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Physics help

Blogs > Ftrunkz
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Ftrunkz
Profile Blog Joined April 2007
Australia2474 Posts
November 06 2009 02:33 GMT
#1
Studying for end of year exams and im in my home town (2 hours away from uni and anyone i know who could help me) so figured i'd ask for help here. I have a feeling the question is way more basic than im making it and im just owning myself, i really dont know.

[image loading]


any help really appreciated, answer is: 7.17*10^-8 C

*
@NvPinder on twitter | Member of Gamecom Nv | http://www.clan-ta.com | http://www.youtube.com/user/ftrunkz | http://www.twitchtv.com/xghpinder
StorrZerg
Profile Blog Joined February 2008
United States13919 Posts
November 06 2009 02:37 GMT
#2
k so you have the answer so?
Hwaseung Oz fan for life. Swing out, always swing out.
Ftrunkz
Profile Blog Joined April 2007
Australia2474 Posts
November 06 2009 02:42 GMT
#3
I cant work out how to get the answer... It's a previous year's exam question, so i have the answers, not solutions tho =[.
@NvPinder on twitter | Member of Gamecom Nv | http://www.clan-ta.com | http://www.youtube.com/user/ftrunkz | http://www.twitchtv.com/xghpinder
Hot_Bid
Profile Blog Joined October 2003
Braavos36390 Posts
November 06 2009 02:43 GMT
#4
fyi, blog stays open because he's actually asking for help understanding the process of the work rather than trying to find a fast answer
@Hot_Bid on Twitter - ESPORTS life since 2010 - http://i.imgur.com/U2psw.png
Cloud
Profile Blog Joined November 2004
Sexico5880 Posts
Last Edited: 2009-11-06 02:48:26
November 06 2009 02:45 GMT
#5
I'm guessing you can calculate the horizontal force of each balloon (it's direction pointing toward the other balloon) by calculating the horizontal component of the vector caused by the weight of the 5 g object (its direction is the same as the rope, from the balloon towards the object, with a magnitude of 5 grams multiplied by gravity).

Then you say that force is equal to the magnetic force calculated with the Lorentz equation or something.

To be honest, I saw this shit like 5 years ago so I can't point to you exactly what to do.
BlueLaguna on West, msg for game.
Freyr
Profile Blog Joined July 2004
United States500 Posts
Last Edited: 2009-11-06 02:55:26
November 06 2009 02:54 GMT
#6
You know the vertical component of the force on each balloon (the buoyancy of two balloons is counteracting the gravitational force on the weight), and you know the angle thanks to pythagoras.

Use this information in conjunction with Coulomb's law (electrostatics, not magnetism) and the wonders of trigonometry to solve for the charge (which is equal in each balloon).

Essentially what Cloud said, minus the magnetism bit.
GOB
Profile Joined September 2009
50 Posts
November 06 2009 02:59 GMT
#7
[image loading]
Come On!
micronesia
Profile Blog Joined July 2006
United States24753 Posts
Last Edited: 2009-11-06 03:18:39
November 06 2009 03:02 GMT
#8
Just got back from parent teacher conferences (lol) so I'll edit something in in a minute.

edit: The steps I used to do it symbolically (not using a calculator right now)

1) Split the triangle into two right triangles
2) Call the hypotenuse L (which is 100) and one of the legs d/2 (since the entire base of the triangle is 30, d/2 is 15... it's probably smarter to define the entire base of the triangle as 2d but oh well)
3) Find the third leg by using the pythagorean theorum
4) Set up free body diagrams for all three objects... just to check: The mass has a downward force of gravity m*g, and two tension forces each going at an angle that you can calculate by using the triangle from above.
5) Solve for T using this free body diagram.
6) The free body diagram for a balloon should be an upward buoyant force (no downward gravity force since we just assume the buoyant force is that much weaker to compensate), a lateral electrostatic force repelling the opposite balloon, and tension going down at an angle and away from the electrostatic force.
7) Add up the forces in the x direction using Newton's second law. Plug in what you know for T, cos of the angle you can find by using your triangle from earlier, and the other force is simply Coulomb's law kQ^2/d^2.
8) Solve for Q.
ModeratorThere are animal crackers for people and there are people crackers for animals.
Ftrunkz
Profile Blog Joined April 2007
Australia2474 Posts
Last Edited: 2009-11-06 03:22:19
November 06 2009 03:06 GMT
#9
thanks for responses, i've atleast been approaching the question in the right way then :D.

wall of possibly incomprehensible working coming, if anyone can decipher and manage to see where i've gone wrong or where my understanding is completely messed up it'd be greatly appreciated =]:

the force due to the charges of the balloons on each other is:
F= (Q^2 * K) / r^2 = (Q^2 * 8.99 * 10^9) / .36 = 2.50*(10^10) * Q^2

so that force must equal the horozontal force from the tension in the strings.

