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Please catch the upward mass before it hits the pulley!!! |

I downloaded the atwood MSExcel sheet and there is no picture where there should be one...on the pic tab. | Stuart, Jay Fri. 27, Sep 2002, 21:28 j.stuart@attbi.com |

That page is for you to put your sample data picture there. :) | Douglas Mon. 30, Sep 2002, 13:04 |

When it says, state two equations, should we use the 2 equations from the instructions or do you mean y=mx+b? | Anonymous Fri. 04, Oct 2002, 15:45 |

When you have a plot, y = mx + b is for mathematics. In science we need to be completely descriptive. What is the physical quantity that represents y? What is the constant of proportionality? What are its units? What is the physical quantity that represents x? What is the y-offset in all the x data? What are the units of this offset? You must ask this of each series that is plotted on a graph. For Atwood's Machine, you plotted (hopefully) two sets of y data against the same x data. You should describe both of these completely as mentioned above and as exemplified in many of the prelabs up to this point as well as in Data & Graphs and MSExcel. For Data & Graphs you had: Displacement = 0.0465 [ms ^{-2}] * Time^{2} + 0.012 [m];Displacement _{theory} = 0.049 [ms^{-2}] * Time^{2}.Please let me know if this does not answer your question. | Douglas Sat. 05, Oct 2002, 07:40 |

Dear Douglas: Is it appropriate to ask what might be meant by Q. 2 of postlab: "list each measurement-type name, its average, and uncertainty with reason"? Could you give any hints on how I can approach answering this question? Thank You!! | Anonymous Wed. 01, Oct 2003, 21:15 |

Hint: Please log in to the key for Data and Graphs PostLab to see the same type of question and how it should be answered. You measured "Mass". What "scale" did you measure? Was it 20 grams or 200 kg? Was it 3.2 milligrams or 43 tons? How well did you know that measurement and why? So, here is a sample where I chose to give a range instead of an average: Mass; 0.0810 [kg] - 0.1404 [kg] ± 0.1 [g] Uncertainty is due to resolution of mass balance. Notice my precision for the 0.0810[kg] matches the precision of the uncertainty 0.1[g] You are right, you also measured velocity with the SmartPulley. You may instead use "Acceleration" in this lab; why? simply because I said so. It would be best to go back even further than velocity to the actual spoke pick-up on the pulley; but we only have 1:50 hr to do the lab... There was a note on the board to measure acceleration five times for one trial. (Repeat the drop five times). You could then use MSExcel to calculate the Standard Deviation =stdev(cell1,cell2,cell3...) That would be your uncertainty. The reason for the uncertainty is your standard deviation from repeated measurements of one trial. If you didn't repeat one of your trials five times, then you may open your Science Workshop file and take one run as your sample. Choose one that is not as "straight" as the others. Find the slope of different sections of the data. Do this five times for the one run. Use the stdev of these values for your plus-minus uncertainty. Each lab will have a different method for finding the uncertainties. I am sorry if your instructor has not covered this in class. I am sorry if this short discussion hasn't helped. Hopefully, next week in class, you can encourage your instructor to discuss this in class. :) | Douglas Thu. 02, Oct 2003, 06:53 |

Douglas, Can you help me answer the question#6 from postlab? | Anonymous Thu. 25, Mar 2004, 12:44 |

View on web site for best viewing. What is the equation for the plot? Δm g = Σm a + 2f _{p}The best fit line tells you the value for Σm and the value for 2f _{p}.You know Δm g = (m2-m1) g; and you are given m1 and m2. All you need to solve for is a. | Douglas Thu. 25, Mar 2004, 14:03 |

I don't understand what u mean by effective operational mass in #5 of the postlab. I don't understand the question. Thanks! | Oladiran, Dami Sun. 28, Mar 2004, 22:56 damioladiran@yahoo.com |

What is the mass of the system, according to how it behaved.That is, NOT according to how it should have behaved, but how it actually did behave. i.e. Total Mass as measured by the mass balance (should have behaved.) Total Mass as found by the plot of measured data (actually behaved.) Thus called the effective, operational mass. | Douglas Mon. 29, Mar 2004, 08:19 |

