Mechanical Engineering Statics Problems: Assignment Solution

This document presents a comprehensive solution to a mechanical engineering assignment, focusing on statics problems. It meticulously solves various fundamental problems, including calculations of moment vectors, resultant forces, and the application of Varignon's theorem. The solutions involve detailed step-by-step calculations, vector analysis, and equilibrium conditions, providing a clear understanding of the underlying principles. Furthermore, the document addresses problems related to force application points and the determination of resultant force locations. The assignment covers a range of topics within statics, making it a valuable resource for students studying mechanical engineering and related fields. References to relevant research papers are also included, enhancing the academic rigor of the solutions.

MECHANICAL ENGINEERING 1
MECHANICAL ENGINEERING
By Name
Course
Instructor
Institution
Location
Date

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
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