Solutions of Vectors - Mechanical Engineering

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= 9cos β=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
FR= √226.668473

MECHANICAL ENGINEERING 3
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
Additional of system of a coplanar Force with cables
F1= 650N, F2= 625N, F3= 650N

MECHANICAL ENGINEERING 4
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
FRy= 373.578N
Part C