Group3_2_ch4

Kaila Solomon, George Souflis, Michael Solimano, Nicole Kloorfain

toc Gravity and Laws of Motion Lab 11/16
task A - Michael task B - Kaila task C - Nicole task D - George

__Objectives:__
 * Find the value of acceleration due to gravity.
 * Determine the relationship between acceleration and incline angle.
 * Use a graph to extrapolate extreme cases that cannot be measured directly in the lab.
 * Determine if mass has an effect on the acceleration down an incline.

__Hypothesis:__ We believe that without air resistance, the value of acceleration due to gravity is 9.8m/s/s. We believe this to be true because we measured this in the free fall lab. We also think that as the incline angle is increased, the value of the acceleration will as well. We think that this will happen because at a 0 degree angle, the acceleration will be 0, and as the angle is increased, it will become closer to being in free fall, so the value will go from 0m/s/s to 9.8m/s/s in a relatively regular fashion. Additionally, we think that an object with a greater mass will accelerate faster. This is because it has more inertia, and will therefore resist change more easily than an object with a smaller mass.

__Methods and Materials:__ First take a ball and find the mass of it using an electronic balance. Using a ruler or tape measure, set up the ramp to a height of .15m and measure the distance to the end of it. Place the ball at the top of the ramp and allow it to roll down as a stopwatch is used to measure the time it takes to reach the bottom. Record this time and repeat for several trials. Find the average time and use it to calculate the acceleration. Still at the same angle, change the distance and observe what happens. Next, change the angle and repeat the process, performing several trials before moving on to a new angle. Develop data tables and an acceleration vs sin(theta) graph and analyze and draw conclusions.

__Procedure:__ <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Below is a sample video of rolling the ball down the ramp at .15m. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">media type="file" key="rolling ball down ramp.m4v" width="300" height="300"

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">__Sample Calculations and Data:__

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">__Group 3 Lab Data__ <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">This graph shows the acceleration as the height of the ramp changes. This data will show us the relationship between ramp height, or the sin of theta, and acceleration.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Link to document :

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">__Graph:__ <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Picture of Lab Set up <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">__Calculations:__

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Sample of Finding an Angle: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Sample of finding the acceleration: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Percent Error <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">__Analysis:__

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Percent Error <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The slope of our graph is equal to the gravity in our experiment which is 8.1348m/s/s. It is close to 9.8 m/s/s which is the actual acceleration due to gravity. Our percent error was 16.99% which is a big percent error but our slope is actually very close to the theoretical value.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">This is how we find the acceleration due to gravity

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Class Data

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Slope values: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Class Values of acceleration@15 cm <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The amount of mass the object obtains has no correlation on the acceleration. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The x component of weight force is causing the ball to roll down because the y component is balanced with the normal for therefore the unbalanced force is the x component which creates the motion and equals the acceleration. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">This is the acceleration calculated from Newton's second law and it is close to our calculated average acceleration but not perfect equality because it is .919 m/s/s comparing to 1.225 m/s/s.
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Mass (kg) || <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Acceleration@15cm (m/s/s) ||
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">0.00898 || <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">1.039 ||
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">0.016 || <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">1.056 ||
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">0.028 || <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">0.83 ||
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">0.226 || <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">0.9197 ||
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">0.5352 || <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">1.048 ||

