Sol Cap 02 - Edicion 8

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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
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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.