HW_11 - PROBLEM 4.15 A follower ABCD is held against a...

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Unformatted text preview: PROBLEM 4.15 A follower ABCD is held against a circular cam by a stretched spring, which exerts a force of 21 N for the position shown. Knowing that the tension in rod BE is 14 N, determine (a) the force exerted on the roller at A, (b) the reaction at bearing C. SOLUTION Note: From f.b.d. of ABCD Ax = Acos60° = i 2 Ay = Asin60° = Afi 2 (a) From f.b.d. of ABCD +3 EMC = 0: (3(40 mm) — 21N(40 mm) +14N(20mm) =0 3A=%N or A = 28.0N 2d 60°4 (b) From f.b.d. ofABCD i. 217x = 0: C, +14N +(28N)cos60° = 0 -. CI = —28N or cx = 28.0N «- +1214} = 0: Cy — 21N + (28 N)sin60° = o Cy = —3.2487 N or Cy = 3.25 N 1 Then C = C; + C3 = (28)2 + (32487)2 = 28.188 N C _ and 0 = tan—1 —y =tan_1( 32487] = 6.6182° Cx —28 or C = 28.2 N 76.62° 4 PROBLEM 4.22 Determine the maximum tension which can be developed in cable AB if the maximum allowable value of the reaction at C is 1000 N. SOLUTION Cmax = 1000 N 2_2 2 c_q+g Cy = ./(1ooo)2 — c3 From f.b.d. of pedal 1.212: =0: Cx-Tmax =0 C, = Tmax +‘) 2MB = 0: Cy(0.4 m) — Tmax [(0.18 m)sin60°] = o Cy = 0.38971Tmax (3) Equating the expressions for Cy in Equations (1) and (3), with Cx = Tmax from Equation (2) (1000)2 — T5,“ = 0.389711Tmax T3“ = 868,150 :rmax = 931.75 N or Tmax = 932 N { PROBLEM 4.62 Eight identical 20 x 30-in. rectangilar plates, each weighing 50 lb, are D c - it held in a vertical plane as shown. All connections consist of frictionless A a 2 a 4 pins, rollers, or short links. For each case, answer the questions listed in Em. P6.1 The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions assuming that the magnitude of the force P is 100 N. Problem 4.61, and, wherever possible, compute the reactions. 8 z SOLUTION 1. Three non-concurrent, non-parallel reactions (a) Completely constrained 4 (b) Determinate 4 (c) Equilibrium 4 From f.b.d. of plate +‘) EMA = 0: c(30 in.) — 50 lb(15 in.) = o C = 25.0 lb 14 21. 2F, = 0: A, = 0 +1213 =0: Ay—501b+251b =0 Ay=251b A=25.01bi4 2. Three non-current, non-parallel reactions (a) Completely constrained 4 (b) Determinate 4 (c) Equilibrium 4 From f.b.d. of plate _+. 2F, = o: B = 0 4 +‘) 2MB = 0: (501b)(15 in.) — D(30 in.) = o n = 25.0 lb T4 +t 2;; = o: 25.01b—501b+C = o C=25.01bT4 PROBLEM 4.62 CONTINUED 3. Four non-concurrent, non-parallel reactions DX (a) Completely constrained 4 (b) Indeterminate 4 (C) Equilibrium 4 50 [5 From f.b.d. of plate +3 2M» = 0: Ax(20 in.) — (so 1b)(15 in.) Ax = 37.5 lb —+ 4 —+> ZFX =0: Dx+37.51b=0 Dx = 37.5 lb «— 4 C 4. Three concurrent reactions (a) Improperly constrained 4 D x (b) Indeterminate 4 (c) No equilibrium 4 50 lb 5. Two parallel reactions (0) Partial constraint 4 (b) Determinate 4 (c) Equilibrium 4 From f.b.d. of plate J) EMU = 0: C(30 in.) — (50 lb)(15 in.) = o C = 25.0 lb 74 +T2Fy =0: D—501b+251b=0 D = 25.0 lb f4 6. Three non-concurrent, non-parallel reactions (a) Completely constrained 4 (b) Deterrninate 4 (c) Equilibrium 4 From f.b.d. of plate +‘j 2MB = 0: 3(20 in.) — (50 lb)(15 in.) = 0 B = 37.5 lb m» 4 1.23:0: Dx+37.51b=0 Dx=37.51b—~ +l2Fy20: Dy—501b=0 Dy=50.01bl or D=62.51b b§ 53.1°4 PROBLEM 4.62 CONTINUED 7. Two parallel reactions (a) Improperly constrained 4 (b) Reactions determined by dynamics 4 (c) No equilibrium 4 8. Four non-concurrent, non—parallel reactions ((1) Completely constrained 4 (b) Indeterminate 4 (c) Equilibrium 4 From f.b.d. of plate +‘) 2MB = 0: 3(30 in.) — (50 1b)(15 in.) = 0 B = 25.0 lb 14 +1213 = o; Dy —501b+25.01b =0 1)y =25.01b l4 1.25:0: DX+C=0 ...
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This note was uploaded on 02/12/2010 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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HW_11 - PROBLEM 4.15 A follower ABCD is held against a...

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