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">__Discussion Questions:__ <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">1. Is the velocity for each ramp angle constant? How do you know? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Velocity is not constant for each ramp angle because the acceleration and time changes for each angle. As the ramp approaches a vertical free fall, the acceleration increases are not the same for all ramp angles. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">2. Is the acceleration for each ramp angle constant? How do you know? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Acceleration values change for each of the different angles therefore it is not constant. The acceleration increases as the ramps height from the table increases thus the angles do not have a constant acceleration. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">3. What is another way that we could have found the acceleration of the ball down the ramp? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">I would have found the acceleration by graphing the velocity and finding the slope of the tangent line to this point on the graph which would be the acceleration. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">4. How was it possible for Galileo to determine //g//, the rate of acceleration due to gravity for a freely falling object by rolling balls down an inclined plane? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Galileo would realize the higher the ramp was, the higher the acceleration would be by rolling the balls down the ramp. He would realize that the closer ramp was to a vertical free fall the accelerations increased and approached 9.8 which is when sin of theta equals one. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">5. Does the mass of an object affect its rate of acceleration down the ramp? Should it affect the motion of an object in free fall in the same matter? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">No, the mass does not affect the rate of acceleration. Our classes observations prove this because there is not a correlation or pattern between mass and acceleration. In the equation to find acceleration, mass cancels out and is not needed to find acceleration which proves its lack of effect. After doing our last unit on free falling objects, we have found that mass does not affect the time our acceleration of free falling objects.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">__Conclusion:__

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">After analyzing the data the results indicate that the acceleration due to gravity is 8.1348 m/s/s (9.040 m/s/s when the y-intercept was not set to zero). This result was extrapolated from the acceleration vs. sin(theta) - since W x =∑F and therefore Wsin(theta) = m*a, and therefore m*g*sin(theta) = m*a, and thus a = g*sin(theta). This theoretical background allowed us to assume that when sin(theta)=1 (meaning there is a 90 degree angle), the resulting acceleration will be equivalent to acceleration due to gravity. In this respect our data failed to show that our hypothesis was correct because from prior knowledge we know that g should equal 9.8 m/s/s. We determined this because according to the theoretical background provided, the slope of the linear equation graphing acceleration vs. sin(theta) should be equal to the acceleration due to gravity. This was not the case due to several sources of error that will be further discussed later.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Next we found that an increasing angles directly relates to an increasing acceleration; that is, the higher the incline, the greater the acceleration value. For example, At an incline height of .15 m, the acceleration was on average 0.9197 m/s/s. Then, at an incline height of .20 m, the acceleration value increased to an average 1.340 m/s/s. As we predicted, and later extrapolated from the graph, the acceleration value approaches 9.8 m/s/s as the incline approaches a 90 degree angle (which would be freefall). In sum, our hypothesis was correct for this portion.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Lastly we compared our data with the results of other groups to determine if mass has a significant effect on acceleration. Contrary to our hypothesis, we found that there is in fact no significant difference in acceleration between a massive ball and a miniscule one rolling down a ramp. For example, a ball with a mass of 0.226 kg was found to have an acceleration of 8.1348 m/s/s. A ball with a mass of 0.00848, significantly less massive, was found to have an acceleration of 8.789 m/s/s. Clearly mass has no bearing on acceleration down an incline. However, we maintain the hypothesis that a more massive object will slow to a stop more slowly than a small one (on a flat plane).

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">After comparing our results to the theoretical results, we found that we had a 16.99% error. This is fair considering the many sources of error which contributed to our limited result. I think much of the error occurred in the times. Reaction time became an issue especially when the trials happened so quickly that it was difficult to start and stop on time. This affected our acceleration values and thus our graph. Similarly our measurements were not exact when it came to the heights, which affected our sin(theta) values (however slightly). Another large contributor as a source of error was friction. Friction minimally affects rolling objects, but when paired with the other sources of error, it may have offset our results to the point where they ended up at. To address these errors, there are several solutions. With a device such as a slow-motion video camera we could playback each trial and determine the exact time for each without estimation. Further, a ruler/tape measure with smaller increments of measurement could be used to ensure exact heights and distances. Friction cannot really be eliminated, but reduced through polishing surfaces and/or adding a liquid such as oil between the surfaces to fill in micro gaps in the surfaces. Drivers should be aware of these tendencies of objects when driving down hills. The steeper the hill, the faster the car will accelerate; therefore, the car runs a heightened risk of illegally going above the speed limit.

=<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Newton's Second Law Lab = <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Task A- Kaila Solomon <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Task B- Nicole Kloorfain <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Task C- George Soufflis <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Task D- Michael Solimano

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Objective: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What is the relationship between system mass, acceleration, and net force?

