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course Phy 241
10/12 1
`qx001. What were your counts for the 50 cm descent of the Atwood system?****
5sec , 1.5sec
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`qx002. What were the two accelerations?
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4cm/s^2, 44.44444cm/s^2
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`qx003. Why did the systems accelerate?
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The mass of one side was greater than the mass of the other, therefore the weight which is the gravitational force that the earth exerts on the mass was higher on one side. That force pulled the side with the greater weight to the ground. When the paper clip was added to the one side it increased the mass which increased the weight which increased the gravitational force and caused the one side to accelerate faster in the second trail.
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`qx004. Suppose the large paperclips all had mass 10 grams, the small clip a mass of 1 gram. What then was the net force accelerating the system on the first trial, and what was the net force on the second?
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Because m*a(vector) =Fnet(vector), we can convert the mass from grams to kilograms and easily find Fnet
1st : Fnet = 4cm/s^2*.03kg = .12N
2nd : 44.4cm/s^2*.031kg = approx 1.3778N
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.. if uncertainty +-1%
`qx005. Given the masses assumed in the preceding, what is the force acting on each side of the system? What therefore is the net force on the system?
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Smaller mass side of 1st = -.3N, larger mass = .42N, Fnet = .12N
Smaller mass side of 2nd = -.3N, larger mass = 1.6778N, Fnet = 1.3778N
Each side had only 3 large clips. Masses would have been .03 kg and .031 kg.
You based your answer on the .12 N net force calculated in the preceding. We can work with that.
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`qx006. Based on your counts and the resulting accelerations, do you think the ratio of the masses of the large to small paperclips is greater than, or less than, the 10-to-1 ratio assumed in the preceding two questions?
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I believe that the ratio of 10-to-1 is greater than true ratio. Maybe more like 4-to-1 or even 3-to-1.
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`qx007. If the mass of each larger clip is M and the mass of a smaller clip is m, what would be the expressions for the net force accelerating the system? What would be the expression for the acceleration of the system?
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Fnet = [(3M+m)-3M]a(vector) = ma(vector), where m is mass of smaller paper clip.
a = Fnet/(3M+m)-3M = Fnet/m, where m is mass of smaller paper clip.
the denominator is the total mass being accelerated, which is all the mass, M + m
so a = m a / (M + m)
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`qx008. If the mass of the each of the larger clips is considered accurate to within +-1%:, would this be sufficient to explain the acceleration observed when 3 large clips were hung from each side?
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At +-1% I’m not sure we should have gotten a accel as high or that high of Fnet for the system.
??????How to go about testing to see that your observations that you calculated are within that +-1% range??????????????????????????????
To do this right would get into some statistics, but for starters you could assume that the mass of 1 side is greater than the mass of the other by 1% of the mass of a single clip, see what the acceleration is, and get a ballpark idea of what that much extra mass would be.
Then you could ask yourself how much mass difference you would reasonably expect if six clips, each with a mass up to 1% higher or lower than the average clip mass, were randomly divided into groups of 3.
That quickly gets into more statistics than we want to mess with at this point, but you can still make reasonable ballpark assumptions.
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... sample the accelerations for random divisions of the six large clips ... predict what the distribution of masses would look like ...
Magnet and Balance
Everyone was given a small magnet and asked to achieve a state where the balance was in an equilibrium position significantly different from that observed without the magnet. It was suggested that the length of the suspended clip beneath the surface of the water should differ by at least a centimeter.
... assuming 1 mm diam ...
`qx009. Describe in a few lines your efforts to achieve the desired result. What worked, what didn't, what difficulties presented themselves, etc.?
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I finally got it to work only after I helped to stabilize my balance after magnet was placed where it needed to be. The most difficult aspect to me was that there seemed to be some point at which my balance would go from being hyposensitive to being hypersensitive. This made it very hard to control the reaction of the beam.
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`qx010. How much difference was there in the length of clip suspended in the water? If you didn't actually measure this, give a reasonable estimate.
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It should have been a little greater than the 1cm, maybe around 1.5cm?
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`qx011. How did you adjust the magnet? If you wanted to quickly increase or decrease the length of the suspended paper clip beneath the surface by 1 millimeter, using only what you had in front of you during the experiment, how would you go about it?
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Trail and error seemed to be the only method that was available. I tried to find the point at which the magnet started to have an effect. I would continuously reset my balance and make minor adjustments to the magnet until the desired result was obtained.
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`qx012. Assuming the diameter of the suspended clip to be 1 millimeter, by how much did the buoyant force on the suspended clip change? How much force do you therefore infer the magnet exerted? If you have accurate measurements, then use them. Otherwise use estimates of the positions of various components as a basis for your responses.
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1st we find volume displaced, so from cross section we find area
A = `pi*(.5millimeter)^2 = approx .7854millimeter^2 = .07854cm^2
Assuming clip raised 1cm we mult by length of paper clip raised to find volume
V = .07854cm^2*1cm = .07845cm^3, which is the same as .07854grams
When balanced the weight or gravitational force is mass *g(9.8m/s^2)
Weight = m*g = .07854g(which is .0007854kg)*9.8m/s^2 =approx .0077N
Boyuoyant force is equal to weight force to stay level so that
Boyouyant force – Weight force = .0077N-.0077N = 0N
There will be one less cm of displaced water but I am not sure how to calculate force needed to find force change??????
The clip displaced an additional .0077 N of water, so the buoyant force changed by .0077 N.
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"
&#Your work looks good. See my notes. Let me know if you have any questions. &#
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