the verticle componant of the strings combined is mg of the little 5g triangle thing.
therefor the verticle componant on one of the strings is F=(.005 * 9.8) / 2 = 0.0245

then i need to work out the triangles, doing so ended up getting me a 1:0.8 ratio (whole string vs verticle componant). So the tension force in the string is 0.0245 * (1/.08) = 0.03

thennnn i need to work out the horozontal componant, which from the diagram in the question is a ratio of 100:60, so 0.03 * .6 = 0.018375 (i left out a lot of figures earlier, adding them now)

so, that 0.018375 = electrostatic force = 2.5 * 10^10 Q^2

solving for Q gave me 8.57*10^-7, unfortunatly not the answer in the... answers... =[.
@NvPinder on twitter | Member of Gamecom Nv | http://www.clan-ta.com | http://www.youtube.com/user/ftrunkz | http://www.twitchtv.com/xghpinder
gyth
Profile Blog Joined September 2009
657 Posts
November 06 2009 03:08 GMT
#10
I get 5.6e-7C
Are you sure your answer is right?
The plural of anecdote is not data.
paper
Profile Blog Joined September 2004
13196 Posts
November 06 2009 03:09 GMT
#11
On November 06 2009 11:59 GOB wrote:
[image loading]


two circles and your first reaction is mouse eyes?
Hates Fun🤔
NukezaFlyin
Profile Blog Joined May 2009
United States10 Posts
November 06 2009 03:12 GMT
#12
I think your ratios are wrong.
randombum
Profile Blog Joined April 2007
United States2378 Posts
November 06 2009 03:14 GMT
#13
Lol Micronesia always finds these physics threads so quickly. Do you do a search for them whenever you log on or something?
Lemonwalrus
Profile Blog Joined August 2006
United States5465 Posts
November 06 2009 03:14 GMT
#14
Whenever one is made someone pings him on irc.

Seriously.
Ftrunkz
Profile Blog Joined April 2007
Australia2474 Posts
November 06 2009 03:19 GMT
#15
On November 06 2009 12:08 gyth wrote:
I get 5.6e-7C
Are you sure your answer is right?

I'd THINK they would be, since they are probably the ones used to mark the exam in 2007. But it's not outside the realm of possibility =o. If someone else confirms your answer i'll gladly assume its right over the ones givin, haha

and to the guy that said my ratios are wrong: That's probably likely, since i dont know where the hell else i'm going wrong =[
@NvPinder on twitter | Member of Gamecom Nv | http://www.clan-ta.com | http://www.youtube.com/user/ftrunkz | http://www.twitchtv.com/xghpinder
micronesia
Profile Blog Joined July 2006
United States24753 Posts
November 06 2009 03:19 GMT
#16
On November 06 2009 12:14 randombum wrote:
Lol Micronesia always finds these physics threads so quickly. Do you do a search for them whenever you log on or something?

I would have been the first response except I was driving home from parent conferences when this thread was made :p

btw I updated my earlier post to show some suggested steps
ModeratorThere are animal crackers for people and there are people crackers for animals.
gyth
Profile Blog Joined September 2009
657 Posts
November 06 2009 03:20 GMT
#17
Its not a 60:80:100 triangle
The plural of anecdote is not data.
Ftrunkz
Profile Blog Joined April 2007
Australia2474 Posts
Last Edited: 2009-11-06 03:29:41
November 06 2009 03:24 GMT
#18
CRAP figured out where i went wrong i think, as others said its my triangles, i shouldnt of been using 60:100 it should be 30:100 since 60 is horozontal distance between the balloons, not the horozontal distance to the triangle. _should_ be able to get the answer now.

Thanks heaps for the speedy replies everyone much appreciated :D
edit: wait, now im getting gyth's answer, 5.6*10^-7C... lol. I guess the answers were wrong? Either way the questions killed me enough over the past hour or two that i'll assume that answer is right and move on, haha... and yay dragoon.
@NvPinder on twitter | Member of Gamecom Nv | http://www.clan-ta.com | http://www.youtube.com/user/ftrunkz | http://www.twitchtv.com/xghpinder
micronesia
Profile Blog Joined July 2006
United States24753 Posts
Last Edited: 2009-11-06 03:49:32
November 06 2009 03:49 GMT
#19
On November 06 2009 12:08 gyth wrote:
I get 5.6e-7C
Are you sure your answer is right?

Hm I got the same thing using google to calculate everything (which sucks so I have no confidence in this).

