6-37, 6-53, 6-58

# 6-37, 6-53, 6-58 - 6—37 A structural member is loaded and...

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Unformatted text preview: 6—37 A structural member is loaded and supported as shown in Fig. P6-37. 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. P6-53. 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° 6-53 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° ...
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## 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.

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6-37, 6-53, 6-58 - 6—37 A structural member is loaded and...

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