# Mechanics Fundamentals

User Generated

HFRE_ERZBIRQ_YRTNY_831

Engineering

## Description

I have in homework on Mechanics Fundamentals and I need help for this with it is asking for, no exert work.
thanks.

### Unformatted Attachment Preview

1. A particle is shot from the ground with an initial speed of 150 m/sec and an elevation angle of 60° with no acceleration in the z direction. (a) Formulate its trajectory, i.e. (horizontal displacement) and y (vertical displacement) vs. time history, for the entire flight period, assuming no air resistance and constant gravitational acceleration g = 9.8 m/s. Find its point of landing. (b) Find the eight alternative points of landing if you use a combination of the following values: 170 m/s and 130 m/s for the initial speed; 65 and 55° for the elevation angle and a constant lateral acceleration = +1 m/s and -1 m/s2 in the z direction. (c) What is the radius of a circle, with its center at the landing point found in part (a), that encloses all those alternative points of landing found part (b)? 2. A 8 lb rocket is launched from the ground with elevation angle of 85° with zero initial speed. The time history of its motor's thrust force is shown below. The thrust force stays at zero for t2 0.4 second. P, Iь 16 0.4 0.2 1, sec Assuming no air resistance, constant gravitational acceleration g = 32.2 fts? a kinetic friction coefficient k = 0.3 between the rocket and launch rail and a particle model for the rocket, what is the trajectory, or x (horizontal displacement) and y (vertical displacement) vs. time history of this rocket if it leaves the launch rail at the end of entire trust period i.e. t = 0.4 second. Use its position at t = 0 as the origin (x= 0 and y = 0). What is the length of the launch rail? 3. Redo problem #2 above if the launch rail length is set at 1 foot. You should try to draw free body diagrams and formulate the equations of motion for different phases of this problem first, assuming constant gravitational acceleration g = 32.2 ftus?, a particle model for the rocket and thrust in the direction of velocity, before you try to develop analytical solution for some of them and comment on the level of difficulty. Notes: This is an individual, not a team, assignment. Each student should submit his/her own solution The submittal deadline is the beginning of class on Tuesday, September 12. Presentation is worth 30% of the assignment points. Staple your submittal in the upper LH corner.
Purchase answer to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Here are the answers, please tell me if you have any problem.

1.
Initial velocity: 𝑣0 = 150 𝑚/𝑠𝑒𝑐
Elevation angle: 𝛼 = 60°
Initial height: ℎ = 0 𝑚
Gravitational acceleration: 𝑔 = 9.8 𝑚/𝑠 2
(a) The horizontal displacement: 𝑥 = 𝑣0 𝑡 cos 𝛼 = 150𝑡 cos 60° = 75𝑡
𝒙 = 𝟕𝟓𝒕
The vertical displacement:
1
1
𝑦 = ℎ + 𝑣0 𝑡 sin 𝛼 − 𝑔𝑡 2 = 0 + 150𝑡 sin 60° − × 9.8 × 𝑡 2
2
2
𝒚 = 𝟕𝟓√𝟑𝒕 − 𝟒. 𝟗𝒕𝟐
The particle landed when y =0
∴ 𝑦 = 75√3𝑡 − 4.9𝑡 2 = 0 ∴ 75√3 − 4.9𝑡 = 0 ∴ 𝑡 =

75√3
= 26.51 𝑠
4.9

Then 𝑥 = 75𝑡 = 75 × 26.51 = 1988.32 ≈ 1988 𝑚
Then, the point of landing: (𝑥, 𝑦) = (𝟏𝟗𝟖𝟖, 𝟎)
1

1

2

2

(b) The landing point is: (𝑥, 𝑦) = (𝑣0 𝑡 cos 𝛼 , 𝑣0 𝑡 sin 𝛼 − 𝑔𝑡 2 , 𝑣0𝑧 𝑡 + 𝑎𝑧 𝑡 2 )
Where: initial velocity in z-direction: 𝑣0𝑧 = 0 𝑚/𝑠
𝑎𝑧 is the lateral acceleration in z-direction.
So, the landing point when y =0 is:
1

1

2

2

(𝑥, 𝑦, 𝑧) = (𝑣0 𝑡 cos 𝛼 , 𝑣0 𝑡 sin 𝛼 − 𝑔𝑡 2 , − 𝑎𝑧 𝑡 2 )
1- 𝒗𝟎 = 𝟏𝟕𝟎 𝒎/𝒔 , 𝜶 = 𝟔𝟓°, 𝒂𝒛 = +𝟏 𝒎/𝒔𝟐 :
1
2 × 170 sin 65°
∴ 𝑦 = 170𝑡 sin 65° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 31.44 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (170 × 31.44 × cos 65° , 0, × 1 × (31.44)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟐𝟐𝟓𝟗, 𝟎, 𝟒𝟗𝟒)

