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A ’w PROBLEM 10.17 Derive an expression for the magnitude of the couple M required to
maintain the equilibrium of the linkage shown. SOLUTION Have
x5 : lsinﬂ
5x3 = 10051960
y“, = [0050
514 = —l sina966
Virtual Work: 5U = 0; M69 — P519; + P5y4 : 0
Mé‘ﬂ — P(lcos€56) + P(7lsin¢966) = 0 M = Pl(sin6 + 9056) 4 PROBLEM 10.20 A couple M of magnitude 75 lb'ft is applied as shown to the crank of
the engine system. Knowing that AB = 2.5 in. and BC : 10 in,
determine the force P required to maintain the equilibrium of the system when (a) 0 = 60°, (11) 9 = 120°. SOLUTION Bsin(0 + ¢) P From the analysis of Problem 10.19, M 2 A
cos¢ Now, with M = 751bft = 9001bin.
(a) For 6 : 60° sin¢ = Zlégsin 60°, ¢ = 12504" (900113411) = (2.5 MW
c0512.504° (P) or P 2 368.5 lb
P = 3691b > 4 (b) For 6 2 120° sin¢ = %sin120°, ¢ : 12.504° sin(120° + 12.504°)
c0512.504° (P) (900 lein.) = (2.5 in.) or P=476.7 1b
P : 4771b + 4 PROBLEM 10.29 Two rods AC and CE are connected by a pin at C and by a spring AE. The
constant of the spring is k, and the spring is unstretched when 0 : 30°.
For the loading shown, derive an equation in R63, 1, and k that must be satisﬁed when the system is in equilibrium. s rwmwmmwwTE P SOLUTION
yE = 10059 6yE = —lsin6l§t9 Unstretched length = 21 x = 2(ZIsin6l) = 4lsint9 5x = 4Zcos€50
F = k(x — 21)
F = k(4lsin€ — 21) Virtual Work:
6U = 0: Pb‘yE —F6x = 0 P(—lsin¢9§0) — k(4lsint9 — 21)(41c05050) = 0
—Psinl9 — 8k1(2sin0 —1)cos6 = 0
or i=(l—23in9)césg L=—;23ing‘
8k] sml9 8k] tam9 ...
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This note was uploaded on 05/18/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.
 Spring '08
 Jenkins
 Statics

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