Solutions to the Fundamental Problems-Mechanical Engineering

Only chapter 4b, 8 items

MECHANICAL ENGINEERING 1
MECHANICAL ENGINEERING
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MECHANICAL 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
The magnitude (Mc)1=540 (-0.444i+0.444j+0.777k )

MECHANICAL ENGINEERING 3
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

MECHANICAL ENGINEERING 4
Obtain the resultant force
FR= {150 lb
ft ×6 ft }+ {1
2 ×50 lb
ft ×6 ft }+600 lb
FR= 1650 lb
Part B
From the Varignon´s theorem , we can obtain the location of the resultant force from point A
FR (d)= {150 lb
ft ×6 ft }(3 ft )+ {1
2 × 50 lb
ft ×6 ft }(2
3 ×6 ft)+(600 lb)(9 ft)
FR(d) =8700 lb-ft
d= 8700
FR = 8700
1650
d= 5.27
Fundamental Problem 4.3
Calculate the distance between the point of application of the force and point o
r= 4+( 3 cos 450-1)
r= (4+1.121)ft
r= 5.121 ft
Obtain the moment of force about point O
Mo= Fr
Mo= 670×5.121

MECHANICAL ENGINEERING 5
Mo= 3431.07 lb . ft
Mo= 3.43 kip.ft
Fundamental Problem 4.29
Part A
(FR)0 = F1+F2
= ¿
=(−280 i+180 j−230 k )
=(−280 , 180 ,230 k )
Part B
(Mo) R = M1+M2
M1 = F1×roe
M1= | i j k
1.5 2 1
−280 180 220|
= i(440-180-j( -330+280) + k(-270 +560)
= ( 260i +50j+290k )
M2 = |i j k
0 2 1
0 0 −450|= -900i
(Mo)R= ( 260i +50j+290k ) -900i
(Mo)R = -640i +50j+290k
(Mo)R = (-640, 50, 290)
Fundamental Problem 4.25

MECHANICAL ENGINEERING 6
FRx = 230-110( 3
5 ¿=164 lb
FRy= 170-110 ¿)= 82 lb
Part A
FR=√¿ ¿
FR=√¿ ¿
FR= 183.357 lb
Part B
Tan θ= FRy
FRx
θ=tan−1( FRy
FRx )
θ=tan−1( 82
164 )
θ=26.56 00
Part C
∑ MA
∑ MA = 110( 3
5 ¿ ( 4 )−110 ( 4
5 ) ( 6 )+ 170(3)
∑ MA = 246lb.ft
Fundamental Problem 4.37
Part B
RA+RB = ( 6×4.5)+( 9×9)+(3×4.5)
RA+RB= 121.5 = ∑ F

MECHANICAL ENGINEERING 7
∑ M =0; ∑ F ×d=27× 0+81 ×6.75+13.5 ×13.5
∑ F ×d=729
d= 729
121.5 =6
The distance between the A and resultant line of action
d=6-2.25
d= 3.75 m
Fundamental Problem 4.33
∑ Fx=0
Rx+F2 ( 4
5 ¿= 0
Rx+16 ( 4
5 ¿=0
Rx= -12.8 kN
Apply equilibrium condition for forces along y direction
∑ Fy=0
Ry+R2( 3
5 )=F 1
Ry+16( 3
5 )=20
Ry= 10.4 kN
Calculate the magnitude of resultant force
R ¿ √ Rx2 + Ry2
R ¿ √ (−12.8)2 +10.42
R=16.49
Part B

MECHANICAL ENGINEERING 8
θ=ta n−1
( Ry
Rx )
θ=tan−1
( 10.4
−12.8 )
θ=−0.8125
θ=−39.093
θ=180−39.093
θ=140.91
Part C
Apply the equilibrium condition for moments about point A.
∑ MA =0
F2 ( 3
5 ) ( 6 ) +R ¿
16 × ( 3
5 ) ( 6 ) +16.49 ¿
9. 6+10.397d- 25.6-40=0
10.39d= 16
d= 1.539
Fundamental Problem 4.31
∑ F ∝=0
∑ ❑ Fy=?

MECHANICAL ENGINEERING 9
R= √ ¿ ¿
R= √ ¿ ¿
R=∑ Fy
R=600+300+600
R= 1500lb
Reference
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