University of Portsmouth Engineering Analysis Coursework 3

This document presents a solved coursework assignment for Engineering Analysis, specifically for the School of Civil Engineering and Surveying at the University of Portsmouth. The assignment covers three main sections: vectors, differentiation and integration, and applications. Section A includes problems on vector operations, such as finding the angle between vectors and resultant forces. Section B focuses on differentiation and integration, involving the application of these concepts to solve related problems. Section C applies these mathematical principles to real-world scenarios, including calculations involving moments of force and area determination. The solution provides detailed steps and calculations for each question, demonstrating the application of core engineering analysis principles. The document is a valuable resource for students studying engineering analysis, offering worked solutions to aid in understanding and problem-solving.

Section A
Question 1⃗
a=3i+j ; ⃗ b=2i+j+3k⃗
a .⃗ b= |a||b|cosƟ
7= √ 10∗ √ 14 cosƟ
Ɵ=cos−1
( 7
√ 140 )
=cos−1 ( .591607 )
=53.72880
Approximately = 540
Question 2
F1=15N in direction j
F2=XY N in the direction of i+2j+k
Where XY=86
F1=15j ; F2=86(i +2 j + k)
F1+F2=86 i+187 j+86 k
Magnitude of this resultant= √ 862 +1872 +862
=223.071N
That is approximately = 223.1N
Question 3⃗
a=3i+j⃗
b=2i+j+3k
Vector perpendicular to these two vectors is the cross product of the two vectors.
That is we have to find ⃗ a ×⃗ b,

=⃗a ×⃗ b=
|i j k
3 1 0
2 1 3|
=3i-9j+k
Unit vector = ⃗ a×⃗ b
|⃗a×⃗ b|
= 1
√91(3i-9j+k)
Section B
Question 4
W= sin 2t
cos 3 t
= dw
dt = 2cos 3tcos 2 t+3 sin 2 tsin 3t
( cos 3 t)2
Where t =86
= dw
dt at t=86
=.078217
Question 5
= y= ( x2 −5 )4
= dy
dx =4 (x2−5)3 2 x
d2 y
dx2 =48 x2 ¿
Question 6
∫
0
π
sin 3 t + 3
2 √ t dt

=−cos 3 t
3 +3 √ t , 0 , π
=−cos 3 π
3 +3 √ π− (−1
3 +0 )
= 2
3 +3 √ π
Section C
Question 7
F=200i-100j-200k
Position vector ⃗ r=2i+j+k
Moment of force is the cross product of the two vectors.
That is ⃗ r ×⃗ F
=
| i j k
2 1 1
200 −100 −200|
=-100i+600j-400k
Magnitude= √ 1002 +6002 +4002
=728.01Nm
Question 8
: y=x4
Area of the channel = ∫
a
b
x4dx
81=x4
X=3, -3
=∫
−3
3
x4 dx
=97.2 cm2

There for required area =81*6-97.2
=388.8 cm2
Question 9
y= x
XY ∗105 ( 7∗104−103 x2 +3 x4 ); XY=86
= x
86∗105 ( 7∗104 −103 x2 +3 x4 )……………………………………………………………1)
Putting y=0
x
86∗105 ( 7∗104 −103 x2 +3 x4 ) =0
7∗104 −103 x2 +3 x4=0
Let x2=t
So the given equation becomes,
3t2−10 t+7∗104=0
t=1.667
That is x2=1.667
x=1.2909m
Putting this value of x in equation 1)
We get y= 1.2909
86∗105 ( 7∗104 −103∗1.667 +3(1.667)4 )
=0.0102m
There fore the deflection is 0.0102m.