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Sol Cap 02 - Edicion 8

COSMOS: Complete Online

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 100. uuur Cable AB: AB =−( 4m) − ( 20m) + ( 5m) i j k ( ) ( ) ( ) 2 2 2 AB = − 4 m + − 20 m + 5 m = 21m uuur T AB TAB AB = TAB = ⎡−( 4m) − ( 20m) + ( 5m) ⎤ AB 21 m ⎣ i j k ⎦ uuur AC = 12 m i − 20 m j + 3.6 m k Cable AC: ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 AC = 12 m + − 20 m + 3.6 m = 23.6 m uuur T AC 1770 N ( 12 m ) ( 20 m ) ( 3.6 m AC = TAC = ⎡ − + ) ⎤ AC 23.6 m ⎣ i j k ⎦ ( 900 N) ( 1500 N) ( 270 N) = i − j + k uuur i j k Cable AD: AD =−( 4 m) − ( 20 m) + ( 14.8 m) Now... ( ) ( ) ( ) 2 2 2 AD = − 4 m + − 20 m + 14.8 m = 25.2 m uuur T AD TAD AD = TAD = ⎡−( 4 m) − ( 20 m) + ( 14.8 m) ⎤ AD 25.2 m ⎣ i j k ⎦ T AD ( 10 m) ( 50 m) ( 37 m) = ⎡− − − ⎤ 63 m ⎣ i j k ⎦ R = T + T + T and R = R; R = R = 0 AB AC AD j x z 4 10 Σ Fx = 0: − TAB + 900 − TAD = 0 (1) 21 63 5 37 Σ Fy = 0: TAB + 270 − TAD = 0 (2) 21 63 Solving equations (1) and (2) simultaneously yields: T = 1.775 kN ! AD T = 3.25 kN ! AB Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System Chapter 2, Solution 101. 2 2 ( ) ( ) 2 2 ( ) ( ) ( ) ( ) ( ) TAB ⎡ ( 450 mm) ( 600 mm) ⎤ d = − 450 mm + 600 mm = 750 mm AB d = 600 mm + − 320 mm = 680 mm AC 2 2 2 d = 500 mm + 600 mm + 360 mm = 860 mm AD T AB = − + 750 mm ⎣ i j ⎦ T = − 0.6i + 0.8 j T AB ( ) AB TAC ⎡( 600 mm) ( 320 mm) T AC = 680 mm ⎣ j − k ⎤⎦ T ⎛15 8 ⎞ AC = ⎜ − TAC ⎝ ⎟ 17 17 ⎠ T TAD AD = ⎡( 500 mm) + ( 600 mm) + ( 360 mm) ⎤ 860 mm ⎣ ⎦ T ⎛25 30 18 ⎞ AD = ⎜ + + TAD ⎝ ⎟ 43 43 43 ⎠ W =−W j At point A: Σ F = 0: T + T + T + W = 0 i component: k component: j component: From Equation (1): From Equation (3): AB AC AD 25 − 0.6TAB + TAD = 0 43 or T AB = ⎛5⎞⎛25⎞ ⎜ T ⎝ ⎟⎜ ⎟ 3⎠⎝43⎠ 18 18 − TAC + TAD = 0 17 43 17 18 or TAC = ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟TAD (2) ⎝ 8 ⎠⎝43⎠ 15 30 0.8TAB + TAC + TAD − W = 0 17 43 15 ⎛17 18 ⎞ 30 0.8TAB + ⎜ ⋅ TAD + TAD − W = 0 17 ⎝ ⎟ 8 43 ⎠ 43 255 0.8TAB + TAD − W = 0 (3) 172 ⎛5⎞⎛25⎞ 6kN= ⎜ TAD ⎝ ⎟⎜ ⎟ 3 ⎠⎝ 43 ⎠ or T AD = 6.1920 kN AD 255 0.8( 6 kN) + ( 6.1920 kN) − W = 0 172 (1) ∴ W = 13.98 kN ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

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