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Physics Q+A 2009


The Law

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How about this one

 

Boxes.jpg

 

Boxes A, B, and C each have a mass of 10 kg. If the boxes are acceelerated upward at 2 m/s^2 what is the tension, T2?

 

So, for this I said box B is my system. Forces upward is T2, forces downward are W of box B and Tension T3 (due to weight of box C)... So, I said:

 

mg + T3 + ma = T2

 

 

The one issue I have now, is whether ma corresponds to the mass of box B or if I have to factor in box B and C. I said box B only and did the calculation:

 

(10)(10) + (10)(10) + 10(2) = T2

220 = T2

 

 

The answer turns out to be 240N. Is this a mistake, or am I doing something wrong? I know had I factored in the weight of box B and C, then it would have been 240N... but I don't understand why I would have to do that since box B is my system. Helpp lawie is confused...

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u did it the longer way..u coulda simplified the diagram before solving. but anyway, if u do it this way you have to remember that T3 is not equal to the weight of Box C cuz remember the whole system is accelerating upwards. if everything was at rest, T3 would be 100N. with an upwards acceleration of 2, we have:

 

T3 - mg = m 2

T3- 100 = 20

T3 = 120N

 

now if u plug this value of T3 into your equation, it should give u 240N for T2.

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u did it the longer way..u coulda simplified the diagram before solving. but anyway, if u do it this way you have to remember that T3 is not equal to the weight of Box C cuz remember the whole system is accelerating upwards. if everything was at rest, T3 would be 100N. with an upwards acceleration of 2, we have:

 

T3 - mg = m 2

T3- 100 = 20

T3 = 120N

 

now if u plug this value of T3 into your equation, it should give u 240N for T2.

 

Ah, thanks I didn't factor in the acceleration. What's the easier way?

 

This topic (and energy) are definitely the hardest topics for me in the whole physics section. I suck at them.

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optics is the worst!

 

the easier way is if u forget about T3 and just combine Box B and Box C.

so then u have

T2 - mg = ma where m is the combined mass of boxes B & C

T2 - 200 = 20*2

T2 = 240N

 

It's not tooo bad if you use the EK method! Took me forever to get it down last time though.

 

:eek: YOU CAN COMBINE THEM?! How come? lol... seems sketchy to me!

 

 

Sigh... I need to get back on familiar soil and do some chem. Physics is gross.

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"The earth is ~ 80 times more massive than the moon. The avg distance between the earth and the moon is just less than 400 000km. If the radius of the earth is 6370 km, the center of gravity of the earth-moon system is located:

 

A) Center of the earth

B) Just beneath the earth's surface

C) Just above earth's surface

D) Exactly between the two

 

Answer: B"

 

I got it down to B and C because I know in a non-uniform density, the centre of gravity is closer to the more dense object. I don't really know how to figure out between B + C, is there an easy way to do something like this? I don't really get EK's answer.

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Well you've correctly deduced that it should not be A or D for obvious reasons. Next, we should intuitively reason that it should be located closer to the Earth because the Earth is much more massive.

 

The easiest way to go from there is first to look at the problem if both were the same mass. In this case, the sum of the masses is 2m (but we can ignore the 'm' because we've reasoned that it should be closer to Earth). So you have a total mass of 2 and a total distance of 400,000Km. So divide the two and get 200,000 Km as your center of mass if both were equal.

 

Now take this a step further and your total mass is now 81. Divide these: 400,000/81 and get ~4900km. So your center of mass is just below the surface of Earth.

 

If you could not make the bolded assumption above, then you'll have to determine the fraction of mass the moon represents to the whole system, then multiply by the distance (400,000km) to get the center of mass (because as the Earth increases in mass from parity, the center of mass gradually shifts toward the Earth's center). So, you would start by saying that the moon is 1 mass, and the Earth is 80 masses. Then find the fraction by: moon / earth + moon = 1 / 81 (This is where the mass cancels to give a ratio, which we did not have to worry about before because of the assumption).

 

Then multiply this by 400,000: 400,000 X 1/81 = ~4900km.

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Well you've correctly deduced that it should not be A or D for obvious reasons. Next' date=' we [b']should intuitively reason that it should be located closer to the Earth because the Earth is much more massive[/b].

 

The easiest way to go from there is first to look at the problem if both were the same mass. In this case, the sum of the masses is 2m (but we can ignore the 'm' because we've reasoned that it should be closer to Earth). So you have a total mass of 2 and a total distance of 400,000Km. So divide the two and get 200,000 Km as your center of mass if both were equal.

 

Now take this a step further and your total mass is now 81. Divide these: 400,000/81 and get ~4900km. So your center of mass is just below the surface of Earth.

 

If you could not make the bolded assumption above, then you'll have to determine the fraction of mass the moon represents to the whole system, then multiply by the distance (400,000km) to get the center of mass (because as the Earth increases in mass from parity, the center of mass gradually shifts toward the Earth's center). So, you would start by saying that the moon is 1 mass, and the Earth is 80 masses. Then find the fraction by: moon / earth + moon = 1 / 81 (This is where the mass cancels to give a ratio, which we did not have to worry about before because of the assumption).

 

Then multiply this by 400,000: 400,000 X 1/81 = ~4900km.

 

 

 

Wow, thanks so much for the great explanation. I fortunately was able to make the assumption above and the way you worded it makes so much sense. Thanks buddy. :)

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Wow, thanks so much for the great explanation. I fortunately was able to make the assumption above and the way you worded it makes so much sense. Thanks buddy. :)

 

Law, my offer still stands. Free tutoring for temporarily borrowing the keys to your bimmer. You gain a valuable asset for your mcat prep in exchange for a few hours lost away from your car. How can you turn down such a great deal. :D

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Law, my offer still stands. Free tutoring for temporarily borrowing the keys to your bimmer. You gain a valuable asset for your mcat prep in exchange for a few hours lost away from your car. How can you turn down such a great deal. :D

 

I would love a few hours of physics tutoring actually. After I cover some more things though and know my weak spots more. :P We'll have to find another method of payment though, I will NEVER give away the keys to my lady. :P

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oh Law

 

you remind me so much of myself! Those types of physics problems always stump me too...which is probably why I got at 8 in PS.:(

 

Unlike you however, I'm choosing to live in denial of having to study physics again and re-write...it's so much easier to focus on the masters research!

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any more questions?? je suis tres bored :P

 

Probably later tonight. :)

 

oh Law

 

you remind me so much of myself! Those types of physics problems always stump me too...which is probably why I got at 8 in PS.:(

 

Unlike you however, I'm choosing to live in denial of having to study physics again and re-write...it's so much easier to focus on the masters research!

 

There's hope Potter. Trust me. I suck at physics and the last time I wrote I managed to get an 11 in PS (up from a 9... and up from a FOUR the first time I ever did a PS section lol!). So hang in there... focus on the research, but don't keep your eyes of the prize. :)

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any more questions?? je suis tres bored :P

 

Je m'ennuie...

 

Don't let stuff get too rusty though' date=' otherwise it will be much harder to come back to testing mode.[/quote']

 

I'm with you, its a hell of a lot less work to keep it remembered than it is to relearn it.

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Probably later tonight. :)

 

 

 

There's hope Potter. Trust me. I suck at physics and the last time I wrote I managed to get an 11 in PS (up from a 9... and up from a FOUR the first time I ever did a PS section lol!). So hang in there... focus on the research, but don't keep your eyes of the prize. :)

 

11 on PS is impressive Law. The first time I wrote it I got 11 on PS too. And I consider myself kinda good at physics so good on ya! :)

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