HW19-Solution

# HW19-Solution - HW19 Solution Terry Cooke Cal Poly San Luis...

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Unformatted text preview: HW19 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 8, Solution 39 FBD block B: (a) Since P 2.69 lb to initiate motion, equilibrium exists with P 0 (b) For Pmax , motion impends at both surfaces: Block B: Fy 0: N B 10 lb FAB cos 30 NB 10 lb s NB 0 3 FAB 2 0.3 N B 0 (1) Impending motion: Fx FB 0 : FB FAB FAB sin 30 2 FB 0.6 N B (2) 20.8166 lb Solving Eqs. (1) and (2): FBD block A: Then Block A: Fx 0: FAB sin 30 NB 10 lb 3 (0.6 N B ) 2 12.4900 lb FAB NA 0.6 N B 0 NA 1 FAB 2 1 (12.4900 lb) 2 6.2450 lb Impending motion: FA sNA 0.3(6.2450 lb) 1.8735 lb Fy 0 : FA P FAB cos 30 FA P 10 lb 0 3 FAB 10 lb 2 3 1.8735 lb (12.4900 lb) 10 lb 2 2.69 lb P 2.69 lb Chapter 8, Solution 41 Sense of impending motion: MB 0: 2W 3N A 4 sNA 0 MB 0: 1.5W 4 NC 3 s NC 0 NA 2W (3 4 s ) (1) (3) 0 Fx NC 1.5W (4 3 s ) (2) (4) 0 Fy : N AB Fx 0: Bx Bx W s N BA sNA Fy : N BC 0: Bx Bx NC s N BC W s NC s N BC s N BC NA s N BA NC NC NA (5) NA s N BA (6) Equate (5) and (6): Substitute from Eqs. (3) and (4): NA s (W 2 s) sNA) sW NC N C (1 s (W 2 s) s NC ) sW N A (1 Substitute from Eqs. (1) and (2): 2W 34 (1 s 2 s) sW 1.5W (1 4 3s 2 s 2 s 2 s) sW 2 34 s 1.5 43 s 1 Solve for s: s 0.0949 Chapter 8, Solution 45 FBD pin C: FAC FBC P sin 20 P cos 20 0.34202 P 0.93969 P Fy 0: N A W NA W FAC sin 30 0 W 0.171010 P or FBD block A: or For impending motion at A: Then or Fy NA FA s 0.34202 P sin 30 0 Fx 0: FA FA FA FAC cos 30 0.34202 P cos 30 sNA 0.29620 P :W 0.171010 P 0.29620 P 0.3 P 1.22500W 0: N B W NB W FBC cos 30 0 W 0.81380 P 0.93969 P cos 30 FB 0 Fx 0: FBC sin 30 FB 0.93969 P sin 30 0.46985 P FBD block B: For impending motion at B: Then or Thus, maximum P for equilibrium FB s NB NB FB s :W 0.81380 P 0.46985 P 0.3 P 1.32914W Pmax 1.225W ...
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## This note was uploaded on 11/07/2010 for the course ME 211 taught by Professor Staff during the Fall '05 term at Cal Poly.

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