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Unformatted text preview: ENGINEERING MECHANLCS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 488 A plate is loaded with a
system of forces as shown
in Fig. P4h88. Exprsss
the resultant of the force
system in Cartesian vector form. 150 N 150 N
Fig. P488 ‘ SOLUTION An examination of Fig. P4ﬁ88 indicates that the force system consists of
a system of three couples in a plane. A scalar analysis yields: With counterclockwise moments positive: For forces A and B: M1 = FAd1 = 250(0.150) = 37.50 N'm For forces C and D: M2 = ”chz : ~350(0.150) = 52.50 N°m For forces E and F: FEdB = 150(0.320) % 40.00 N'm +M2+M3
37.50 » 52.50 + 48.00 = 33.00 Nm = 33.0 N'm
E 2 33.0 Nm 5 = 33.0 E Nm 226 ENGINEERING MECHANICS _ STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4w91 Three couples are applied
to a bent bar as shown in
Fig. P491. Determine the magnitude of the resultant .‘u couple C and the direction
angles associated with the resultant couple vector. ﬁg. P491 SOLUTION The scalar components of the resultant couple C are: Cx = ~80(3) = —240 ft‘lb
CY = ~?5(4) = "300 ft'lb
C = ~100(8) = "800 ftlb 2 The resultant couple C eXpressed in Cartesian vector form is: C = —240 i  300 j u 800 E c a (~240)2 + (—300)2 + (—800)2 = 887.5 ft’lb E 887 ft'lb Ans.
1 Ex ~1 —z40 o 9X = cos 6“ = cos §§?T§ = 105.7 Ans.
—1 Cy —1 —300 0 9y 2 cos 6— =cos 887.5 = 109.8 Ans.
1 CZ 1 “300 o 92 = cos 6 2 cos §§?T§  154.3 Ans. 12'2‘? D. STURGES ENGINEERING MECHANKCS  STATICS, 2nd. Ed. W. F. RILEY AND L. 4107 'Replace the SOD—lb force
shown in Fig. P4~107 by
a force at point A and a
couple. Express your ahswer in Cartesian vector form. Fig. P41 07 SOLUTION
F A F : 300(cos 30° 3 + sin 30° 3)
= 259.8 f + 150.0 3 1b e 200 i + 150 0 3 1b F = (20 — 10 tan 30°) 3 —10 j in. = 14.223 3 , 10 3 in. A F x F = (14.220 1 — 10 3) x (250.3 3 + 150.0 3)
= 4732 E in.°lb e 4.73 E in.'kip 4108 Replace the 450*N
force shown in Fig.
P4—108 by a force
at point A and a
couple.’ Express
your answer in
Cartesian veetor n " ﬂMmm“
form. ' Fig. 24300
SOLUTION
FA = ﬁ = 450(cos 58° I * sin 58
= 233.5 E  331.0 3 N a 239 i — 382 3 F = (0.200 + 0.400 cos 329) i + 0.400 sin 32° 3
= 0.3392 3 + 0.2120 3 m EA = F x F a (0.5392 3 + 0.2120 3) x (230.5 3  331.0 3) = —250.3 E Nm a —250 ﬁ_Nm ...
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 Spring '10
 Jones
 Statics

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