Solutions of Vectors - Mechanical Engineering

Only Chapter 2a 10 items

MECHANICAL ENGINEERING 1
MECHANICAL ENGINEERING
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MECHANICAL ENGINEERING 2
Cartesian Vectors
Given that F1= 9kN, F2= 9kN, α =6 60, β=3 50 , γ =11 40, θ=3 00 .
Now from the figure, the angle by force 2 with z axis is given by;
γ ´=90−¿
Now, we will find the components of two forces separately for F1;
F1x= F1cos α=9 cos 66
Fix= 3.66kN
F1y= 9cosβ=9 cos 35=7.372
F1z= F1cos γ=9 cos 114=−3.66 kN
For F2
The components F2 in xy is given by.
F2xy= F2sin γ ´=9sin 53.13 00 =7.2 kN
Then,
Fx= F2xy cos θ=7.2cos 35=5 .8978 kN
F (2y) = F (2xy) = sin θ=7.2sin 35=4.1297 kN
Fz= F2cos γ ´=9 cos 53.1300=5.4 kN
Part A
FRx = F1x+F2x = 3.66kN+5.8978 kN = 9.5578kN
Part B
FRy = F1y+F2y= 7.372+4.1297= 11.50171kN
Part C
FR2= F12+F22= -3.66+5.4 = -1.74 kN
Part D
FR= √ FR x2 +FR y2 +FR z2
FR= √ 9.5 72 +11.501712+ 1.742

MECHANICAL ENGINEERING 3
FR= √226.668473
FR= 15.055kN
Part E
Angle of resultant force with x-axis
θ ´ ´=co s−1
( FRx
FR )=co s−1
( 9.5578
15.055 )
θ ´ ´=co s−10.634837
θ ´ ´=5 0.5920
Part F
β ´ ´=co s−1
( FRy
FR )=co s−1
(11.50171
15.055 )
β ´ ´=co s−1 0.763979
β ´ ´=40.1837190
Part G
The angle of resultant force with z is
γ ´ ´=co s−1
{ FRz
FR }= { 1.74
15.055 }=co s−1 0.115576
γ ´ ´=83.36 30

MECHANICAL ENGINEERING 4
Additional of system of a coplanar Force with cables
F1= 650N, F2= 625N, F3= 650N
Let the angle made by F1 with the x axis be 0
Then sin θ=3
5 and cos θ= 4
5
Now,
F1= F1cos i – F1 sin θj
F1= (650 × 4
5 −650 × 3
5 ) N
F1= ( 520 i−390 j ) N
F2= ( 625 cos 30 j+625 sin 30 j ) N
F2= (312i+541.265j)N
F3= (-650 cos 20 i + 650j)
F3= (-610.8 i + 222.313j) N
Part A
Horizontal components force
(FResultant)x = ( 520+312-610.8)N
(FResultant)x = 221.2 N
Part B
Vertical component of resultant force
(FResultant)y= (-390 + 541.265+222.313)N

MECHANICAL ENGINEERING 5
FRy= 373.578N
Part C
Magnitude of resultant force;
FR=√ FR x2 +FR y2
FR=√ 221.22 +373.57 82
FR=434.154 N
Part D
Now, tan θ= FRy
FRx
tan θ= 373
221.2
tan θ=1.686
θ=59.326 90
Position Vectors
Part A
Given that ,
xA = 2.8ft , zA= 2.2 ft, xB= 2.7 ft, yB=1.4 and zB = 2.0ft
The location of A is given by ( 2.8, 0.00, 2.20) ft
Position of A is given by A (2.80i+ 0.00j+ 2.20k)
The vector of A is given by A ( 2.80i+ 2.20k)ft
Part B

MECHANICAL ENGINEERING 6
The vector of point A, rA = 2.8i+2.2k
The vector of point B , rB = ( -2.7+1.4-2.2)ft
rAB = rB-rA = ( -2.7i+1.4j-2k)-( 2.8i+2.2k)ft
rAB = (-5.5i+1.4j-4.2k)
Part C
rA = 2.8i +0.0j+2.2 k
rB= (-5.5i+1.4j-4.2k)
rA-rB = 8.3i+1.4j-2k)
u= 8.31i+1.4 j−2 k
√75.016
u= 8.31i+1.4 j−2 k
8.661
u= 0.958i+0.161644j-0.23k
Vector Dot Product
Part A
Part B

MECHANICAL ENGINEERING 7
Part C
Part D
Part E
A.B.C

MECHANICAL ENGINEERING 8
Part F
Let v1 be (i+j+k)
V1.V1 = (i+j+k) (i+j+k)
V.V1= i2+j2+k2
Vector Addition: Geometry and Components
Part A
Place A´s tail at B´s tip , C´s tail at A´s tail and C´s tip at B´s tip
Part B
Using geometric notation, we have
C2+A2+B2-2AB cos C
Where C = making angle with the opposite side C
And we know that angle θ is an exterior angle which is a supplement of angle C, then we will have
Part C
From cosine laws with ∅ we have
Sin ∅
B = ( Sin c/C = (Sin a/A )
Sin ∅ =¿ ( Sin c /C ) B
∅ = sin -1 [sin π−θ ¿ B/C]
Part D

