- Assume that the block is on the exactly horizontal table
Assume that the wood 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.
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 Coulomp’s Law and the equation of motion. The Coulomp’s 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. TcosA and TsinA 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.
- The reaction force given by the table
- The vertical component of tension T, which is TsinA.
Horizontally:
- The limiting friction between the block and table, which is μR.
- The horizontal component of tension T, which is TcosA.
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=TcosA.
Because R=mg-TsinA, μ(mg-TsinA)=TcosA.
μmg-μTsinA=TcosA
T(cosA+μsinA)= μmg
T=μmg/(cosA+μsinA)
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/(cosA+μsinA)=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 change the degree of A. μmg in this equation is constant, so I should make the (cosA+μsinA) as big as possible to decrease T as much as possible.
Suppose y=(cosA+μsinA)
Use calculus dy/dA= μcosA-sinA
When dy/dA=0 μcosA – sinA = 0
μcosA = sinA
μ = tanA
Now I work out an equation that is μ = tanA. Therefore I know that when tanA is equal to the coefficient of friction for the pair of surface it requires less tension to move the block. In other words, it requires less force to lift the case.
Here is a list for the apparatus will be used in this experiment.
- a wood block
- a pulley
- a string
- a table
- a stand
- a rule
- an electronic scale
8. some weights
The diagram on the next page shows how this experiment works.
First I measured the mass of the block by using the electronic scale. And the mass of the block is 356g that is 0.356kg.
Then I made the sinA equal 0 that means the height from the top of the block to the pulley is zero, and the degree of angle A is zero as well. The reason why I use sin not the degree is it is very hard the measure degree when I did the experiment. And I cannot get an exactly degree of an angle. As compare with using degree, sin is more accurate. When the block is equilibrium, if A is zero degree, the tension on the string is the same as the friction force between the block and the table. So the friction force is the same as the weight of the weights. Then I can use the equation F=μR to calculate the coefficient of friction μ.
At this situation the weight of the weights is 92.56g (to 2 decimal place). Through calculation, the coefficient of friction μ is 0.26 (to 2 decimal place).
As I know theμ, I can give the exactly value of the angle which minimises the force required theoretically. tanA=0.26 so A=14.57º
The graph below shows the theoretical curve.
Then I have tried several angles by changing sinA. When I changed sinA, I did not change the height from the top of block to the pulley, but I changed the length of string between pulley and block, because in this case, change the height of the pulley was more different and would produce more errors.
I measured the force for each angle I have taken three times and then take the average force. The reason why I did this is also reduce the error.
The smallest mass of my weight is about 0.1g, so it is the limitation of the accuracy of this experiment. And there is an out line data appeared which is 91.4. So when I calculate the average mass, I avoid this data. By the way, the average mass is to 2 decimal place. Tension is the mass multiply by g.
Followed is the graph of these data.
Through this experiment, I found the angle which can minimise the force required is about 8.21º. The result is not quite close to the theoretical one which is about 14.57º. I think this is because there are too many variation in the experiment and model as well.
- Discussion of 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 suttle of the weight. Because the electronic scale was very sensitive, even some dust would influence the measurement.
- The change of the situation
Every time when I wanted to regress the block, 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.
There is a graph to show the effects of these variations.
- Comparison between the experimental results and the prediction of the model
In my prediction, the angle which can minimise the force required is about 14.57º. However through the experiment, I gained the result which the angle is about 8.21º. It is far away from the theoretical angle. So the two results do not match each other very well. The graph below shows this clearly.
The experimental line is more curved than the theoretical one. That means it is more changeable. I consider the reason of this is the variations which I mentioned above. Take the wrong reading and change the situation make the difference between each value bigger than it should be.
But the basic sharps of two graphs are the same. And in fact, 6 degree is not a very big angle. Therefore I can say the match of the theoretical results and the experimental results is not too bad.
The other reason why these two results are difference is the assumption I made at the beginning. There are some factors I did not consider when I manipulate 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 dos exist, I need heaver weights to move the block. The reason is I need create 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 consider the easiest point is that make the measurements more accurate. I can use a better rule to take the distance measurement, and use a more sensitive scale. Also I can improve the environment of the experiment to reduce the effect to the weight measurement.
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 smooth in practice. And 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.
