HW15-Solution

HW15-Solution - HW15 - Solution Terry Cooke: Cal Poly San...

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Unformatted text preview: HW15 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 6, Solution 76 We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along the BD. Free body: ABC Attaching FBD at D and resolving it into components, we write MC 0: (400 N)(135 mm) 450 FBD (240 mm) 510 FBD 0 255 N FBD 255 N C Fx 0: C x 240 ( 255 N) 510 Cx 0 120.0 N Cx 120.0 N Fy 0: C y 400 N 450 ( 255 N) 510 0 Cy 625 N Cy 625 N Chapter 6, Solution 93 Free-body: Pipe 2 W (30 lb/ft)(7.5 ft) 225 lb F 8 D 17 225 lb 15 F 120 lb D 255 lb Geometry of pipe 2 By symmetry: Equate horizontal distance: r r CF 4.5 in. CD (1) 8 r 17 25 r 17 CD CD CD 15 17 15 17 25 5 r r 15 3 From Eq. (1): CF CF 5 5 r (4.5 in.) 3 3 7.5 in. Free-body: Member CFG MC 0: (120 lb)(7.5 in.) Gx (16 in.) Gx 0 56.25 lb Gx 56.3 lb Free body: Frame and pipes PROBLEM 6.93 (Continued) Note: Pipe 2 is similar to pipe 1. AE CF 7.5 in. E F 120 lb MA 0: G y (15 in.) (56.25 lb)(24 in.) (225 lb)(4.5 in.) (225 lb)(19.5 in.) (120 lb)(7.5 in.) Gy 0 510 lb Gy 510 lb Fx 0 : Ax 120 lb 56.25 lb Ax 0 176.25 lb Ax 176.3 lb Fy 0 : Ay 510 lb 225 lb 225 lb Ay 0 60 lb Ay 60.0 lb Chapter 6, Solution 95 (a) Free body: Trailer (We shall denote by A, B, C the reaction at one wheel) MA 0: (2400 lb)(2 ft) D(11 ft) 0 D Fy 0 : 2 A 2400 lb 436.36 lb 436.36 lb 0 A 982 lb A 981.82 lb Free body: Truck MB 0: (436.36 lb)(3 ft) (2900 lb)(5 ft) 2C (9 ft) 0 C Fy 732.83 lb 0 C 733 lb 0 : 2 B 436.36 lb 2900 lb 2(732.83 lb) B 935.35 lb B 935 lb (b) Additional load on truck wheels Use free body diagram of truck without 2900 lb. MB 0: (436.36 lb)(3 ft) 2C (9 ft) 0 C Fy 0 : 2 B 436.36 lb 2(72.73 lb) B 72.73 lb 0 290.9 lb C 72.7 lb B 291 lb ...
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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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