MZB125 Introductory Engineering Mathematics - Rocket Problems

This assignment solution covers several engineering mathematics problems. The first question involves calculating the distance of a rocket from the origin and the angle of its trajectory. The second question focuses on torque calculations related to a force and weight acting on a point. It derives expressions for torque and determines conditions for negative resultant torque. The third question uses Gaussian elimination to solve a system of linear equations related to cost analysis and determines the most cost-effective batch size. Finally, the fourth question explores the properties of a hypocycloid, deriving expressions for velocity and acceleration, and determining the relationship between acceleration and distance. The document is available on Desklib, a platform offering a range of study tools and solved assignments for students.

Running head: ENGINEERING MATHEMATICS 1
Engineering Mathematics
Name
Institution

ENGINEERING MATHEMATICS 2
Question 1
Part a
a= [ 3
5
10 ], b= [ 7
11
30 ] ,c =
[10
24
55 ]⃗
OP= [ 3
5
10 ]+ [ 7
11
30 ]+ [10
24
55 ]= [20
40
95 ]
Distance of the rocket from origin ¿ √ 202 +402 +952= √ 11025=105
Part b
tanθ= Z−component
Radial component = 55
√102 +242 =2.11538
θ=tan−1 2.11538=64.699°
Part c
cosθ=⃗ a .⃗ c⃗
¿ a|.|⃗c∨¿ ¿
cosθ= ( 3 ~i+5 ~j+10 ~k ) . ( 10 ~i+22 ~j+55 ~k )
¿ ( 3 ~
i+ 5 ~
j+10 ~
k )∨.∨ ( 10 ~
i+ 22 ~
j+55 ~
k )∨¿= 3 ( 10 ) +5 ( 24 ) +10(55)
( √ 32 +52 +102 ) . ¿¿ ¿
¿ 700
√ 134 ( √ 3701 ) =0.994
θ=cos−1 0.994=6.28 °

ENGINEERING MATHEMATICS 3
Question 2
Part a⃗
AP=⃗ AO +⃗ OP but ⃗ OP=r ~j
OC=r ,OB=rcos ∅ , BC =h
BC=OC −OB=r −rcos ∅=r ( 1−cos ∅ ) =h
h
r = ( 1−cos ∅ )
cos ∅=1− h
r …(equation1)
sin2 ∅ =1− ( 1− h
r ) 2
= h
r ( 2− h
r )
sin ∅ = √ h
r ( 2− h
r ) …(equation2)
So, ⃗ AO=⃗ AB+⃗ BO=−rsin ∅ ~
i+rcos ∅ ~
j …(equation 3)
∴⃗ AP=−rsin ∅ ~i+rcos ∅ ~j+ r ~j=−rsin ∅ ~i+r (1+ cos ∅ ) ~j

ENGINEERING MATHEMATICS 4
Substituting cos ∅ and sin ∅ obtained in equations 1 and 2 into ⃗ AP we obtain:⃗
AP=−r √ h
r (2− h
r ) ~i+r (1+1−h
r ) ~j⃗
AP=− √hr (2− h
r ) ~
i+r (2−h
r ) ~
j⃗
AP=− √2 hr −h2 ~
i+r (2− h
r ) ~
j
Part b
Force ⃗ F=Fcos θ ~i−Fsinθ ~j
Torque by force ⃗ F about point A¿⃗ AP×⃗ F⃗
AP ×⃗ F=τ⃗ F / A =
| ~
i ~
j ~
k
− √2 hr −h2 r (2− h
r ) 0
Fcosθ −Fsinθ 0
|
¿ ~
i ( 0 ) − ~
j ( 0 ) + ~
k ( Fsinθ √ 2 hr−h2−Fr ( 2− h
r ) cosθ )
τ⃗ F / A= ( Fsinθ √ 2hr −h2−F ( 2r −h ) cosθ ) ~k
Part c
Torque of ⃗ W about point A is⃗
W =−mg ~
j
Then, from equation 3

ENGINEERING MATHEMATICS 5⃗
AO=r (−sin ∅ ~i+cos ∅ ~j )⃗
AO ×⃗ W =τ⃗ W / A =
| ~
i ~
j ~
k
−rsin∅ r cos ∅ 0
0 −mg 0
|
¿ ~
i ( 0 ) − ~
j ( 0 ) + ~
k (mgrsin ∅ )
¿ mgrsin ∅ ~
k
From equation 2, mgrsin ∅ ~
k =mgr √ h
r ( 2− h
r ) ~
k
τ⃗ W / A =mg √ 2 hr−h2 ~
k
Part d
The resultant torque about point A ¿ τ⃗ F / A +τ⃗ W / A
¿ ( Fsinθ √2 hr −h2−F ( 2 r −h ) cosθ ) ~k +mg √2 hr−h2 ~k
Resultant Torque= ( ( Fsinθ+ mg ) √ 2 hr−h2−F ( 2 r−h ) cosθ ) ~
k
The equation has two z components. That is, ( Fsinθ+mg ) √ 2 hr−h2 and F ( 2 r−h ) cosθ
For the resultant torque to be negative, the latter component should be larger than the first
component. That is,
F ( 2 r−h ) cosθ> ( Fsinθ +mg ) √ 2 hr−h2
Part e