1-We didn't have enough time to determine the amount of friction the pulley system generated. Without actually going back and doing it, is there a value you can provide so that I can get a theoretical value? 2- Which equation is which? What is the primary equation and which one is the theoretical? Would the theoretical be the delta mg= the sum of m's*a + 2f? | Nguyen, Lyna Tue. 30, Mar 2004, 22:51 stln74@aol.com |

Your Q2) The primary, theoretical equation (that's singular!) in words is:Applied Force = Newton's Second Law + Friction Δmg = Σm a + 2fBut, don't put in your values (just yet) for Σm and 2 f!!!! That wouldn't be theory any more.Your Q1) If you did the plot of Δmg vs. a, according to this theory equation, just referenced, the slope should be Σm and the vertical intercept should be 2f. Where f is the friction. So, I guess you measured friction without even knowing it. :)Thanks for the great questions. | Douglas Wed. 31, Mar 2004, 05:04 |

Our graph of applied force vs. accceleration is not linear and we don't understand why. Would you give us an idea as to what might be the problem with our data. If we didn't finish the uncertainty portion of our graph does this factor into our problem? | Rodriguez, Karen Sat. 02, Oct 2004, 15:35 karenjrdez@aol.com |

Karen, No, the uncertainties shouldn't cause the graph to be non-linear. What about your NSL vs. Acceleration. Is it linear? If only the Applied Force (Δm g) is non-linear, and if instead it appears quadratic, then there is a good chance you used Δm a instead of Δmg.Sorry, but that's the best I can do without actually seeing your data. | Douglas Sat. 02, Oct 2004, 15:42 |

Douglas, #8 in the post lab of atwoods machine, you ask for the slope of our plot, and what would happen to the slope if the total system mass was increased and why? You must restate the question within your paragraph. Are you asking if we used tons/slugs/etc. instead of kg? I'm not sure if it would make any difference, could you help me out on this please? thanks. | Anonymous Tue. 28, Jun 2005, 00:25 |

Johnny, No, I'm not asking about a different unit of mass, but a different value. Say your total mass was 0.225[kg]. What would happen to the plot of Delta(m)g vs. Acceleration, if you increased the system mass...to let's say...0.275[kg]? | Douglas Tue. 28, Jun 2005, 05:54 |

Douglas: I am not 100% clear on question #8 on the prelab. I'm not sure how you find the mass of the system from the graph. | McWilliams, Jill Thu. 12, Oct 2006, 00:44 jillybean5@msn.com |

If we plot Applied Force (Δ M * g) versus Acceleration, what is the slope? What is the vertical intercept? Here is our base equation: Applied Force = Newton's Second Law Force + Friction, We can rewrite the equation as such: Δ M * g = Σ M * Accel. + friction. Again, if we plot (Δ M * g) versus Acceleration, then what is the slope? What is the vertical intercept? slope = Σ M vertical intercept = friction in bearings. I hope this helps a little, :)oug | Douglas Thu. 12, Oct 2006, 13:13 |

Douglas, On question 5, we are given two lines. The dotted one seems to be the theoretical equation, and the solid is the linear fit, so neither of them seem to go with the actual data. And the actual points that are on the graph are going to be hard to get because we will be dealing with such small numbers, it will be hard to get a precice reading. If I am going in the wrong direction, please help me. Thanks | Anonymous Thu. 15, Mar 2007, 11:51 |

Neither of them seem to go with the actual data??? Hmmmm. The dotted line goes through the origin; it is likely theoretical. There are two linear equations given on the graph. If the linear equation (y=mx+b) on the graph represented this dashed line, then it would have a b=0 and R ^{2} = 1. Neither of these is the case for the y=mx+b given on the plot. The dashed line must therefore by theoretical (as compared to the solid line) and is described by the equation y=mx with R^{2} = 1.We are interested in the operational mass, or the effective mass, or the mass that is found for how the system behaves rather than how it is measured on the mass balance.The theoretical data was described by using the mass balance value of the system mass. But the system "behaved" or "operated" with a different mass value (different from the mass balance measurement) which appears on the plot by analyzing the real data's slope. The real data points appear linear, though some scatter is seen; and the solid line appears to describe those data points. R ^{2} = 0.9987 is pretty linear but does show scatter.If we plot ΔMg vs. acceleration, then what does the slope of the trend-line of this real data represent? (The vertical intercept represents a force offset, likely bearing friction.) | Douglas Fri. 16, Mar 2007, 07:40 |