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Methods and Materials: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">First, set up the drop and pull system. Place the track on the table an inch or two away from the edge. Put the dynamic cart on the track with the string hanging towards the edge of the table. Clamp the super pulley on the edge of the table. Attach the photogate to the pulley and then plug in the USB to a computer. Place the string from the cart on the wheel of the pulley, let the mass hanging be on the other end hanging of the table. There are small little weights that add up to 25 pounds. Place all 25 pounds on top of the dynamic cart. Let the cart roll on the ramp as well as the mass hanging falling off the edge; record the velocity and then repeat 3 times. Because the wheel is hooked up to the computer studio, the sensor will calculate the velocity and will make a velocity time graph. The slope of the velocity time graph is acceleration therefore finding the line of best fit will provide accurate acceleration from the results collected. After the three trials, the weights must be distributed throughout the drop and pull so one at a time. Take one weight off the cart and place it on the hanging mass object and then complete three trials of recording the movement of the string. Repeat this process until all of the 25 pounds are on the hanging mass.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Procedure: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Below is the video sampling our method of this process.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Hypothesis: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The graph acceleration v net force is going to be an increasing positive linear because the mass has to remain constant. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The graph mass v acceleration will be a linear line that is decreasing negatively because the force has to remain constant.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Data Tables and Graphs: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Data Table 1: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">r <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Data Table 2: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Graph of data table 1: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Graph of data table 2: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Video:

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">media type="file" key="Movie on 2011-11-30 at 08.55.mov" width="300" height="300"

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">link to excel document: [|newton's second law.xlsx] <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Analysis:

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">(ignoring friction)

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Sample Calculations:

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">For this experiment we plotted two graphs - an acceleration vs. net force graph and an acceleration vs. mass graph. We found the former to be a positive linear function and the latter to be a hyperbolic function (only in the first quadrant). The relationship in the former was found to be directly proportion - that is, acceleration is directly proportional to net force. In the latter, there was found to be an inversely proportional relationship - that is, acceleration is inversely proportional to mass. Looking at the equation ∑F=ma, these relationships are clearly manifested.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">In the acceleration vs. net force graph, slope is the reciprocal of the total mass. The percent error turned out to be 9.39199%. Slope should equal 1/m because of a=∑F/m, which can be rewritten as a=∑F(1/m), where acceleration is y and net force is x. Further, the y-intercept means friction over total mass. This value should be negative because friction takes away from the acceleration of the cart.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">In the acceleration vs. mass graph, the power on the x is -7.214. It should be -1, which means there was over a 700% error. The leading coefficient is 0.0045. This represents net force.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Friction would lower the acceleration, so a bigger would be needed to create the same acceleration. So, the slope was too small if claiming no friction was present. Friction can be a source of error because it changes the slope of the graphs and makes it harder to determine the relationships between the components of the graph. With friction, the equation would be a=(hanging weight)(1/total mass)-(friction/total mass), or in our graph, y=1.6873x-0.2741. a = (0.015)(9.8)(1/0.537) - 0.2741 = 0.000357 m/s/s

