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course PHY 201
10/08 8:30pm
`qx001. What were your counts for the 50 cm descent of the Atwood system?****
For the 1st run with 3 paper clips on each side, my count was 17 which corresponds to 2.55s.
For the 2nd run with an extra small clip on 1 side, my count was 8 which was 1s.
I suspect your counts are twice as long as you are figuring. No problem for the purposes of this analysis.
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`qx002. What were the two accelerations?
****
1st run
v0=0
`ds=50cm
`dt=2.55s
Vave=`ds/`dt=50cm/2.55s=19.6cm/s
vF=39.2cm/s
a=(vF+v0)/`dt=(39.2cm/s+0cm/s)/2.55s=15.4cm/s^2
2nd run
v0=0
`ds=50cm
`dt=1s
Vave=`ds/`dt=50cm/1s=50cm/s
VF=100cm/s
A=(vF+v0)/`dt=(100cm/s+0cm/s)/1s=100cm/s^2
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`qx003. Why did the systems accelerate?
****
The systems accelerated because the position of the clips was higher on 1 side than the other allowing the force of gravity to act greater on that side.
If the other side had been higher would it have accelerated in that direction?
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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?
****
1st run with 3 large paperclips
F=ma
F=10g*15.4cm/s^2
F=154g/cm/s^2
2nd run
F=ma
F=11g*100cm/s^2
F=1100g/cm/s^2
Good, but there were three large paperclips, not 1, on each side. Had there been just 1, your results would be correct.
We can of course set the system up with just one large clip on each side and see what happens.
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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?
****
Side 1 has a mass of 11g and side 2 has a mass of 10g.
A=net force/mass
A=1N/21g=1000g/m/s^2/21g
You do have 1000 g m/s^2; that isn't 1 N.
And it's g * m/s^2, not g / m/s^2.
A=47.6m/s^2
right acceleration, anyway (assuming the 10 g and 11 g masses)
So, for side 1:
F=ma=11g*47.6m/s^2=
523.6g/c/s^2=.52N
g * cm/s^2, and you get 523 dynes; that isn't .52 N, though
Side 2 has a mass of 10g
A=net force/mass
A=1N/21g=1000g/m/s^2/21g
A=47.6m/s^2
So, for side 2:
F=ma=10g*47.6m/s^2=
476g/m/s^2=.48N
To find net force on the system, I took the difference in the 2 masses on each side and used the following formula:
Fnet=ma
=1g*47.6m/s^2
=47.6g/m/s^2
=.05N
good, but your Newton isn't correct, and watch your unit g * cm/s^2
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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?
****
I think it is less than a 10-to-1 ratio. The accelerations for the 1st run was 15.4cm/s^2 and the 2nd run was 100cm/s^2. This ratio is about a 6-to-1 ratio.
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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?
****
Fnet=[(3M+m)-3M]*a
Fnet=[(3M+m)-3M]*g, not Fnet=[(3M+m)-3M]*a.
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`qx008. If the mass of 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?
****
I am not exactly sure how to determine the answer to this question. However, I would say yes because the system did accelerate with the 3 clips on each side because the system was not in equilibrium at the start.
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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.?
****
The 1st thing we did was measure the distance from the table to the top of the metal bracket on both sides of the balance and got 9cm. At a distance of 3cm from the top of the bracket, we didn’t get any response. We then placed the magnet above the metal bracket at a distance of 2.5cm and got very little response. Next, we moved the magnet to 2cm away from the top of the bracket and got a very “quick” response.
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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.
****
I didn’t actually measure this but my reasonable measurement was that the paper clip moved about 3 or 4 centimeters.
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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?
****
We placed the magnet on a meter stick above the top of the metal bracket and lowered the magnet on the meter stick until we got the system to respond to the magnet.
If we wanted to move the paper clip by 1 millimeter, we would just get the magnet probably about 2.5cm to 3cm away from the top of the metal bracket. This is where the system started to respond very lightly and I think could have been about 1mm.
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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.
****
First, we need to find the area of the circle. Using 1mm in diameter and the formula (pie)R^2, we get:
(3.14).5cm^2=.785cm^2=area
that would be .785 mm^2
Next, we find the volume displaced. Assuming the paperclip moved 3cm with the magnet about 2cm away from the top of the bracket, we multiply the displacement by the area to get:
3cm*.785cm^2=2.36cm^3=volume
To find the force the magnet exerted, we use:
F=ma
F=2.36cm^3*980m/s^2
g = 980 cm/s^2, not 980 m/s^2
you don't get a force by multiplying a volume by the acceleration of gravity
the mass of 2.36 cm^3 volume of water is 2.36 grams, and you would get a force by multiplying this by g.
F=2312.8cm^3*m/s^2
the mm^3 would make a big difference; you would end up with about 2.3 dynes
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I’m not sure if on this problem, I calculated the wrong things. I’m also not sure what units to express the answer in either.
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you're good. See my note.