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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 = (1501bft/rad) + (126011))[911 sin78.7°
0119‘ 12
= 1179.6 ftlb > o a = 78.7°, Stable 4
2
1—1: = (150 lbft/rad) + (1260 1b)[lg ft]sin323.8°
d0 12
: —470 ﬁlb < 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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 Spring '08
 Jenkins
 Statics

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