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Conclusion <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Was the purpose satisfied? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The purpose of our lab was satisfied. In the lab, we attempted to give experimental credence to the equation Net Force = mass * acceleration. In doing this, we utilized a cart with a string attached to it, which then was attached to a pulley system. The speed of the wheel on the pulley was measured, and was expected to vary based on our systems total weight, and the changing hanging weights that were used. The first test that we conducted was analyzing the effect of net force, or the hanging weight, on the acceleration of the cart. As we did not want to change the total mass, we replaced the weights from the cart by adding them to the hanging weight, keeping the total weight the same. When our data was graphed, we attained the equation y=1.6873x-0.2741. This line shows that as the hanging weight, or the net force, increases the acceleration also increases. The line is based on the equation acceleration=Net Force * 1/m, derived from y=mx+b. When analyzing this equation in regards to the equation a= (net force)/(mass), we can see that the slope equaled 1/(total mass), and the x value was our Net Force. The y intercept of the graph is thus also equal to friction/mass. Our equation had a negative y intercept, which makes sense because friction should have a negative effect on the acceleration. This equation also had an R squared value of .93, which showed that the line of best fit was an accurate representation of our data. This data was consistent with our hypothesis, where we thought that this graph would have a positive, linear relationship. This data had a percent error of %9.39, taken from our experimental value of 1/.537 (because m = 1/total mass) against our actual slope of 1.6873. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The next graph displays the correlation between the total mass of the system and the acceleration. We hypothesized that this graph would have a linear, negative correlation. In this test, we kept our hanging weight the same, while increasing the weight on the cart, and thus changing the total weight of the cart and hanging mass. Our graph showed that as the weight of the cart increased, but not the net force, the acceleration became slower. This was accompanied by the equation y=.0045x(^-7.214), which had a power fit. Because the graph had a power fit and was not linear, our hypothesis was not entirely accurate. This equation was based on the equation y=Ax(^-b). Our A value (.0045) was equal to the Net Force of our experiment. The percent error for this graph was calculated by comparing our exponent, -7.214 (experimental), to the expected exponent of 1. The calculations shown above yielded a percent error of over 700%. This graph had an R squared value of .95, again showing that it was a strong fit.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">There were many sources of error inherent in this lab, which led to our percent errors of %9.39 for acceleration v net force, and over %700 for acceleration v mass. One source of error was in the string that connected the cart and the hanging mass. The string needed to be parallel between the cart and the top of the wheel. While we came close to parallel, a slight angle of elevation or decline would have affected our results for the velocities. Also, the hanging weight did not hang freely from the wheel, but rather came into contact with a part of the vice that attached it to the table. This could have caused a friction, which could have negatively affected the validity of our results. Also, when the data was collected by //datastudio//, it also included when the cart stopped, although we tried to avoid this. We thus were forced to choose the data points that were representative of our velocity, which could have led to further data in the slopes of our velocities, or our accelerations.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Had we redone this lab, we could have addressed some of our error. Had a tool been available, we could have measured the exact angle of our string, to assure that it was parallel to the track. This would have assured that the data was not skewed because of a slope in the line. Also, if a different means of measuring the data on //datastudio// were available, we could have accurately taken out any extraneous data points. By doing these, we could have further cut down our error. This lab had many real applications, as it itself was a pulley system. Pulleys are commonly used today, and these calculations could be used to find the acceleration based on differing masses or net forces of pulley systems. As an example, when pulling an object with a crane, one must know the force that the object has on the crane, as well as the necessary total mass and forces necessary to move the object up or down.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Coefficient of Friction Lab
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Task A: Nicole Kloorfain <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Task B: George Souflis <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Task C: Michael Solimano <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Task D: Kaila

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Objective: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Between the surface of the track and the friction cart, what are the coefficients of static and sliding friction? What is the relationship between the friction force and the normal force?

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Hypothesis: The coefficient of frictions will be between 0 and 1.The relationship between friction force and normal force will be directly proportional because of the equation f=µN. According to this, friction will increase as normal force increases.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Methods and Materials:

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">First, set up a track. Have a cart with a string attached, and various masses to add to the cart. The masses should be relatively large (500g-1000g) to see significant jumps in Tension force between trials. Next, hook a Force meter to the string and put in the USB link into a computer. Make sure to open data Data Studio and "Create Experiment." This will track tension over time after setting the y-axis to Force-Pull Positive. The experiment is now set up to begin trials. Add 500 g to the cart (which should also be massed and added to total mass). Start the run and "zero" the sensor. Holding the string parallel to the track, pull with a slow constant speed. If all was done correctly the graph should spike and then level out. The max tension value will be used for static friction, and the mean of the leveled out portion of the graph will represent sliding friction.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">FBD: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Data:

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Class Data: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Calculations: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%; line-height: 0px; overflow: hidden;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Sample % Difference <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Static friction <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Percent Difference = (I average – individual trial I)/(average) x 100 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Percent difference = (I .1741 - .1623 I)/(.1741) x 100 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Percent difference = %6.77

