This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 6—37 A structural member is
loaded and supported as
shown in Fig. P637. The member has a uniform
cross section and weighs
203 lb.
reactions at supports A and B. Determine the SOLUTION A free—body diagram of the beam is
shown at the right. The reaction at support A is represented by
force components Ix and Ky. The reaction at B is a vertical force 3. The weight 9 of the bean acts
through the center of gravity G of the beam and is directed toward
the center of the earth. Lid1 + dez + Lad3
6(3) + 4(6) + 3(7.5} Ax + 200  300 = 0 100 lb = Ay{2)  203(2.962) — 200(2)  IOOIT}  300(2) = 0 1158.0 lb = 1158 lb /(100)2 + (1153)2 = tan—1 1162 = tan 100 A = 100 i « 1153 J 1162.3 1b 3 1162 lb —35.03° a —35.1° lb = 1162 lb s 35.1° A = 8(2) — 203(4.962}  200(2) — 100(9)  300(2) = 0 B = 1466.0 lb = 1466 lb 3 = 1466 1b = 1466 lb T Determine the force
exerted by the cable at
B and the reaction at
support A of the bar
shown in Fig. P653. SOLUTION From a free—body diagram
for the bar: + C EMA = T cos 200(3)
+ T sin 20°13 tan 30°}
800(6) = 0 140?.02 lb 3 140? lb Ans. A  T sin 20° X A  1407.02 sin 20° = o X 431.2 lb 8 481 1b = AY + T cos 20°  800 = A? + 140?.02 cos 20°  300 = 0 522.1 lb 8 —522 lb {(481.2}2 + (—522.1)2 = T10.03 lb 3 710 lb _ 1 522.1 _ _ o _ 0
tan 481.2 — 4?.33 H 47.3 lb = 710 lb 3 47.3° 653 A wagon with a mass of 3500
kg is held in equilibrium on
an inclined surface by using
a cable as shown in Fig.
P6—58. Determine the force
in the cable and the forces
exerted on the wheels at A and B by the inclined surface. SOLUTION W = mg = 3500(9.80?) = 34.325 N From a free~body diagram
of the cart: + /’I 2Fx T cos 38°  W sin 300 T cos 33°  34.325 sin 30° 21,780 N 3 21.8 kN Ans. = 3(4) + 3 sin 30° {1.5)  3 cos 30° (2} — T cos 33° (3 tan 33°) — T sin 33° (5} 3(4) + 34,325 sin 30° {1.5)  34,325 cos 30° (2) — 21,730 cos 33° {3 tan 33°} — 21.230 sin 33° {5) = 0 35,245 N a 35.2 kN = 35.2 kN 3 60° A + B  w cos 30°  T sin 33°
A + 35,245 — 34.325 cos 30° — 21.730 sin 33° = o
7330 N a 7.33 kN ” = 7.39 RN 5 60° ...
View
Full
Document
This homework help was uploaded on 04/16/2008 for the course ENGR 1100 taught by Professor Anderson during the Fall '06 term at Rensselaer Polytechnic Institute.
 Fall '06
 Anderson

Click to edit the document details