MECHANICAL ENGINEERING 9
To manipulate these vectors using vector components, we must first choose a coordinate system.
In this casechoosing means specifying the angle of the x-axis. They axis must be perpendicular
to this and by convention isoriented radians counterclockwise from the x-axis.
There is only one unique coordinate system in which vector components can be added.
Statement is FALSE
Video Solution Problem - Addition of Coplanar Force Systems
Vector along P = 40 cos 60i+40 sin 60 j
Vector along F = 60 cos 135i+ 60 sin 135j
Resultant force = P+F
= 40 cos 60i+40 sin 60 j + 60 cos 135j+60 Sin 135j
= ( 40 × 1
2 + 60× −1
√2 ¿ i+ j¿
R
→
=¿ = ( 20-30 √2+ j ¿+30 √ 2)
Magnitude= √ ¿ ¿
Magnitude = 80.3 lb
Angle = tan-1 ( ( 20 √ 3+ 30 √ 2
20−30 √ 2 )

MECHANICAL ENGINEERING 10
Angle = 106.20
Interactive Figure: Vector Addition of Forces
PART A
Ropes F1, F2 and F3 because all the ropes combine the forces at point 0
Boat Statics
Part A
The equilibrium force condition provides the total of all forces on object is zero when it is not in
acceleration. And this can be expressed as below
∑ F=0
is sum of all the forces on the object.
The components of a force are as follows:
Fx= Fsin θ
Fx is the x components of the force , and Fy is the component of force along y . Thus F is the magnitude
of force and θ is the angle between the x axis and force is in anti-clockwise. The sign convention is as
follow;
Treat for all the forces in upwards or towards the right as positive and all forces in downwards or
towards the left as negative. Drawing the boat free diagram and rope illustrating of the forces FAB, FAD
and FAF are the forces acting on the boat due to the rope which is running to A from B, to A from D from
F. The component of the force FAB along x and y components are FABx and FABy. The force component
FAD along y and x and FAFy.

MECHANICAL ENGINEERING 11
Use the x component equation of a force and then substitute FAD for F, and FADx for Fx in the below
equation.
For the above free diagram, a boat owner pulls her boat into the dock there are six bollards to which to
tie the boat. Owner has three ropes and can tie the boat from the boat's center A to any of the bollards.
One rope runs to A from B, another to A from D, and a final rope from A to F.
Employ the equation
∑ F=0
The 3 forces acts on the boat whilst it is immobile, FABx and FADx towards the left to the negative and FAF
towards right so positive. Substitute in the above equation.
The magnitude of force provided by the third rope in terms of θ 2 F AB cos θ

MECHANICAL ENGINEERING 12
The force magnitude given by the 3rd rope in terms of θ is 2 F AB cos θ
The 3 forces act on the boat whilst it is immobile FABx and FADx towards left so negative and FAF towards
right so positive. Thus, the resultant force acting on the boat along the horizontal direction is
The equilibrium force condition provided that total forces on the boat on an object is 0 as it is not
accelerating ( it is 0). Since the boat is not accelerating in the horizontal direction. The net force in the
horizontal direction is 0. That is
∑ F=0
F AF−F ABx=0
The magnitude of force given by the third rope of θ is 2 F AB cos θ
Video Solution Problem - Force Resultant in 3-D
X components of F1, Fx=5 cos 60 = 2.5 KN
y component of of F1 ,
fy= 5 Cos45 = 3.54 kN
z component of F1
F2= 5cos 60 = 2.5kN
F1= Fxi+Fyj+Fzk
= ( 2.5i+3.54j+2.5k) kN
F= -2j kN
Option A is correct
Cartesian Vector
Given that F= 135N, θ=21∧∅ =69
Part A
Fxy = F sin ∅ = 135 × sin69=¿ ¿126.033
Part B
Fx=Fxy cos θ= ( 135 sin69 ) cos 21
Fx= 117.66 N

MECHANICAL ENGINEERING 13
Part C
Fy –Fxy sinθ=¿
Fy= -45.166 N
Part D
Fz = F cos ∅ =135 cos 69=48.379 N
Part E
α =co s−1( Fx
F ¿=co s−1( 117.66
135 )
α =29.3601
Part F
β=co s−1
( Fy
F ) =co s−1
( −45.166
130 )
β=110.33
Part G
γ=co s−1
( Fz
F )=co s−1
( 78.24
130 )
γ=52.997

MECHANICAL ENGINEERING 14
References
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from kinematics data and inertial properties of the body by using inverse dynamics technique.
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locally aggregated tensors. In 2011 18th IEEE International Conference on Image Processing
(pp. 669-672). IEEE.
Pollard, J.P., Porter, W.L. and Redfern, M.S., 2011. Forces and moments on the knee during
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Scheinerman, E.R. and Tucker, K., 2010. Modeling graphs using dot product representations.
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