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Kinetic friction <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Percent difference = %14.42

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The above calculations compare the class data to our data

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Compare the slope of the graphed lines to calculated mu average

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Finding mu: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">friction = (mu)(Normal) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">mu = friction/weight because N = W

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Average of above calculation: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">(static) = .136 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">(kinetic) = .0898

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The following differences were solved using the same percent difference equation as is above. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Percent Difference Static: %16.2 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Percent Difference Kinetic: %7.14

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%; line-height: 0px; overflow: hidden;">media type="file" key="Movie on 2011-12-07 at 08.37.mov" width="300" height="300"

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%; line-height: 0px; overflow: hidden;">Discussion Questions


 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Why does the slope of the line equal the coefficient of friction? Show this derivation.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">According to this equation, the equation f = (mu)(N) matches the equation, y = mx + b. Thus, the coefficient of friction is equal to m, or the slope. Also, b should equal zero, as any b would have been error in our lab.


 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Look up the coefficient of friction between your material and the aluminum track. Discuss if your measured results fall within the range of theoretical values. Be sure to cite your source!

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">I found that the coefficient of friction between aluminum and a mild steel, which should be similar to our material, was .61 when static and .47 when sliding. Our static friction coefficient was .1623, and .0967, which are rather far from the expected amounts.


 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What variables affected the magnitude of the force of friction? What variables affected the magnitude of the coefficient of friction?

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The force of friction was affected by the surface on which we slid the cart. Had the surface changed, either rougher or smoother, it would have opposed the motion to a differing level, and thus would have changed the magnitude of friction. Also, because friction equaled tension, a greater tension force would have raised the friction. The coefficient of friction is effected by the surface used. Had the surfaces changed, the coefficient of friction would have changed.


 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">How does the value of coefficient of kinetic friction compare to the value for the same material’s coefficient of static friction?

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The coefficient of static friction was higher than that of the kinetic friction. This is because it took a greater amount of force to move the object when it was at rest. This is because objects tend to stay at rest if already at rest, and thus it would take a greater force to move them. Also, if in motion, an object will tend to stay in motion, so it will take less force to keep it moving. This accounts for the differences in slopes, or coefficients of static friction.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">**Conclusion:** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Our hypothesis was that the coefficient of frictions would be between 0 and 1. We also said that the coefficient would increase as the mass increased in a linear relationship. This was confirmed with our results. The reason for this happening was because a larger mass would take more tension to move, therefore creating a greater amount of friction. After gathering our data, we found that the coefficient for static friction was .1623 and that the coefficient for kinetic friction was .0967, supporting our hypothesis that the coefficients would be between 0 and 1. These values are the slopes of the lines that we graphed. We then calculated the percent difference for the coefficient of static friction to be 6.77% and the percent difference for kinetic friction to be 14.42%. These are not too far off from the class average, but are far enough to indicate sources of error.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">There could have been multiple sources of error that contributed to the results we got. One aspect that was very important to properly executing this lab was making sure that the string that was attached to the cart stayed parallel to the track at all times. Although we could be very close in this, it is difficult or impossible for someone to hold it where there is no angle at all while pulling the cart. Similarly, it is difficult for the person who is pulling the cart to keep a completely steady hand. Wavering up, down, left, or right is something that would have skewed the results. It is also hard maintain a constant speed when pulling the cart. For a person to have to keep the cart moving without even the slightest change in speed is very difficult to do. These sources of error are also indicated in the r 2 values that we got, which was around .897 for static friction and .98 for kinetic friction. This shows that while our values are not significantly off, they were affected by outside influences.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">This situation can be applied to everyday circumstances. If you ever need to know the coefficient of friction and how it will affect something as a whole, such as friction between tires and the ground, what we learned in this lab can be applied to help figure this out. This lab also provided a visual that makes understanding the concepts of friction and the coefficient of friction better, which allows one to be able to relate it to things that they see in the real world.