#$&*
course Phy 241
October 8 around 11am.
Here are questions related to the lab activities for the 10/04 class:Atwood System:
The Atwood system consists of the paperclips suspended over the pulley. A total of six large clips connected by a thread were suspended, three from each side of the pulley. The system was released and, one side being slightly more massive than the other due to inconsistencies in the masses of the clips, accelerated from rest, with one side descending and the other ascending. The system accelerated through 50 cm in a time interval between 4 and 6 seconds; everyone used their 8-count to more accurately estimate the interval. Then a small clip was attached to the side that had previously ascended. This side now descended and the system was observed to now descend is an interval that probably lasted between 1 and 2 seconds.
If you weren't in class you can assume time intervals of 5 seconds and 1.5 seconds. Alternatively you can wait until tomorrow and observe the system yourself; the initial observation requires only a couple of minutes.
`qx001. What were your counts for the 50 cm descent of the Atwood system?
****
4.76 seconds for the 3 large paperclips (0.17 seconds for 1 count)
1.7 seconds for the added small paperclip
#$&*
`qx002. What were the two accelerations?
****
4.41 cm/s^2
34.6 cm/s^2
#$&*
`qx003. Why did the systems accelerate?
****
The systems accelerate because there was one side that had more mass than the other which caused one side to accelerate downward and the other one upward and the acceleration due to gravity.
#$&*
`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?
****
Fnet=m*a
264.6 g*(cm/s^2)
2110.6 g*(cm/s^2)
#$&*
.. if uncertainty +-1%
`qx005. Given the masses assumed in the preceding, what is the force acting on each side of the system? What therefore is the net force on the system?
****
132.3 g*(cm/s^2)
I don’t know if I’m doing what this is asking for, but what I did was find the net force of one side. So, instead of saying 60g of paperclips total, one side would be 30g of paperclips. The other system with the added small paperclip, one side will be 31g. For example:
Fnet=(30g)*(4.41cm/s^2)
=132.3 g*(cm/s^2)
I think I need to use 9.807cm/s^2 somewhere though.
Net force on the system:
Fnet=(30g)*(34.6cm/s^2)
=1038 g*(cm/s^2)
Fnet=(31g)*(34.6cm/s^2)
=1072 g*(cm/s^2)
1072-1038=34.6 g*(cm/s^2)
What am I doing wrong? Or is this right?
Based on the 1000 cm/s^2 approximation for g:
If there are 30 g on one side of the system then the force on that side is 30 g * 1000 cm/s^2 = 30 000 g cm/s^2, or 30 000 dynes, or .30 Newtons.
If there are 31 g on the other side of the system then the force on that side is 31 g * 1000 cm/s^2 = 31 000 g cm/s^2, or 30 000 dynes, or .31 Newtons.
The two forces have opposite effects, so the magnitude of the net force would be the 1000 dynes = .01 N difference.
#$&*
`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?
****
Less than 10-to-1 ratio
you observed a difference on the order of 2000 g cm/s^2 = 2000 dynes
what does this tell you?
#$&*
`qx007. If the mass of each larger clip is M and the mass of a smaller clip is m, what would be the expressions for the net force accelerating the system? What would be the expression for the acceleration of the system?
****
6 large paperclips:
Fnet=(6M)(a)
A=(Fnet)/(6M)
6 large+1 small paperclips:
Fnet=(6M+m)(a)
A=(Fnet)/(6M+m)
you have mass 3 M on one side and 3 M + m on the other. The resulting gravitational forces work against, not with, one another.
#$&*
`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?
****
Yes
#$&*
... sample the accelerations for random divisions of the six large clips ... predict what the distribution of masses would look like ...
Magnet and Balance
Everyone was given a small magnet and asked to achieve a state where the balance was in an equilibrium position significantly different from that observed without the magnet. It was suggested that the length of the suspended clip beneath the surface of the water should differ by at least a centimeter.
... assuming 1 mm diam ...
`qx009. Describe in a few lines your efforts to achieve the desired result. What worked, what didn't, what difficulties presented themselves, etc.?
****
What I did was use a meter stick and held it up right next to the balance. I then took the small magnet and started it at a position where it didn’t “interact” with the balance and kept moving it down toward the balance until we reached a point where the magnet “grabbed” it and the balance remained stable. One difficulty was figuring out which side of the magnet was attracted to the balance, without messing up the entire balance.
#$&*
`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 measure this but when the magnet was 2.5 cm away from the balance, there was a slight movement, not much at all, so my estimate would be less than 2 mm.
#$&*
`qx011. How did you adjust the magnet? If you wanted to quickly increase or decrease the length of the suspended paper clip beneath the surface by 1 millimeter, using only what you had in front of you during the experiment, how would you go about it?
****
The magnet was approx. 2.5 cm away from the balance for it to start a slight movement toward it.
You would think, if you wanted to increase, for example, the length of the suspended paperclip by 1 mm, you would have to get the opposite side to go down that much too. I don’t know how you would do that perfect with the magnet, though.
#$&*
`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.
****
I don’t think I did my experiment right because I didn’t measure the clip before and after the magnet was in play. However, here’s what I DO know: The buoyant force on the clip should’ve changed by the water displaced by the object. You also, have an upward force (the buoyant force) and a downward force (gravity and force of magnet).
#$&*"
Good. See my notes, especially about the Atwood system.
#$&*