mechanics assignment friction

BACHELOR OF ENGINEERING IN MECHATRONICS

Assignment 3 Dynamics

Student name: Barry Kearney

Supervisor: Jim Duffy

Project title: Lab 3 projectiles / Tractive Force

date submitted: 16/11/2013

Introduction: Part A

This lab was based on projectile motion and it was to prove the theory that was covered in lecture 5 to be correct. When dealing with projectile motion, it is the theory that when an object has been fired from its starting point into the air, it will come under the influence of gravity and is attracted to ground with an acceleration of g m/s squared. In the lab a projectile launcher was used to project two steel balls, one in the horizontal direction and one in the vertical direction. The ball that was launched in the vertical direction was ball 1 and the ball that was launched in the horizontal direction was ball 2. The purpose of this experiment was to investigate projectile motion through the use of a vertical acceleration apparatus which shows the independence of vertical acceleration from the horizontal velocity.

Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown obliquely near the earth 's surface, and it moves along a curved path under the action of gravity only. The path followed by a projectile motion called its trajectory. Projectile motion only occurs when there is one force applied at the beginning of the trajectory, after which there is no force in operation apart from gravity.

Introduction: Part B

Part B of the lab was on Tractive Forces. Tractive force means the force available at the contact between the drive wheel tyres and road is known as 'tractive effort‘ or ‘tractive force’. As used in mechanical engineering the term tractive force can either refer to the total traction a vehicle exerts on a surface, or the amount of the total traction that is parallel to the direction of motion. The published tractive force value for any vehicle may be theoretical—that is, calculated from known or implied mechanical properties—or obtained via testing under controlled conditions. The example that was taken in the lab was of a train of 3 parts that were coupled together by couples (T1) and (T2). The purpose of this lab was to prove the theory covered in lecture 6 was correct and to see the relationship between force, mass and acceleration in tractive forces which comes from Newton’s 2nd law. We know that force = mass x acceleration and we also were giving the conditions to which the train was under.

Table 1, Part A: recorded and calculated data
Measured time and distance for the vertical ball and the horizontal ball projected from projectile launcher.

Test
1st Ball (vertical)
2nd Ball (horizontal)

Time of flight (t)
(s)
Distance (s)
(m)
Time of flight (t)
(s)
Distance (s)
(m)

1
0.5
0.93
0.6
1.38

2
0.4
0.93
0.51
1.46

3
0.43
0.93
0.56
1.36

4
0.35
0.93
0.57
1.34

5
0.60
0.93
0.68
1.39

6
0.35
0.93
0.50
1.40

7
0.40
0.93
0.54
1.45

8
0.28
0.93
0.50
1.31

9
0.30
0.93
0.47
1.46

10
0.30
0.93
0.47
1.32
Average values

0.391
0.93
0.54
1.387

Table 2, Part A:
Calculated Horizontal velocity, acceleration due to gravity, the % difference in the value of gravity, and the Vertical velocity.

Horizontal velocity (Vh) (calculated)
2.57 m/s
Acceleration due to gravity, g (calculated)
6.38 m/s squared
% difference in the value of g
-34.96%
Vertical striking velocity (Vv) (calculated)
3.83 m/s

Calculations Part A:

(Horizontal velocity) S = V x T therefore V= S = 1.39 = 2.57 m/s T 0.54

(Acceleration due to gravity) Sv = UvT – 1 g(t)squared 2
Therefore

g = 2 x (Sv – UvT) = 2 x (0.93 – (0 x 0.54)) = 2 (0.93) = 1.86 = 6.378 = 6.38 m/s squared T squared 0.54 squared 0.2916 0.2916

(% difference in the value of g)

% difference = Calculated – g x 100 = 6.38 – 9.81 x 100 = - 34.96 % G 9.81

(Vertical striking velocity)

V = U + G x T

V = 0 + (9.81)(0.391)
V = 0 + 3.83571
V = 3.83 m/s

Discussion part A

In this lab that was completed it was shown that the theory behind projectile motion is correct. It was proven that both balls came under the influence of gravity once they left the projectile launcher and that they were both attracted to ground. The two balls were launched from the same vertical height but the ball number 2 that was travelling in the horizontal direction travelled a further distance than ball number 1 in the vertical direction. Even though ball number 2 travelled a further distance the two balls will hit the ground at the same time as they both come under the same force of gravity however this was not shown in our table 1 (Fig 1) because their was human errors such as, two people starting the stop watches at different times, the person pressing the trigger mechanism was releasing the balls faster sometimes than other times even though we would start the stop watches on the count of 3. The other factors that had to be taken into consideration is, if the projectile launcher was at any sort of an angle due to the work bench not been balanced or level or an even surface. However the readings that were taken were still very close to each other so this gave us a good idea to what figures that we would be dealing with. In this experiment the initial velocity of each ball was 0 m/s. To calculate the acceleration due to gravity we manipulated the equation to find (g) gravity.

