HW20 - ENGINEERING MECHANICS u STATICS, 2nd. Ed. w. F....

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Unformatted text preview: ENGINEERING MECHANICS u STATICS, 2nd. Ed. w. F. RILEY AND L. D. STURGES 7»86 Determine all forces acting on member ABCD of the frame shown in Fig. P7—86. .5. w Lifleflmemww = I’50mm 50mm 50mm 50mm Fig. P7435 Ilfi’v'fjflx‘frv '. ‘ SOLUTION From a free—body diagram for the complete frame: + Q 2MB = 600(50) - M200) = 0 A = 150 N = 150 N T 3”! + we ZFX = D = 0 D = 0 X X + T EF = 150 u 600 + D = 0 Y Y D = 450 N Y ‘ 'fi = 450 N 1‘ Ans. c Member CE is a twowforce member; therefore, the line of action of force 6 is known as shown on the free—body diagram for member ABCD: H10 150 N + c 2MB 2 0 sin 45° (100) + 450(150) — 150(50) = c c a -848.5 N C e 849 N a 45° Ans. B — (-848.5) cos 45° z o X “600 N = 600 N 9* 3 = 600 N em ‘2‘? i ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES I 7—98 Determine all forces acting on member ACE of the frame shown in Fig. P7—98. The diameter of the pulley at E is 120 mm. The mass of body W is 100 kg. SOLUTION From a free—body diagram for the complete frame: w = mg = 100(9.807) 2 980.7 N + Q 2MB = Ax(0.210) — 980.7(0.795) = 0 A 3713 N = 3713 N “a X H B + A z Bx + 3713 2 0 X X X ~3713 N = 3713 N +- “I'- l M fit} I! 03 H — 980.7 = O B Y B = 980.7 N = 980.7 N T From a free-body diagram for the pulley: + T SF : E - 2(980.7) = 0 Y Y Ey = 1961.4 N 5 1961.4 N T From a free-body diagram for bar ACE: + q ENC : —Ay(0.375} — 1961.4(0.360) = 0 A “1882.9 N 3 1883 N ¢ ‘9 + —9 SF 3 3713 + C = 0 C x x x ll ~3713 N z 3713 N e“ H + T SF 2 A + C - 1961.4 Y 5’ Y 3844 N = 3844 N ? t! 3 ("1882.9) + Cy‘- 1961.4 = O C d“? 2.. ' ENGINEERING MECHANICS * STATICS, 2nd. Ed. w. F. RILEY AND L. D. STURGES 17~93 (Continued) A = ViAxiz + (Ay)2 = {(3713}3 + (—1882;2 m 4163 N a 4150 N a «1 "1883 _ _ 8A * tan M3713 «- 26.89 E = 4160 N g 26.90 Ans. c m /{cx}2 + (Cyiz = /(-3713}2 + (3344)? = 5344 N a 5340 N _ ~1 3344 m 0 ac - tan :ggfg - 134.01 6 m 5340 N a 46.0° Ans. E a 1961 N Jr Ans. 2'43 a ENGINEERING MECHANICS - STATICSs 2nd. Ed. ‘ W. 7*99* Determine all forces acting on member ACD of the frame shown in Fig. P7w99. The diameter of the pulley at D is 8 in. The weight of body W is 500 lb. SOLUTION From a free-body diagram for the pulley at D: +4213“) —500cos45°=0 X D = 353.6 lb 3 353.6-lb m4 X . o w +szymny—5oo—5oo $11145 —0 500)}; By = 853.6 lb = 853.6 lb T From a free-body diagram for bar BC: + 9 EMS = 500 sin 45" (10.343 1 + Gyms) C —228.8 lb 2 228.6 lb ¢ Y From a free—body diagram for bar ACD: + C EMA = Cx{16) + 228.6(16) + 353.6(28) " 853.6(28) = 0 646.4 lb 3 646.4 lb +— + l M ’5?! II A - C - 353.6 X X X A “ 645.4 ~ 353.6 3 0 X A z 1000 1b = 1000 lb *6 I! + T 2FY 2 Av + 228.6 - 853.6 2 O A? = 625 1b a 625 115 T 9145"” NEINEERING MECHANICS “ STATICS, 2nd. Ed; W. F. RILEY AND L. D. STURGES §fi+99 (Continued) / 2 2 2 ,2 ,, - A = y{93) T {A ) z /(1009) + (925; 9 1199.2 lb u 11:9 lb . _ -3 625 _ 0 8A ~ tan T656 « 32.00 K m 1179 1b a 32.0° Ans. 2 2 2 2 c = (0x) + (Cy) = (646.4) + (228.6) = 685.6 lb 9 686 lb 9C a tanwl _§ig‘i a 160.520 C m 686 1b 9 19.480 Ans. 2 2 2. 2 9 m /(DX} + (Dy) z /(~353.6) + (~853.6) a 923.9 lb 9 924 lb 9D : tanwi :ggg'g = —112.50° fi = 924 lb 9 57.50 Ans. 6‘95 ...
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This note was uploaded on 07/05/2010 for the course ENGR 120 taught by Professor Jones during the Spring '10 term at Florida State College.

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HW20 - ENGINEERING MECHANICS u STATICS, 2nd. Ed. w. F....

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