Douglas, Can yuo give me a hint on how to determine the data equation for question #7 | Galindo, Cesar Wed. 28, Mar 2007, 11:15 cdgalind@ouray.cudenver.edu |

You plotted Applied Force versus Acceleration and obtained a best fit line using MSExcel. Let's say that output was y=0.2367x+0.2 The y is the Applied Force. The x is the Acceleration. The 0.2367 has units of kg. The 0.2 has units of N. Rewrite the equation using that information. The second equation you need to write is for the theory, or "NSL," or Newton's Second Law. It will look similar to the above equation that you create, only this new equation will have a slope equal to the average ΣM from the data table; and of course the intercept will be null. | Douglas Thu. 29, Mar 2007, 08:35 |

on our final x-cell chart page how many equations are we supposed to have? 2 or 3? and what are they? i have theory (physics equation) and a 'best fit" equation for our data's line, is there one that i'm missing? | Anonymous Thu. 06, Mar 2008, 14:51 |

Two equation are mandatory: Applied Force = ... NSL Force = ... Many people naturally have a third equation which is the output from the Spreadsheet's Linear Regression of the Applied Force (y = ...) | Douglas Fri. 07, Mar 2008, 09:54 |

I am confused about question #7, how do I find the Y-intercept. I am guessing that the acceleration I should use is the average of the trial that was repeated times. Is the y-intercept the standard deviation? | Griffith, Holly Sun. 09, Mar 2008, 00:35 holly.griffith@cudenver.edu |

The equations to list are supposed to be from the Force vs. Acceleration plot; not from the Velocity vs. Time plot. | Douglas Mon. 10, Mar 2008, 16:28 |

Hi Douglas Do I need a NSL Force for the plot of Force vs. Acceleration? If it is necessary, how can I get the equation? | Anonymous Sun. 09, Mar 2008, 21:59 |

Yes, you do need to list the NSL equation on the F vs. A plot. You built the N(ewton's)S(econd)L(aw) data using Avg. Total Mass * Acceleration. Therefore, you know precisely the equation that created the NSL data for the plot. Also note: Do *not* apply a trend line to this data, rather reformat the data to appear as it is, a line whose equation you know. | Douglas Mon. 10, Mar 2008, 16:31 |

I am not sure if I am calculating the Total Mass correct. It say Total Mass= sum M = Average (M1+M1). So what I did was add all of the Total Masses together so I got a total of 1.5365. Is this correct? | Anonymous Tue. 06, Oct 2009, 17:16 |

The Total Mass is the values representing the system mass for each trial. The Total Mass column should show fluctuation due to uncertainty in measurement. The Average Total Mass needs to be used in the NSL Force calculation in order to create perfectly linear data. Find the Average Total Mass by summing the individual total masses and dividing by the number of total masses you used. | Douglas Fri. 09, Oct 2009, 12:13 |

hello Sir writeup 1-I have some trouble to find "percent discrepancy from eachother" I did (=stdev(D12:D18)/D$20*100) but it seem not correct for me! please could you give me hint for that? 2-Stardard Deviation should I include the average acceleration? like (=stdv(B32:B36) or (B32:B37)? Sorry about that! thank you! | konan, Arnaud Fri. 05, Mar 2010, 12:52 karnaudmichel@yahoo.fr |

You have two values of Total Mass. One is the Average from the data table. The other is the slope of F vs. a. |Avg Mass - Slope| / Avg Mass * 100% | Douglas Mon. 08, Mar 2010, 09:06 |