When dealing with projectiles, we use the same equations as linear motion but the (a) for acceleration is replaced or substituted with (g) for gravity. The acceleration due to gravity was 6.38 m/s squared. In theory this acceleration should have been 9.81 m/s squared but due to the human errors that occurred during the experiments there was a difference of -3.43 m/s squared these % errors came from miscalculating of the time taken for the balls to hit the ground and the distance travelled by the horizontal ball. When the steel ball number 2 is projected from the projectile launcher in the horizontal direction, the time it takes for the steel ball to hit the ground is independent of its initial horizontal velocity, the steel ball will continue to move in the horizontal direction with the same horizontal velocity in which it was projected from the projectile launcher with because there is no acceleration so it stays at a constant velocity. The distance that the steel ball number 2 travels in the horizontal distance before it hits the ground is dependent on the time of flight and the horizontal velocity that it was projected with.

Projectile motion only occurs when there is one force applied at the beginning of the trajectory, after which there is no force in operation apart from gravity, this was proven in the experiment as ball number 1 was let fall from a height with no other force applied and ball number two was projected with a horizontal velocity from the projectile launcher and both balls were attracted to ground as they came under the influence of gravity.

We found the value of acceleration using the average vertical height in which the ball was projected from and used the average horizontal time in which it took ball number two to hit the ground as ball number two was projected with an horizontal velocity it still should hit the ground at the same time as ball number one does as there both under the same force of gravity. If our measurements and calculations were 100% we should have got an acceleration of 9.81 m/s squared.

The horizontal component of the velocity of the object remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration due to gravity is constant. It is important to note that the Range and the Maximum height of the Projectile do not depend upon mass of the projected body. The Range and Max Height are equal for all those bodies which are thrown by same velocity and direction. Air resistance does not affect displacement of a projectile; this is why we do not take the mass of the balls into consideration or the mass of any objects when dealing with projectiles. This experiment proves and supports the theory behind projectile motion to be correct.

Conclusion part A
We do not take the mass of the balls or bodies into consideration when dealing with projectile motion as the air resistance does not affect the displacement of the projectile.
The range and height are equal for all bodies which are thrown by the same velocity and direction.
There was a small difference in calculating the acceleration due to gravity, this was because of the different readings and human errors that took place during the experiment.
Both balls come under the influence of the same gravity and are attracted to ground and should hit the ground at the same time.
In theory both balls should hit the ground at the same time, but because there were two people using stopwatches to record the times taking there was going to be a difference in the readings and calculation.
The horizontal distance ball number two travels before it hits the ground is dependent of the time of flight and the horizontal velocity of projection.
Ball number two will travel at the same horizontal velocity because there is no acceleration or any other force applied.
The vertical component of the velocity will increases linearly because the acceleration due to gravity is constant, so it picks up speed as it is falling from a height.
The experiment has supported and proved the theory to be correct.

Table 3: Part B
Calculated data for deceleration and the force in each couple of the train (T1 and T2)

Deceleration
A (m/s2)
Force in coupling 1
T1 (N)
[Tension or compression
Force in coupling 2
T2 (N)
[Tension or Compression]
All brakes function normally
-4.409
0
0
The brakes of car A fail
-2.776
-32654.77 Compression
16334.13 Tension
The brakes of car B fail
-2.776
16334.135 Tension
16334.135 Tension
The brakes of car C fail
-2.776
16334.135 Tension
-32654.77 Compression

Calculations for Table 3

10Mg = 10,000Kg

Tan = 5 = 0.05 The Tan inverse of .05 = 2.86* degrees 100

To find the force of each car individually coming down the slope

Mass x g x Sin2.86* = 10000 x 9.81 x Sin 2.86* = 4894.77 N

To find the resistance force which is equal to 0.5 times the normal force we find the normal for of each car individually and multiply the answer by 0.5.

Mass x g x Cos 2.86*= 10,000 x 9.81 x Cos 2.86*= 97977.81N

Braking force = 97977.81 x 0.5 = 48988.905 N Braking force = Resistance force

(1) –R + MxgxSin2.86 = MAx

Because all the brakes function normally we can take the 3 cars and treat them as 1, so we multiply the mass by 3 and multiply the resistance by 3.