Page | 1

2- 𝒗𝟎 = 𝟏𝟕𝟎 𝒎/𝒔 , 𝜶 = 𝟔𝟓°, 𝒂𝒛 = −𝟏 𝒎/𝒔𝟐 :
1
2 × 170 sin 65°
∴ 𝑦 = 170𝑡 sin 65° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 31.44 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (170 × 31.44 × cos 65° , 0, × (−1) × (31.44)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟐𝟐𝟓𝟗, 𝟎, − 𝟒𝟗𝟒)
3- 𝒗𝟎 = 𝟏𝟕𝟎 𝒎/𝒔 , 𝜶 = 𝟓𝟓°, 𝒂𝒛 = +𝟏 𝒎/𝒔𝟐 :
1
2 × 170 sin 55°
∴ 𝑦 = 170𝑡 sin 55° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 28.42 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (170 × 28.42 × cos 65° , 0, × 1 × (28.42)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟐𝟎𝟒𝟐, 𝟎, 𝟒𝟎𝟒)
4- 𝒗𝟎 = 𝟏𝟕𝟎 𝒎/𝒔 , 𝜶 = 𝟓𝟓°, 𝒂𝒛 = −𝟏 𝒎/𝒔𝟐 :
1
2 × 170 sin 55°
∴ 𝑦 = 170𝑡 sin 55° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 28.42 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (170 × 28.42 × cos 65° , 0, × (−1) × (28.42)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟐𝟎𝟒𝟐, 𝟎, −𝟒𝟎𝟒)
5- 𝒗𝟎 = 𝟏𝟑𝟎 𝒎/𝒔 , 𝜶 = 𝟔𝟓°, 𝒂𝒛 = +𝟏 𝒎/𝒔𝟐 :
1
2 × 130 sin 65°
∴ 𝑦 = 130𝑡 sin 65° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 24.04 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (130 × 24.04 × cos 65° , 0, × 1 × (24.04)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟏𝟑𝟐𝟏, 𝟎, 𝟐𝟖𝟗)
𝒎

6- 𝒗𝟎 = 𝟏𝟑𝟎 , 𝜶 = 𝟔𝟓°, 𝒂𝒛 = −𝟏 𝒎/𝒔𝟐 :
𝒔

1
2 × 130 sin 65°
∴ 𝑦 = 130𝑡 sin 65° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 24.04 𝑠
2
9.8
Page | 2

So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (130 × 24.04 × cos 65° , 0, × (−1) × (24.04)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟏𝟑𝟐𝟏, 𝟎, −𝟐𝟖𝟗)
7- 𝒗𝟎 = 𝟏𝟑𝟎 𝒎/𝒔 , 𝜶 = 𝟓𝟓°, 𝒂𝒛 = +𝟏 𝒎/𝒔𝟐 :
1
2 × 130 sin 55°
∴ 𝑦 = 130𝑡 sin 55° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 21.73 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (130 × 21.73 × cos 55° , 0, × 1 × (21.73)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟏𝟔𝟐𝟎, 𝟎, 𝟐𝟑𝟔)
8- 𝒗𝟎 = 𝟏𝟑𝟎 𝒎/𝒔 , 𝜶 = 𝟓𝟓°, 𝒂𝒛 = −𝟏 𝒎/𝒔𝟐 :
1
2 × 130 sin 55°
∴ 𝑦 = 130𝑡 sin 55° − × 9.8 × 𝑡 2 = 0 ∴ 𝑡 =
= 21.73 𝑠
2
9.8
So, the landing point is:
1
(𝑥, 𝑦, 𝑧) = (130 × 21.73 × cos 55° , 0, × (−1) × (21.73)2 )
2
(𝑥, 𝑦, 𝑧) = (𝟏𝟔𝟐𝟎, 𝟎, −𝟐𝟑𝟔)
(c) The landing point in part (a) is: (1988, 0,0)
The farthest points from the center point (1988, 0,0) are: (1321, 0, −289)
and (1321, 0, 289) and they have equal distance from the center.
The radius of the circle is the distance between the points (1988, 0,0)
and(1321, 0, 289):
𝑟 = √(1321 − 1988)2 + (0 − 0)2 + (289 − 0)2 = 𝟕𝟐𝟕 𝒎

Page | 3

2.

x
𝐹𝑡ℎ
𝐹𝑡ℎ
y
𝐹𝑁

𝑊 cos 𝜃

𝜇𝑘 𝐹𝑁
𝑊 sin 𝜃

𝑊

85°

𝐹𝑁 = 𝑊 cos 𝜃 = 8 cos 85° = 0.6972 lb
- For 𝟎 < 𝒕 ≤ 𝟎. 𝟐
𝐹𝑡ℎ − 𝜇𝑘 𝐹𝑁 − 𝑊 sin 𝜃 =
∴𝑎=

𝑊
8
𝑎 ∴ 16 − 0.3 × 0.6972 − 8 sin 85° =
𝑎
𝑔
32.2

32.2
× (16 − 0.3 × 0.6972 − 8 sin 85°) = 31.48 𝑓𝑡/𝑠 2
8

To get the displacement:
1
1
𝑑1 = 𝑣0 𝑡 + 𝑎𝑡 2 = 0 × 0.2 + × 31.48 × 𝑡 2 = 𝟏𝟓. 𝟕𝟒 𝒕𝟐 𝒇𝒕
2
2
At t= 0.2 s, 𝑑1 = 15.74 × 0.22 = 0.6296 𝑓𝑡
The horizontal displacement: 𝑥1 = 𝑑1 𝑐𝑜𝑠𝜃 = 15.74 𝑡 2 cos 85° = 𝟏. 𝟑𝟕 𝒕𝟐 𝒇𝒕
The vertical displacement: 𝑦1 = 𝑑1 sin 𝜃 = 15.74 𝑡 2 sin 85° = 𝟏𝟓. 𝟔𝟖 𝒕𝟐 𝒇𝒕
- To get the final velocity at the first interval which is equal to initial velocity of the
seco...

### Review

Anonymous
Really useful study material!

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4