Solutions to the Fundamental Problems-Mechanical Engineering

MECHANI
CAL ENGINEERING 1
MECHANICAL ENGINEERING
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MECHANI
CAL ENGINEERING 2
Fundamental Problem 4.23 - Enhanced - with Hints and Feedback
The coordinate of point 0, ( x0,y0,z0) = (0,0,0) ft
The coordinate of point A, ( xA, yA,ZA) = ( -2,2.3.5) ft
The coordinate of B, ( xB,yB,zB) = ( 1.5, -2,0) ft
The unit vector eoA is
eoA = ( xA−x 0 ) i+¿ ¿
eoA= (−2−0 ) i+¿ ¿
eoA= −2i+2 j+3.5 k
√20 .25
eoA= 1.5i−2 j+ ok
4 .5
eoA= -0.444i+0.444j+0.777k
The unit vector eoB is
eoB = ( xA−x 0 ) i+¿ ¿
eoB= ( 1.5−0 ) i+¿ ¿
eoB= 1.5i−2 j+ ok
√ 6.25
eoB= 1.5i−2 j+ ok
2.5
eoB= 0.6i-0.8j
The momemnt vector ( Mc)1 is
(Mc)1= -0.444i+0.444j+0.777k

MECHANI
CAL ENGINEERING 3
The magnitude (Mc)1=540 (-0.444i+0.444j+0.777k )
The magnitude (Mc)1= -239.7i + 239.7j+ 419.58k.
Moment vector (Mc)2 is
= 300 (0.6i-0.8j)
= 180i -240j ( lb. ft)
Moment vector (Mc)3 is
(Mc)3 = - (Mc)3k
(Mc)3 =-300k (lb.ft)
The negative sign is because of which the ( Mc) 3 acting in negative z direction
The resultant coule moment acting on pipe is
(Mc)R= ( Mc)1+(Mc)2+ ( Mc) 3
( Mc)R = (-239.7i + 239.7j+ 419.58k) + (180i -240j) -300k
( Mc)R = (59.7i-0.3j+119.58k ) l.b ft
Fundamental Problem 4.40
Part A
From the diagram below