 ft-lb > o a = 78.7°, Stable 4 2 1—1: = (150 lb-ft/rad) + (1260 1b)[lg ft]sin323.8° d0 12 : —470 fi-lb < o .9 = 324°, Unstable 4 W 10 61—192 = (1501b~ft/rad) + (1260 110(5 11151113791 2 493.5 ft~lb > 0 6' 2 379°, Stable { PROBLEM 10.75 45” mm/\ Angle H is equal to 45° after a block of mass m is hung from member AB “fig as shown. Neglecting the mass of AB and knowing that the spring is unstretched when 19 = 20°, determine the value of m and state whether the equilibrium is stable, unstable, or neutral. SOLUTION- Potential Energy .3 13 Have V = ghgp + ”'33 7: where xSP=r(t9—00), r=100mm, 190=ZO°=§rad yB = Lwcosfi, LAB = 450 mm Then V = % kr2 (6 — 602 + mgLAB c050 and fl = krz (19 — 00) — mgLAB sine d0 dZV W = krZ — mgLABcosfi With k = 800 N/m, 0 : 45° Equilibrium: ‘11 = 0: (800 N/m)(0.1m)2 1 — 5 — m(9.81m/sz)(0.45 m)sinl = 0 d0 4 9 4 Then m=1.11825kg m=1.118kg< Stability sz 72' Now = (800 N/m)(0.1m)2 — (1.118 kg)(9‘81m/s2)(0.45 m)cosZ 6192 : 4.51] > 0 Stable 4 PROBLEM 10.92 Two bars are attached to a single spn'ng of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown. SOLUTION P H ' Spring: 5 = £sin¢ = Esme $— 3 3 EM :7 F For small values of 4i and 0: ¢ = 26 — L 2L 1 g V =P —cos¢+—cos6’ +—ks2 3 3 2 24. -4559 3 a _ PL 2 7(00529 + 20056) + ék£g3£8inflj dV — E(—25in2l9 — Zsinfi) + gkLZsinficosfl fi‘3 = —‘:—L(2sin20 + 25in6) + gkLzsinZH 2 (1—1: = —E(400520 + 2cosl9) + 11an2 00526 d0 3 9 2 When 6:0: d—Z=—fl+ikfi d6 3 9 ' 2 For stability: ”1—1: > o: —2PL + ikL2 > 0 d6 9 OSP<%kL4 ...
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