There is a very large packing case, far too heavy to lift, and I want to drag it along the floor by a piece of rope. I want to find that how does the force I have to use vary with the angle that the rope makes with the horizontal.
Mechanics 2 Coursework
* Introduction
There is a very large packing case, far too heavy to lift, and I want to drag it along the floor by a piece of rope.
I want to find that how does the force I have to use vary with the angle that the rope makes with the horizontal.
And also I want to find the angle which minimises the force required.
First I will work out the angle theoretically, in other words, I will make a model of this situation. Then I will do a simple experiment to find what the degree of the special angle is practically. Finally I would like to compare the experiment results with the prediction of the model.
* Assumptions
As do an experiment, I cannot make the environment to be perfect and when I make the model, it is impossible for me to consider every factory in this case. Therefore I need make some assumptions.
. The string used in this experiment is light and inextensible.
The rope used in this particular experiment is thin and about 1.5m long. So for this experiment, the mass of the rope is extremely small and it can be ignored. The string itself will extend when there are several forces acting on it. Hence it will create a tension on the string. However it is very small as well. Therefore I will ignore it when I model the experiment. The tension on the string would have the same magnitude as the weight has been put in the end of the string.
2. The pulley used in the experiment is smooth
In this case, I assume that the coefficient of friction of this pulley is zero. So there is no friction force between pulley and string.
3. There is no air resistance.
In such situation, the packing case would be actually moving very slowly, so the air resistance will be very small; and also it's very complicated to model the air resistance for this object. Therefore it's worth to assume that the air resistance in this situation is negligible.
4. Assume the wood block I used in the experiment is a particle rather than a body.
In this situation the block is only used for providing a force. So it doesn't matter whether it is a body or not, while if consider the block is a body, I need consider toppling and sliding in this experiment. However it dos not improve anything. Therefore I assume the block is a particle.
5. Assume that the block is on the exactly horizontal table
Assume that the block is on the horizontal ground, so the friction acting on the block would have the same magnitude as the horizontal component of the tension on the string.
* Manipulating the Model
I am going to use some mechanics theories to set up several equations to manipulate this model. And I also will use these equations to predict the results of the experiment.
The main theories I will use here are the Kinetic Friction Law and the equation of motion. The Kinetic Friction Law says if the object is on the point of sliding or sliding, the frictional force between the object and the surface is given by:
F =µR
And the equation of motion shows that resultant force acting on an object equals to the mass of this object multiply the acceleration of this object. The equation is given by:
F=ma
The diagram below shows the actual experiment I have done.
According the assumption I made, T should equal to m'g. Tcos? and Tsin? are the horizontal and vertical components. g is the acceleration of gravity, which is 9.8 N/kg. Followed is the list of all the force which is act on the block.
Vertically:
. The weight of the block, which is mg.
2. The reaction force given by the table
3. The vertical component of tension T, which is Tsin?.
Horizontally:
. The limiting friction between the block and table, which is µR.
2. The horizontal component of tension T, which is Tcos?.
If the block at rest, just move or sliding at a constant speed, I can say it is in equilibrium. So the final force on the block should be zero. In this particular case, we only interested in the equilibrium on horizontal. Because the block is putting on the table, vertically it should be always equilibrium. For horizontal, friction force should ...
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3. The vertical component of tension T, which is Tsin?.
Horizontally:
. The limiting friction between the block and table, which is µR.
2. The horizontal component of tension T, which is Tcos?.
If the block at rest, just move or sliding at a constant speed, I can say it is in equilibrium. So the final force on the block should be zero. In this particular case, we only interested in the equilibrium on horizontal. Because the block is putting on the table, vertically it should be always equilibrium. For horizontal, friction force should equal the horizontal component of T. Therefore we can obtain an equation µR=Tcos?.
Because R = mg-Tsin?, µ(mg-Tsin?) = Tcos?.
µmg-µTsin? = Tcos?
T(cos?+µsin?) = µmg
T = µmg/(cos?+µsin?)
Since there is one string link the block and the weights, the tension which is acting on the block is the same as it acting on the weights. Therefore the tension T is the same as the weight of these weights, which is m'g. Hence the equation above can be written as follows.
T = µmg/(cos?+µsin?) = m'g
By this equation, I will make the prediction of the results of this experiment. The purpose of the experiment is that find the minimum T through changing the degree of ?. µmg in this equation is constant, so I should make the (cos? + µsin?) as big as possible to decrease T as much as possible.
Suppose y = (cos?+µsin?)
Use calculus dy/dA = µcos?-sin?
When dy/dA = 0, µcos? - sin? = 0
µcos? = sin?
µ = sin? / cos?
µ = tan?
Now I work out an equation that is µ = tan?. Therefore I know that when tan? is equal to the coefficient of friction for the pair of surfaces, it requires less tension to move the block. In other words, it requires less force to lift the block.
* Conducting the experiment
Optional Extra Experiment
I will now be conducting an experiment with the sole purpose of finding the value for µ.
The below diagram will show the experiment I have done.
