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Students will able to learn about Newton 2nd law and also will be able to do the procedure of this Atwood’s Machine lab experiment.
Typology: Lab Reports
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Qty Equipment Part Number 1 Mass and Hanger Set ME‐ 1 Photogate with Pully ME‐6838A 1 Universal Table Clamp ME‐9376B 1 Large Rod ME‐ 1 Small Rod ME‐ 1 Double rod Clamp ME‐ 1 String
Newton’s 2 nd^ Law (NSL) states that the acceleration a mass experiences is proportional to the net force
applied to it, and inversely proportional to its inertial mass (𝑎 ൌ ி ). An Atwood’s Machine is a simple device consisting of a pulley, with two masses connected by a string that runs over the pulley. For an ‘ideal Atwood’s Machine’ we assume the pulley is massless, and frictionless, that the string is unstretchable, therefore a constant length, and also massless.
Consider the following diagram of an ideal Atwood’s machine. One of the standard ways to apply NSL is to draw Free Body Diagrams for the masses in the system, then write Force Summation Equations for each Free Body Diagram. We will use the standard practice of labeling masses from smallest to largest, therefore m 2 > m 1. For an Atwood’s Machine there are only forces acting on the masses in the vertical direction so we will only need to write Force Summation Equations for the y‐direction. We obtain the following Free Body Diagrams for the two masses. Each of the masses have two forces acting on it. Each has its own weight ( m 1 g, or m 2 g ) pointing downwards, and each has the tension ( T ) in the string pointing upwards. By the assumption of an ideal string the tension is the same throughout the string. Using the standard convention that upwards is the positive direction, and downwards is the negative direction, we can now write the Force Summation Equation for each mass.
In the Force Summation Equations, as they are written here, the letters 𝑇, 𝑔, and 𝑎 only represent the magnitudes of the forces acting on the masses, or the accelerations of the masses. The directions of these vectors are indicated by the +/‐ signs in front of each term. In these equations the + signs are not actually written out, but they should be understood to be there. Understanding this we can see that m (^1) is being accelerated upwards at the exact same magnitude that m 2 is being accelerated downwards. The
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reason m 2 is being accelerated downwards is due to m 2 having a larger weight than m 1 , and therefore there is a greater downwards acting force on m 2 than m 1. To solve for the magnitude of the acceleration that both masses will experience, we can simply use the substitution method by solving one equation for the tension T, then substituting that into the other equation. Let’s use the question for mass 1 to solve for the tension, then insert that into the equation for mass 2, then solve for the magnitude of the acceleration.
Here we see that the magnitude of the acceleration the two masses experience is given by the ratio of the difference of the two masses and the sum of the two masses all times gravitational acceleration. Since that ratio will always be less than 1, the acceleration will always be less than gravitational acceleration. As the ratio gets closer to 1, then the value of the acceleration of the masses approaches the value of gravitational acceleration. However, as the value of this ratio gets closer to zero, then the value of the acceleration approaches zero as well.
Also, comparing this equation to NSL we get.
Here we see that the net force acting on each mass is equal to gravitational acceleration times the difference of the two masses. From the above algebra we can clearly see that 𝐹௧ ൌ 𝑎ሺ𝑚ଵ 𝑚ଶ ሻ as well.
Move and scale the highlight box such that only the data points that represent the moments when the masses were falling are highlighted. (You might have to scale the graph as well.) Click on the down arrow next to the applied selected curve fit icon to open the list of best fit curves. Select the linear fit. The magnitude of the slope of the line is the magnitude of the acceleration the masses experienced. (The slope will be positive or negative depending on the direction of rotation of the pulley.) Record, in the Constant Net Force Table, the experimental acceleration (a (^) ex ) for Run 6.
Run m 1 (kg) m 2 (kg) m 1 +m 2 aex (m/s^2 ) Fnet (N) ath (m/s^2 ) % Error 1 0.0350 0. 2 0.0400 0. 3 0.0450 0. 4 0.0500 0. 5 0.0550 0.
Complete the above chart. Use the experimental acceleration to calculate the net force, and find the % error between the theoretical acceleration and the experimental acceleration for each run.