I'm trying to decide if there is a mistake in this method...

sqrt((2*.005*9.8*.3^3)/(9*10^9*sqrt(1-.3^2)))
ModeratorThere are animal crackers for people and there are people crackers for animals.
AcrossFiveJulys
Profile Blog Joined September 2005
United States3612 Posts
November 06 2009 03:55 GMT
#20
On November 06 2009 11:59 GOB wrote:
[image loading]


Rofl. So I guess the whole face is edible?
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
Last Edited: 2009-11-06 04:06:49
November 06 2009 04:06 GMT
#21
equilibrium so basically
sum all force, 0

forces will be gravity and charge and helium_floating and tension
break tension force on string into 2 parts, horizontal and vertical.


Sooo

Oh the left helium baloon:
4 forces:
1) baloon floating up
2) tension from string pulling down
3) tension from string pulling right
4) electricStatic force pushing right
They sum to 0

On the right balloon
1) same
2) same
3) same but left
4) same but right
They sum to 0

On the mass
1) tension of 2 strings combined, upward
2) Tension of left string going left
3) Tension of right string going right
notice 2,3, cancle
4) Gravity going down
They sum to 0

Unknowns are charge force, helium floating force, rope tension force. 3 unknowns, and you have 3+ equations.
Solve.
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
SanguineToss
Profile Blog Joined August 2009
Canada815 Posts
November 06 2009 04:13 GMT
#22
evan are u a teacher?
StorrZerg
Profile Blog Joined February 2008
United States13919 Posts
November 06 2009 04:13 GMT
#23
Good luck with the test mate, your in physics c? right?
Hwaseung Oz fan for life. Swing out, always swing out.
gyth
Profile Blog Joined September 2009
657 Posts
November 06 2009 04:20 GMT
#24
I think we've agreed on the correct way to solve the problem.
We just don't know why the solution is wrong.
The plural of anecdote is not data.
Ftrunkz
Profile Blog Joined April 2007
Australia2474 Posts
Last Edited: 2009-11-06 04:25:55
November 06 2009 04:24 GMT
#25
On November 06 2009 13:13 StorrZerg wrote:
Good luck with the test mate, your in physics c? right?

umm, first year second semster physics, it's called "Physics 2: Physical Science & Technology" lol. It's just the standard second semestre physics subject at my uni (otherones are like biological physics and advanced).

edit: re-reading the problem, it says the weight at the bottom "weighs" 5g... not that it's "mass" is 5g... could this be our source of error? (ie, force should be 5*9.8, not 0.005*9.8?)
@NvPinder on twitter | Member of Gamecom Nv | http://www.clan-ta.com | http://www.youtube.com/user/ftrunkz | http://www.twitchtv.com/xghpinder
micronesia
Profile Blog Joined July 2006
United States24753 Posts
November 06 2009 04:49 GMT
#26
On November 06 2009 13:24 Ftrunkz wrote:
edit: re-reading the problem, it says the weight at the bottom "weighs" 5g... not that it's "mass" is 5g... could this be our source of error? (ie, force should be 5*9.8, not 0.005*9.8?)

Eh, you can try substituting in the new number and see what you get but, that would be an incredibly stupid thing to do...
ModeratorThere are animal crackers for people and there are people crackers for animals.
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
November 06 2009 05:07 GMT
#27
On November 06 2009 13:13 SanguineToss wrote:
evan are u a teacher?

nope but I tutor math, or used to.
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
Hiphopapotamus
Profile Joined July 2009
United States121 Posts
November 06 2009 06:32 GMT
#28
Yeah, I also get Q = 5.55 * 10^-7 using the below three equations...

arcsin(a) = 30/100

mg = 2Tcos(a)

Tsin(a) = kQQ/r^2
My lyrics are bottomless!
gyth
Profile Blog Joined September 2009
657 Posts
November 06 2009 06:34 GMT
#29
They used 60/(100*100) for r^2.
If it wasn't your teacher that made the mistake you might be able to wiggle some bonus out of it.
The plural of anecdote is not data.
GTR
Profile Blog Joined September 2004
51541 Posts
November 06 2009 06:35 GMT
#30
i'm so happy i dropped physics for my hsc, it's just nothing but maths and theory =[
Commentator
Divinek
Profile Blog Joined November 2006
Canada4045 Posts
November 06 2009 10:36 GMT
#31
On November 06 2009 15:35 GTR wrote:
i'm so happy i dropped physics for my hsc, it's just nothing but maths and theory =[


Which is what's so wonderful about it!
Never attribute to malice that which can be adequately explained by stupidity.
Oh goodness me, FOX tv where do you get your sight? Can't you keep track, the puck is black. That's why the ice is white.
micronesia
Profile Blog Joined July 2006
United States24753 Posts
November 06 2009 11:53 GMT
#32
On November 06 2009 15:35 GTR wrote:
i'm so happy i dropped physics for my hsc, it's just nothing but maths and theory =[

This is so wrong!
ModeratorThere are animal crackers for people and there are people crackers for animals.
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