T = Mg
F = µR
R = mg
F = µmg
On the point of sliding, µmg = Mg
µ = M
m
Therefore the masses M and m need to be calculated. This can be done by varying the mass of the block (m) by adding weights, and testing to see what mass (M) is required to move the block. The results should form a straight-line graph, when M is plotted on the y-axis and m on the x-axis. The gradient of this line would then be equal to µ.
A list of the apparatus used in this side experiment is shown below:-
. A Pulley
2. String (1m long approximately)
3. A Block
4. Metre ruler
5. Clamp Stand, boss & clamps
6. Set of weights of mass 5 grams
Method
. Calculate the mass of the block using an electronic scale.
2. Tie string, approximately 1m in length, to the block.
3. Secure the pulley to the clamp stand using the boss and clamp. Ensure that the pulley is straight.
4. Adjust the height of the pulley so that the string is horizontal between the block and the pulley.
5. Wrap the loose end of the pulley over the pulley.
6. Place the block 0.5m from the centre of the pulley.
7. Tie a mass of 5g to the free end of the string, as M.
8. Increase the value of M 5g at a time, until the block begins to slide.
9. Note down the value of M at which the block slides.
0. Add 50g to the mass of the block (m) and repeat steps 7-9.
1. Repeat steps 7-10, until 350g have been added to m.
Results
Mass of Block (kg)
Mass added to move the block (kg)
0.262
0.125
0.312
0.135
0.362
0.150
0.412
0.160
0.462
0.210
0.512
0.220
0.562
0.245
0.612
0.295
As can be seen in the above theoretical graph, a straight-line graph was produced. This line has the equation:
y= 0.4762 - 0.0156, so µ = 0.4762.
Therefore the optimum angle of this theoretical experiment = tan-1 (0.4762) = 25.46º.
Main Experiment
Here is a list for the apparatus will be used in the main experiment.
. A block
2. A pulley
3. A string
4. A table
5. A clamp stand with a clamp
6. A meter rule
7. An electronic scale
8. Some weights (several sets of 10 gram weights)
The diagram below shows how this experiment works.
To Reduce Experimental Error:-
To take a more accurate value for the angle, we will use the measurements taken in the diagram rather than using a protractor to measure the angle. The main reason for this is to make the experiment more accurate, since human error is involved when using a protractor to measure the angle.
tan ? = h-d / n
? = tan-1 (h-d / n)
Using these measurements which we made with a metre rule, the angle ? can be found more accurately using trigonometry (this is shown in the above diagram). As well as this, the angle ? can be accurately set to a desired angle. If the height of the pulley (h) remains the same, then only the distance at which the block is placed from the pulley needs to be altered.
n = h-d / tan ?
By replacing ? with the desired angle, the distance at which the block needs to be placed from the pulley can be calculated.
The mass required to move the block was 190g, and the mass of the block was 604g. Therefore µ = 190/604 = 0.315 (to 3dp)
Therefore the optimum angle = tan-1 (0.315) = 17.46º (to 2dp).
Method
The angle, at which the block is being pulled, is to be set at 5º intervals, by adjusting the distance between the block and the pulley. Then the mass M needs to be altered to adjust the force that is pulling the block. It is to be increased 5g at a time until the block moves. The mass, which causes the block to move, is to be noted down, as M.
. Calculate the mass of the block, using an electronic scale.
2. Tie a rope or string of which is of a length just over 1m to the block.
3. Calculate the value d, by measuring the height above the table at which the string is tied to the block.
4. Secure the pulley to the clamp stand using the boss and clamp. Ensure that the pulley is straight.
5. Set h to a height that is 0.25m greater than d, to simplify the equations.
6. Wrap the loose end of the string over the pulley.
7. Place the block at distance d from the pulley, calculated using the formula
n = h-d / tan ?,
Therefore the value of ? will then be equal to 15º.
8. Tie a mass of 10g to the free end of the string, call this value M.
9. Increase the value of M by 10g, until the block begins to slide.
0. Note down the value of M at the point when the block begins to slide.
1. Repeat steps 7-10, for values of ? of 30º, 45º, 60º, and 75º.
(These above steps were used to gain the preliminary results)
2. Now we will take more accurate results, by increasing the mass M by 5g, and increasing ? by 5º, in the range of angles where the mass required to move the block is the lowest (in this investigation the range was from 15º to 45º).
3. Repeat step 12, to make results more reliable.
Preliminary results
In our preliminary experiment we tried to find out approximately the optimal angle which is needed to minimise the force required to move the block. An initial set of results was obtained. The measuring of the mass was less accurate, as it was increased by 10 grams each time.
Angle ? (º)
h (m)
d (m)
Mass (g)
5
0.26
0.933
30
30
0.26
0.433
20
45
0.26
0.250
30
60
0.26
0.144
80
75
0.26
0.067
240
The above graph shows us that the optimum angle required to move the block is somewhere in between 15º and 45º. From this, we will now conduct the real experiment where we will take results at 5º intervals between 15º and 45º to gain a more accurate value.
