COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 131.
Free-Body Diagram:
Express tension, weight in terms of rectangular components:
uuur
EF = ( 300 mm ) i + (1350 mm ) j − ( 700 mm ) k
uuur
EF
=T
T=T
EF
=
300 i + 1350 j − 700 k
( 300 )2 + (1350 )2 + ( − 700 )2
6
27
14
Ti +
T j− Tk
31
31
31
(
)
W = − ( mg ) j = − ( 7 kg ) 9.81 m/s 2 j = − ( 68.67 N ) j
ΣM B = 0:
M B + − ( 375 mm ) i + ( 350 mm ) k × ( − 68.7 N ) j
+ ( −100 mm ) i + ( 700 mm ) k × T = 0
or
( M By j + M Bzk ) + ( 375 mm )( 68.7 N ) k + (350 mm )( 68.7 N ) i
i
j k
T
+ −100 0 700 ( mm ) = 0
31
6 27 −14
continued COSMOS: Complete Online Solutions Manual Organization System
Setting the coefficients of the unit vector i equal to zero:
(a)
i:
−
27
T ( 700 mm ) − ( 68.67 N )( 350 mm ) = 0
31
T = 39.422 N
(b)
ΣFx = 0:
Bx +
or T = 39.4 N
6
( 39.422 N ) = 0
31
Bx = − 7.6301 N
ΣFy = 0:
By − 68.67 N +
27
( 39.422 N ) = 0
31
By = 34.335 N
ΣFz = 0:
Bz −
14
( 39.422 N ) = 0
31
Bz = 17.8035 N
B = − ( 7.63 N ) i + ( 34.3 N ) j + (17.80 N ) k
Using the moment equation again and setting the coefficients of the unit vectors j and k to zero:
ΣM B ( y − axis) = 0:
14
M By − ( 39.422 N ) (100 mm ) +
31
6
31 ( 39.422 N ) ( 700 mm ) = 0
M By = − 3.5607 N ⋅ m
ΣM B ( z − axis) = 0:
27
M Bz + ( 68.67 N )( 375 mm ) − ( 39.422 N ) (100 mm ) = 0
31
M Bz = − 22.318 N ⋅ m,
Therefore:
M B = − ( 3.56 N ⋅ m ) j − ( 22.3 N ⋅ m ) k
COSMOS Chapter 4 Solution 131
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