In this experiment, our aim is to determine the unknown tension by using the components of the force vectors in equilibrium.
Method
First, we place a ring with three strings over the center (reference point), and set up the three pulleys. We will have two known masses which are 50 g each hanged from two strings which can be changed by applying different masses, and moving the direction of the force. In addition, the third pulley will help the system to be balanced, equilibrium, and we will put it at exactly 0° and attach an unknown mass to this string. We can say that the system is balanced, equilibrium, when the ring is center, reference point, so it is important to make the necessary adjustments (moving the strings until the ring is centred) according to center. If all the forces are balanced, the central ring cannot be in contact with the centre of the force table.
Hypothesis
Force is a vector quantity that has both magnitude and direction which can be resolved into its two components: x and y-direction. In this experiment, the system is in equilibrium and as Newton’s first law explains, it is the state in which the forces on the system are balanced, the net force (Fnet), the vector sum of x and y components acting on the system, equals 0. In this case, our aim is to find the unknown tension when we suspend the known masses from strings and balance it accordingly. The unknown tension affects the system in negative x-direction. Since the known (hanging) masses are equal to each other, the first tension (T1) and the second tension (T2) are equal, which we refer to as T. In addition, their x-components both indicate the positive x-direction. So, to keep the system constant, Fnet x should always balance the unknown tension (Tu=W=mg, and W is gravitational acceleration on the object caused by the mass of the Earth; the magnitude of it is the object’s mass, m, times the gravitational acceleration, g, mg.). So, we should take the sum of the both strings’s x-components. In other words, T x +T x+ (-Tu) = 0, or to clarify, Tcos() +Tcos()-mg=0. When we derive the equations accordingly we end up with Tu= Tcos() + Tcos()= 2cos(), Tu= 2 Tcos().
Furthermore, because it is an equilibrium, we expect this fact to apply in the y-components as well. So, the magnitude of the first string’s y-component which is (Ty), and the second string’s y-component (Ty) should be equal but they should be in the opposite directions, (Fnet y =0). That is to say, Fnet y = Ty+ (-Ty) = 0, or Fnet y =T sin()-T sin() = 0. As a result, as we assumed and claimed T = T, and we will get the result as ()= () to balance the system.
Save your time!
We can take care of your essay
- Proper editing and formatting
- Free revision, title page, and bibliography
- Flexible prices and money-back guarantee
Moreover, since y=mx+b, we expect Tu= 2Tcos() to be Tu*(1/T)=2cos() . So, in our graph, x-axis will represent 1/T, and y- axis will represents 2cos(). We will examine their relations to find our slope, Tu.
Variables
- Independent Variable. The known masses are the independent variables because we changed it in our experiment to observe its relationship with the dependent variable.
- Dependent (Reasoning) Variable. The dependent variables are the angles since when we applied different known mass, the angles are adapted to protect the equilibrium.
- Controlled Variable. The amount of friction inside the pulley is one of the controlled variables because we are trying to keep the roughness of the surface constant and at approximately 0 (frictionless) to arrange the rest of the system accordingly.
Graph
The Graph of the Relationship between 1/T and 2cos() in order to find the unknown tension: y=1.231x+0.02021. The gradient indicates the unknown tension and its unit is ‘Newton(N)’ since it is a force.
Results and Conclusions
In this experiment, our aim was to determine the unknown tension by using the x and y components of force vectors in equilibrium. Since we were trying to find the unknown tension (Tu), we assigned various masses to two strings and interpret the results we got from there. In our hypothesis, we assumed that, theoretically, since it is an equilibrium T1=T2. In addition, the angles should also be equal in order to balance the system. To prove what we have found, we used our independent and dependent variables as sources for us to derive them and get proper equations and interpretations. We thought that the unknown tension affects the system in negative x-direction. To indicate the forces which balance it, we observed T’s x-components and realized that since they both are in the positive x-direction (opposite direction of Tu) they balance the Tu value. That is the reason why we concluded that Tu= 2Tx which is equal Tu= 2Tx to Tu= 2Tcos(). In addition, we also proved our equilibrium in y- component as well which helped us see 1=2. Then we put the relations to our linear graph equation. Since dependent variable= slope*independent variable, when we derive Tu= 2Tcos() to put it in y=mx+b, we got the final equation as Tu*(1/T)=2cos(). So, we can state that our hypothesis on the general equation was correct.
Additionally, the fact that T1=T2 and 1=2 was a completely true statement unless there were no systematic and random errors. For instance, the center (reference point) was not always at the middle point of the ring. In addition, even though we assumed the pulley as frictionless, we indicate it as a controlled variable, unfortunately, there was a low friction which we can observe from the y=mx+b equation’s b value as 0.02021. This affected the whole system’s accuracy. Moreover, the experimenter’s instrument ‘spirit level’ was a device which was used to check the level of surfaces, and it showed that the surface was not flat. Furthermore, there were some errors which occurred randomly as parallax error, the video resolution was not perfect, thus it was impossible to select the exact points. Random errors affect the measurement in an unpredictable manner and it may be due to carelessness and lack of concentration. Because of the window in the room that the experimenter conducted his experiment, because of the gas flow, the masses swung. As we explained, we assumed 1=2 and to make calculations we took the average value of them, this might also cause an error. Yet, as we explained before, due to errors which affected the system’s accuracy and precision, we had % 4.21 difference compared to the unknown tension’s actual value.
To reduce the amount of error, the window might have been closed, and so the flaw of gas might have been prevented. In addition, the center might have been calibrated more carefully to decrease the percentage error of the experiment. Furthermore, if we had the chance to read the exact points face to face, it would also help us to reduce parallax error.
Citations
- Holz, Stephen. Class interview. 02 December 2020.
- Walker, James. Physics, 4th Edition. Pearson, 2014.
Stuck on your essay?