ENGINEERING MATHEMATICS 6
Cage radius , r=3 m h=1 m, θ= π
6 , g= 9.81m
s2 , F=500 N
We have:
F ( 2 r−h ) cosθ> ( Fsinθ +mg ) √ 2 hr−h2
Substituting the given variables into the inequality we get:
500 ( 6−1 ) cos ( π
6 )>
(500 sin ( π
6 )+ 9.81m ) √2 ×3 ×1−12
500 ( 5 ) √3
2 > ( 500(0.5)+9.81 m ) √5
1250 √3>¿
Dividing both sides by √5 we obtain
968.2458>250+9.81 m
968.2458−250>9.81 m
718.2458>9.81 m
73.2157>m
m<73.2157 kg
The heaviest mass that can be pulled¿ 73.2157 kg
Question 3
2 x+2 y + 4 z=19000 …(1)

ENGINEERING MATHEMATICS 7
2 x+ y +5 z=21000 … (2)
2 x+3 y + 4 z=20000 …(3)
We solve the equations using Gaussian elimination as follows:
Subtracting equation 1 from equation 2 we obtain
( 2 x+ y +5 z )− (2 x +2 y+4 z )=21000−19000
− y +z=2000 …(4)
Then, subtracting equation 3 from equation 2 we get
( 2 x+ y +5 z )− (2 x +2 y+ 4 z )=21000−20000
−2 y + z=1000 …(5)
Subtracting equation 5 from equation 4 we obtain
(− y + z )− (−2 y + z )=2000−1000
y=1000
Substituting y=1000 into equation 5 we get
−2(1000)+ z=1000
z=1000+2000=3000
Substituting y=1000 and y=1000 into equation 1 we get
2 x+2 y + 4 z=19000

ENGINEERING MATHEMATICS 8
2 x=19000−2 y−4 z
2 x=19000−2 ( 1000 )−4 ( 3000 ) =19000−2000−12000=5000
x= 5000
2 =2500
Therefore, x=2500 , y=1000 , z=3000
Part b(i)
Batch
size
Total Cost of Chemicals Total Cost of Preservative Total cost of
production
1 19000 0 19000
5 =19000 ×5 ×75 %=71250 2 ×10000=20000 71250+2000=91250
10 =19000 ×10 ×70 %=133000 5 ×10000=50000 133000+50000=183000
Part b(ii)
Batch 1total cost=19000
¿ batch ¿ cost per batch= 91250
5 =18250
¿ batch¿ cost per batch=183000
10 =18300
The most cost effective batch size is 5 since it has the lowest unit cost.
Question 4

ENGINEERING MATHEMATICS 9
Part a
r ( t ) =( ( R−r ) cost +rcos ( R−r
r t )) , ( ( R−r ) sint −rsin ( R−r
r t ) )
When R=2r ,the position of the hypocycloid will be:
r ( t )=
( ( 2r −r ) cost +rcos ( 2 r−r
r t ) ), ( ( 2r −r ) sint −rsin ( 2 r−r
r t ) )
r ( t )= ( rcost+ rcost ) , ( rsint−rsint )
r ( t )=( 2rcost ,0)
r ( t ) has an x-component only implying that the hypocycloid is a horizontal segment.
Part b
Given that r ( t ) =
( ( R−r ) cost +rcos ( R−r
r t )) , ( ( R−r ) sint −rsin ( R−r
r t ) )
Velocity=r' ( t )
¿ ( − ( R−r ) sint−r ( R−r
r ) sin ( R−r
r t ) ) , ( ( R−r ) cost−r ( R−r
r ) cos ( R−r
r t ))
¿ ( − ( R−r ) sint−( R−r) sin ( R−r
r t ) ) , ( ( R−r ) cost−( R−r )cos ( R−r
r t ) )
Acceleration=r' ' ( t )
¿ (− ( R−r ) cost −( R−r)2
r cos ( R−r
r t ) ), (− ( R−r ) sin t + (R−r)2
r sin ( R−r
r t ))

ENGINEERING MATHEMATICS 10
Part c
When R=2r ,
acceleration=r' ' ( t )
(− ( 2 r −r ) cost−(2 r−r)2
r cos ( 2 r−r
r t )) , ( − ( 2r −r ) sint + (2r −r )2
r sin ( 2 r−r
r t ))
¿ ( −r cost−r cos t ) , ( −r sint +rsint )
¿ (−2 rcost , 0 )
r'' ( t )
r (t) =−2 rcost
2 rcost =−1
Hence, the acceleration is proportion to the distance OP