Post-Lab Question 5. I think I am on the right track?? After reading your 'should have behaved'(Σm a + 2fp) and 'actually behaved'(Δm g)how would I derive a mass amount from this equation: (M2-M1)g = (M2+M1)a + Ff I don't know how to solve this for mass? | Anonymous Tue. 09, Mar 2010, 23:00 |

The actually was greater than the behaved massactual mass because the pulleys have mass.What are the units for the slope of (m2-m1)g versus acceleration? N / (m/s/s) = kg This slope is the effective, opperational, "all-told system mass" --- pulleys and all. I hope this helps. | Douglas Wed. 10, Mar 2010, 07:52 |

Hi Douglas, I'm having an issue with question 2 on the postlab telling us to list each measurement name, it's average, uncertainty etc. I've looked at the forum and also referred back to the Data & Graphs postlab question that is similar to this but I'm still lost. Do we list Mass1, Mass2 and the acceleration? Do we list the Total Mass and Acceleration? Do we pull the largest mass and the smallest mass? Do we use the values we calculated for the last trial that we had to do 5 times? | Anonymous Thu. 11, Mar 2010, 12:17 |

I recommend using Avg. Total Mass because you have a single value and an uncertainty and the rationale for the uncertainty (std. dev.). For Acceleration use the last five runs that used the same mass1 and mass2 for each run. Again, you have an average and standard deviation. | Douglas Thu. 11, Mar 2010, 14:25 |

I'm getting stuck on #6 and #7 on the prelab. 6) Mass1 = 0.106 kg; Mass2 = 0.121 kg; Acceleration = 0.585 m/s2; g = 9.8m/s2. What is the calculated applied force? (Neglect friction.) I'm using f=(m2-m1)*a, and not getting the correct answer. 7. Given: Mass1 = 0.111 kg; Mass2 = 0.116 kg; Acceleration = 0.202 m/s2; g = 9.8m/s2. What is the theoretical Newton's Second Law Force? (Neglect friction.) I'm using f=(m2-m1)*g, and not getting the correct answer. | Trezza, Nicole Wed. 02, Mar 2011, 06:44 nicole.trezza@gmail.com |

The applied force is the driving force. The difference in masses times the acceleration due to gravity is the driving force. It is the force that pulls the system. If the two masses m1 and m2 were equal, there would be no net force. If you shift 10g from mass1 to mass2 the difference in masses would be 20g and that difference in masses would allow the whole system (m1+m2) to accelerate; but not at rate of g. The pulling (or applied force) would be 20[g]*1[kg]/1000[g]*9.8[m/s/s]. Acceleration should be multiplied by Total Mass (M1+M2) to be the Net Force (via Newton's Second Law.) The pulling force (applied force) is ΔMg. That pulling force equals the sum of ΣM*a and 2f (where 2f is the friction in both pulleys.) | Douglas Wed. 02, Mar 2011, 07:45 |

How should we address significant figures throughout the Excel spreadsheet? For instance, our acceleration measurements give us three significant figures (for example, .296,.743, 1.19, etc.), but Excel only seems able to display them with a given number of digits after the decimal (e.g. 0.296, 0.743, 1.190) which implies greater accuracy. I've tried to manually set up each cell with the correct number of digits (by adjusting the number of digits to display after the decimal), but then I can't get the decimals to line up and it looks ugly :) Also, I can't figure out how to indicate that there are only 3 significant figures on the horizontal axis of our plot (Applied Force vs. Acceleration) Any suggestions? | Stephens, Jody Sun. 06, Mar 2011, 15:09 jsteph35@msudenver.edu |

Jody, These are great questions. The key word to remember with MSExcel is display. If not using the round function, MSExcel's display is different than the internally stored value.In this class our significant figures are important but second to uncertainty in measurement and error propagation. So, if we can measure to the nearest 0.001 [kg], then 0.040 [kg] and 1.128 [kg] both show the same measurement resolution yet different significant figures. 0.040 [kg] has 2 significant figures. 1.128 [kg] has four. But they both show a measurement to the nearest 1 gram. To format the numbers on the axes of a chart double-click the axis and format the Number option. Here you can set the display to be a special format. | Douglas Mon. 07, Mar 2011, 07:55 |