-146966.715 + 14684.30 = 30,000 Ax therefore

Ax = -146966.715 + 14684.30 = - 4.409 m/s squared (Deceleration) 30,000

Total Force = 30,000 x -4.409 = -132270 N

Also Fa + Fb + Fc = Total Force

10,000 x -4.409 + 10,000 x -4.409 + 10,000 x -4.409 = -132270N so there will be no forces in T1 or T2

(2) Because only two cars have brakes we multiply the resistance by 2 instead of 3 this time. 2 x 48988.905 = 97977.81 N

-97977.81 + 14684.30 = 30,000 Ax therefore

Ax = -97977.81 + 14684.30 = - 2.776 m/s squared (Deceleration) 30,000

T1+ MxgxSin2.86* = MAx therefore T1 + 4894.77 = 10,000 x -2.776

T1 + 4894.77 = -27760

T1 = -27760 – 4894.77 = -32654.77 N

T1 = -32654.77 N (Compression)

T2 – R + MxgxSin2.86* = MAx therefore T2 – 48988.9 + 4894.77 = 10,000 x -2.776

T2 -44094.13 = - 27760

T2 = -27760 + 44094.13

T2 = 16334.13 N (Tension)

(3) Ax = -97977.81 + 14684.30 = - 2.776 m/s squared (Deceleration) 30,000

T1 – R + MxgxSin2.86* = MAx

T1 – 48988.905 + 4894.77 = 10,000 x -2.776

T1 – 44094.135 = - 27760

T1 = - 27760 + 44094.135

T1 = 16334.135 N (Tension)

T2 – R + MxgxSin2.86* = MAx

T2 – 48988.905 + 4894.77 = 10,000 x – 2.776

T2 – 44094.135 = -27760

T2 = -27760 + 44094.135

T2 = 16334.135 N (Tension)

(4) Ax = -97977.81 + 14684.30 = - 2.776 m/s squared (Deceleration) 30,000

T1 – R + MxgxSin2.86* = MAx

T1 – 48988.905 + 4894.77 = 10,000 x -2.776

T1 – 44094.135 = -27760

T1 = -27760 + 44094.135

T1 = 16334.135 N (Tension)

T2+ MxgxSin2.86* = MAx

T2 + 4894.77 = 10,000 x -2.776

T2 + 4894.77 = -27760

T2= -27760 – 4894.77

T2 = - 32654.77 N (Compression)

Discussion part B

In this part of the lab the experiment was on tractive forces and it was shown that the theory behind lecture 6 is correct. We were given a tractive force question and to work out the deceleration of the train and the force in each of its couplings (T1 & T2). The way to work this out was by Newton’s second law. The only way of solving this equation was understanding Newton’s second law and applying the theory. Newton’s second law details the relationship between force, the mass, and the acceleration. The acceleration of an object is in the direction of the force. If you push or pull an object in a particular direction, it accelerates in that direction. The acceleration has a magnitude directly proportional to the magnitude of the force. If you push twice as hard (and no other forces are present), the acceleration is twice as big. The magnitude of the acceleration is inversely proportional to the mass of the object. That is, the larger the mass, the smaller the acceleration for a given force (which is just as you’d expect from inertia). The mass is a quantitative measure of inertia of a body which is its resistance to rate of change of velocity. For a body with high inertia, the acceleration will be small. For a body with small inertia, the acceleration will be large.[1] The equation for the force is F= Mass x Acceleration, If you rearrange the force equation to solve for acceleration, you can see that if the size of the force doubles, then so does the size of the acceleration (if you push twice as hard, the object accelerates twice as much), and if the mass doubles, then the acceleration halves (if the mass is twice as big, it accelerates half as much).

If a car tried to drive up a slope it will have resistance acting against motion up the slope. Mass x Gravity x Cos* would gives us a friction force of the weight of the car on its tyres against the surface that the car was trying to drive up, and we would also have the resolved body weight of the car pulling it back down the slope. Resolved body weight would = Mass x Gravity x Sin *. When dealing with tractive force motion up a slope and down a slope has its advantages and disadvantages. When driving a truck up a slope the resolved weight of the truck will act against the motion of the truck and will pull the truck back down the slope. If the truck had no hand brake or brakes it would need a traction force at its wheels supplied by the engine just to hold the truck in its same position without moving otherwise the resolved body weight of the truck would pull it down the slope. If the truck was driving down a slope the resolved body weight would act with the motion of the truck and acceleration would not be even required. If the truck had no brakes the resolved body weight would act as a force and push the truck down the slope. The truck would have no resistance to the motion down the slope.

