HW18-Solution

HW18-Solution - HW18 - Solution Terry Cooke: Cal Poly San...

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Unformatted text preview: HW18 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 8, Solution 25 FBD window: T Fx 5 lb 0: N A NA ND ND 0 Impending motion: FA FD MD sNA s ND 0 : (18 in.)W (27 in.) NA (36 in.) FA sNA 0 W 10 lb W NA Fy 0 : FA W FA FD FD 3 NA 2 2 2W 3 4s T W FD T 0 W 2 ND ) Now FA s (NA 2 sNA Then or W 2 2 s 2W 34 s s 0.750 Chapter 8, Solution 30 (a) P 0 MD 0: N A (2 ft) (50 lb)(3 ft) NA 75 lb NA 75 lb 0 Fx 0: N D Fy 0 : FA FA FD FD 50 lb 50 lb sNA s ND 0 But: ( FA ) m (FD ) m 0.40(75 lb) 30 lb 0.40(75 lb) 30 lb Thus: and (b) P 20 lb MD ( FA )m ( FA ) m ( FD )m 60 lb FD 0 25 lb ( FD ) m . FA Plate is in equilibrium 0: N A (2 ft) (50 lb)(3 ft) (20 lb)(5 ft) NA Fx 0: N D NA 25 lb Fy 0 : FA FA FD FD 50 lb 20 lb 30 lb sNA s ND 0 But: ( FA ) m (FD ) m 0.4(25 lb) 10 lb 0.4(25 lb) 10 lb Thus: and ( FA )m ( FD )m FA 20 lb ( FD ) m FD . ( FA ) m Plate moves downward Chapter 8, Solution 32 FBD ABD: MD 0: (15 mm)N A (110 mm)FA FA 15 110 0 0.13636 0 : FA Dx 0 Dx FA A 0 ANA Impending motion: Then or A A 0.1364 Fx FBD pipe: Fy 0: NC NA NC 0 NA FBD DF: MF 0: (550 mm)FC FC (15 mm)N C C NC (500 mm) Dx 0 Impending motion: Then 550 C 15 500 FA NC But So NC 550 C NA and FA NA A 0.13636 C 15 500(0.13636) 0.1512 ...
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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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