I'm having trouble with Question #4 on the postlab. I can fill in some values for the equation, Δmg = Σma + 2f, using the slope and y-intercept from the velocity graph. For example, (m_{2} - m_{1}) * 9.8 = (m_{2}+m_{1}).146 [m/s^{2}]+ .143[N]. But from there I can't figure out how to solve for either m. Any suggestions? | Stephens, Jody Sun. 06, Mar 2011, 16:59 jsteph35@msudenver.edu |

Question #4 deals with reading a Velocity vs Time plot from question #3. Simply read the legend which shows data was properly named by mass values instead of the ambiguous Run #1 type naming. | Douglas Mon. 07, Mar 2011, 08:03 |

Question 8 of prelab, from looking at the previous plea for help from a past year I understand that the slope of the line is the sum of mass but how do I get that to be just mass. I am very lost as to the route. | Hazeman, Angela Fri. 30, Sep 2011, 23:09 angela.hazeman@ucdenver.edu |

Δm * g = Σm * a + 2f If we plot (Δm * g) VERSUS (acceleration), what is the slope? | Douglas Mon. 03, Oct 2011, 12:51 |

Doug, I am confused on number 2 and 6 on the post lab. For #2, say for example we observed acceleration of our velocity vs time graph to be 0.0828 +/- .00076. Is that what I would use as my value and uncertainty? or would I say +/- .0001? And for #6, I'm confused on how to find acceleration of a single mass from a linear trendline. Y=0.2418x + 0.0145. M1= 0.0959kg M2= 0.1109kg. Find accel of mass 1. Can you point me in the right direction? Thank you! | Anonymous Mon. 10, Oct 2011, 10:31 |

Measurement uncertainty for Acceleration came from the five repeated v vs t plots created at the end of the data taking. The Standard Deviation (=STDEV(range)) tells us the uncertainty. If the stdev yields ±0.003874 then you would state Acceleration = 0.083[m/s/s] ±0.004 [m/s/s/] Notice the precision of my measurement 0.083 matches the precision of the uncertainty 0.004. Also note for this example the uncertainty has only one significant figure. Uncertainty teaches us two things: 1) what is the precision of the measurement (to what decimal place can we measure.) 2) at that decimal place, how do we count (by 1's, 2's, 3's, 7's, etc.)? Given y=mx+b and the values for m1 and m2 allows you to use the relationship: Δm g = Σm a + 2f The plot has given you values for Σm and 2f. Solve for acceleration. | Douglas Tue. 11, Oct 2011, 08:43 |

Doug, I am confused by prelab question #6. I used the equation F=(m1 + m2)*accel., however, this is not providing me with the correct answer. I used this same equation for #7 and g the correct answer. I must be missing something for the equation I'm using for number 6, any help would be great, thanks. | Anonymous Fri. 15, Jun 2012, 22:40 |

Thank you for the question. Question 6 says: What is the calculated applied force? Question 7 says: What is the theoretical Newton's Second Law Force? Applied Force and Newton's Force are distinct for this experiment. Newton's force is simply Newton's Second Law adapted to say the sum of the forces is equal to the product of total mass and acceleration. By acceleration we mean the actual acceleration experienced by the system. The applied force, on the other hand, is the pulling force. It is the force resulting from unbalanced forces where m1 and m2 are different values. The applied force is found to be equal to the difference in m2 and m1 multiplied by the acceleration of gravity. Consider two masses of equal value. What is the net pulling force? Zero Newtons. The system is balanced. Consider now the case where m2 and m1 are 210g and 190g respectively. Gravity looks at the system and sees a net weight of 20grams*acceleration of gravity. So we say 0.020 [kg] * 9.8 [m/s/s] = net applied force. Of course the system is not going to move with acceleration of 9.8 [m/s/s]. We use NSL force to find the actual acceleration afforded since total mass is far greater than the 20g difference. I hope this helps. If you want some entertainment, here is a video of my sons playing before a crowd of a couple thousand people this weekend. Have a great day, Douglas https://youtu.be/0WlZFsGelDk?hd=1 | Douglas Sun. 17, Jun 2012, 06:00 |

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