The tractive force required to propel a road vehicle or a train can be taken as the equivalent to the pull in a tow-rope or coupling. A tractive force is an antiquated term for force pulling something. In this experiment this was proved as the force in the couplings T1 and T2 were equal to the force in the cars. In part of the experiment when all 3 cars of the train’s brakes worked each car decelerated at the same time as they all had the same braking force so they all slowed down together at the same rate and came to a stop together therefore there was no force in the couplings T1 or T2. It can be seen from the theory that no force should be in the couplings when they all braked together and all the brakes functioned normally and each car had the same braking force. It was also proven mathematically as well so this proves the theory to be correct. The next part of the lab was when the front car of the train which was car A, its brakes did not work. This meant that car B and car C would have to slow down the 3 cars as they are coupled together but would only have the braking force of two of the cars. When car A’s brakes did not work it took the train longer to decelerate to a stop as it decelerated at -2.776 m /s squared where as in part 1 it decelerated at -4.409 m/s squared so it took the train longer to come to a stop in part two. However this been the case if car B and car C brakes together and car A does not brake then they will have a force in the couplings T1 & T2. As they are slowing down car A is not slowing down so they will have a force pulling them forward so the force needed to stop car A will result across the two couplings T1 & T2. As it could be seen from theory it now needed to be proved mathematically. When the calculations were completed as can be seen from the above calculations, the force in T1 was -32654.77 N (Compression) and T2 had a force of 16334.13 N (Tension). The reason T1 was in compression was because car B and car C were pulling car A back in the direction up the slope and the coupling will always act in the opposite direction so it was acting down the slope so it was in compression. T2 was in tension because Car A Would have been pulling it down the slope so the coupling acting in the opposite direction was acting up the slope so it was in tension. If both these forces from T1 & T2 were added together ignoring the minus sign as it only explains which direction the force is acting we would get a value of 48988.9 Newton’s. This was the force required to bring car A to a stop. The braking force for each car in part 1 was 48988.9 Newton’s and the force across T1 & T2 in part 2 when added up came to 48988.9 Newton’s. The force in the couplings was equal to the force in the car this is what is meant by, the tractive force required to propel a road vehicle or train can be taken as the equivalent to the pull in a tow-rope or coupling because there are no other forces such as an engine etc…

In my opinion this experiment proved the theory behind tractive forces to be correct as it was shown that a train travelling down a slope the resolve body weight will push it down the slope so it poses no resistance to motion, but if a train is travelling up a slope its own resolved body weight will act as a resistance against the motion of the train travelling up the slope as the resolved body weight will pull it down the slope. It was proven that when the train had less braking force meaning less resistance it took longer to stop as it decelerated at a slower speed. It proved that the forces in the couplings were equal to that of the cars.

Conclusion part B

Tractive force, when a vehicle is travelling up a slope the resolved part of the body weight will act against motion and act as a resistance.
When a vehicle is travelling down a slope the resolve part of the body weight will act like a force and push the vehicle down the slope acting with the motion and acceleration.
The force in the couplings is equal to the force of the cars.
The force in T1 & T2 where equal to the braking force of the car whose brakes did not work.
It took the train longer to come to a stop in part 2,3,and 4 because it only had two cars braking which means the train had less resistance trying to stop it so it decelerated slower causing it longer to come to a stop.
The coupling will always act in the opposite direction to the direction of the force that is applied on it.
Newton’s law was also proved as the total force for each car for part 1 was Ft = mass x acceleration which was 10,000 x -4.409 = -44090 Newton’s we can see now that the total force for a car in parts 2,3, and 4 is -27760 Newton’s because the acceleration has changed Ft = 10,000 x – 2.776 = -27760 Newton’s.
The experiment has supported and proved the theory to be correct.

References:

[1] Jim Duffy, Class Notes – Mechanics 1, “Lecture 6 – Kinetics: Force ‐ Mass ‐ Acceleration, Department of Engineering, Institute of Technology Blanchardstown, Dublin, Ireland, Jan. 2013.

References: [1] Jim Duffy, Class Notes – Mechanics 1, “Lecture 6 – Kinetics: Force ‐ Mass ‐ Acceleration, Department of Engineering, Institute of Technology Blanchardstown, Dublin, Ireland, Jan. 2013.