Final Experiment
Angle ? (º)
h (m)
d (m)
Mass (g)
5
0.26
0.933
25
20
0.26
0.687
20
25
0.26
0.536
20
30
0.26
0.433
25
35
0.26
0.357
25
40
0.26
0.298
35
45
0.26
0.250
35
The above graph shows the more accurate experiment. The equation of the graph has been shown and a best fit curve has been added so that we can later compare these results with the theoretical curve.
Now we must find out the optimum angle. This is the point at which the mass required to move the block is minimal, therefore the minimum point on this graph needs to be found. This can be done by differentiating the equation below:-
y = 0.031x2 - 1.3929x +137.26
dy/dx = 0.062x - 1.3929
At the minimum point, dy/dx = 0
Therefore 0.062x-1.3929 = 0
x = 1.3929/0.062
x = 22.47º
*
Variation in the experimental results
The graph of the experimental results is not smooth; some points are a little bit up or down. I think it is because of the errors.
. Measurement
When I was doing the length measurement, the veracity of my reading was dependent on the rule I used. Even if I took my reading three times, I couldn't avoid all the errors of the measurements. Also when I measured the weight of my weights, I didn't measure the inaccuracies of the weights. As the electronic scale was very sensitive, even some dust would influence the measurement. When taking the final measurement for the mass required to move the block, there will be an inaccuracy value of anything up to 5 grams because of the weights used.
2. Change of Position
Every time when I wanted to move the block back into its original position, I couldn't do that is exactly as it was before, in other words, the situation of this experiment had a tiny change every time.
Every time the experimented was repeated in an attempt to attain more accurate results, it was clear that the block was not replaced in its original position and because of this; the tension in the string when a mass was applied was between 5 and 20 grams different from the previous measurement at the exact same angle.
From the graph I have drawn on the previous page, it is clear that the points plotted are not in a perfect pattern along the curve. The variations in the experiment have resulted in a difference between the calculated masses and the actual masses that were required to move the block.
Below is a graph to show the error of the mass measurements. This has been shown by using error bars.
It can clearly be seen that the error bars for each point are crossing through the best fit curve. This shows that the experiment was carried out accurately. Only 1 point is outside of this range; this could have just been down to procedural error or any one of the number of points mentioned in the Assumptions.
*
Comparison between the experimental results and the prediction of the model
I will now show a graph which will compare the experimental results which I have obtained from conducting the experiment and this will be compared with the theoretical results. This graph was constructed using the program Omnigraph.
The blue line shows the results from the experiment, while the black line shows the theoretical results.
The experimental graph plotted which I have plotted is:-
y = 0.031x2 - 1.3929x + 137.26
On this graph the minimum value for M is obtained when ? = 22.5º.
The theoretical graph plotted which we have plotted is:-
y = (0.4762 * 262) / (cos x + 0.4762 * sin x)
On this graph the minimum value for M is obtained when ? = 25.5º.
The degree of error is only 3º which is highly accurate and shows that the experiment was conducted properly and carefully.
The curves on both graphs are very similar in every respect. This would suggest that the results do fit the data. The masses required to move the block are higher in the experimental model than what is suggested in the theoretical model. Between 0º and 40º the difference between the two curves is only 5 grams. However the difference is greater between 40º and 80º, with the greatest difference being around 30 grams. It is not easy to explain why these inaccuracies have come about, but it would be wise to return to the 'Assumptions' for an answer to these inaccuracies.
The experimental model, produced using results calculated, appears to be a good match to the theoretical model, due to reasons explained above.
The calculation used for the theoretical model was:-
M = µm
cos? + µsin?
* Revision of the process
The other reason why these two results are different is the assumption I made at the beginning. There are some factors I did not consider when I manipulated the model and did the experiment. The biggest two errors I think are the friction between string and pulley and the extension of the string.
Since the friction between string and pulley does exist, I need heavier weights to move the block. The reason for this is that I need to create an extra force to cancel the friction. The extension of the string also required the bigger tension, because the extension of the string itself produced the extra resistant force.
In order to improve this experiment, I will consider the easiest point which is to make the measurements more accurate.
* I can use a better ruler to take the distance measurement, and also use a more sensitive scale.
* Also I can improve the environment of the experiment to reduce the effect to the weight measurement.
* I could use more accurate weights to decrease the inaccuracies in the Mass, M. The maximum error in the mass in the experiment was 5 grams, with the use of more accurate weights the error could have been reduced to 1g for example.
* The µ value could have also been inaccurate therefore it would have been wise to take several measurements for the value of µ all along the table to gain the most accurate value as possible.
In order to improve the model, I think the best thing to do is considering more factors than I did in this course. For example, if I did not assume the pulley is smooth, the theoretical result should be more close to the real one, because the pulley does not behave smooth in practice. Also I can consider the extension of the string as well. However that will make the calculation much more complicated than I have done in this coursework. If the mentioned calculations have been done, I can avoid these two biggest factors which make the prediction imprecise, and the prediction should be much better.
Khushpal Grewal