Physics Question?

[taki]

A student performs Coulomb's classic experiment. Shes adds Y coulomb of positive charge to a small metal sphere that is suspended by a thread. Shes adds X coulomb of positive charge to a small metal sphere that is held on a glass rod. She positions the X coulomb sphere as indicated in the diagram and because like charges repel eachother the Y coulomb sphere is pushed out to the right.

 

Calculate the mass of the Y coulomb spere. hint:It is bigger than you might guess.

 

angle=30 degrees

X = 3

Y = 6

l

l 30 cm

l

l

O O

X Y

[taki]

Ok maybe someone know how distance, coulomb force, and mass is related?

[Pyrotex]

Ok maybe someone know how distance, coulomb force, and mass is related?

Let me dust off my favorite Physics text book. What fun!!

[taki]

Can someone tell me if this is right?

 

30 cm = 0.3 m

(0.3m)(tan 30) = 0.17 m (distance)

 

F= (9.0x10^9N.m^2C^-2)(3.00 C)(6.00 C)/ (0.17m^2) = 5.6e12 N

 

(coulomb's law)^^

 

1N = 1 kg*m/s^2

 

F=mg

m = F/g

m = (5.6e12 kg*m/s^2)/(9.8 m/s^2) = 5.7e11 kg

[taki]

Anyone? Maybe someone can tell me if I am to determine the mass, do I only consider the forces in the x-direction. Or the sum of the forces in both directions. I would think F=mg would be enough to relate the coulmb force sphere Y "feels" and mass of the sphere. However I think maybe the tension should be considered. Someone please help.

[taki]

Anyone know about Coulomb's law, mass, distance?

[Jay-qu]

Ok buddy, slow down, your questions will be answered, you just need to give us some time :cup:

 

From what I can surmise of your diagram, the Y sphere has been pushed out and now subtends a 30 degree angle with the vertical - is this correct?

 

Yes 30cm is .3m :shrug:

[taki]

Ok buddy, slow down, your questions will be answered, you just need to give us some time :cup:

 

From what I can surmise of your diagram, the Y sphere has been pushed out and now subtends a 30 degree angle with the vertical - is this correct?

 

Yes 30cm is .3m :shrug:

 

 

 

Yes, sorry the diagram is not so good.

[Jay-qu]

Assuming a few things along the way, this is what I have got to:

[Jay-qu]

LOL :doh:

 

accidently did the wrong one..

 

[math]cos{30}=\frac{mg}{T}[/math]

[math]m=\frac{T}{g}cos{30}[/math]

[taki]

Assuming a few things along the way, this is what I have got to:

 

 

 

YES, that is exactly right. First mistake I made was I used tan instaed of sin (thanks).

 

Now since I can calulate the Coulomb's force on Y, can I relate it to F=mg? I am concerned that maybe I didn't consider all the forces in the X and Y direction, like tension. Then again when I think about "mg" it seems it only has do with the Y direction.

[Jay-qu]

sorry see post #10 :shrug: I was un-needingly complicating the matter!

[taki]

sorry see post #10 :shrug: I was un-needingly complicating the matter!

 

What do you put in for tension "T" (for this particular problem)? From what I can see in your diagram you used the length of the string (.3m) for T. Is that correct?

[Jay-qu]

No, this is the part where I am wrong.. I have been trying to work it out, when I get home I will have a look in my physics book!

 

It is definitely not .3m, for mass to be correct you must know the T in newtons..

 

:shrug:

 

Ill be back

 

J

[ronthepon]

For problems involving three forces in equilibrium (like these) you can use the wonderful Lami's theorem.

 

I'll use JQ's picture as a reference.

 

The forces [math]T[/math], [math]c[/math] and [math]mg[/math] are in equilibrium.

 

By the usage of elementary geometry, you'll figure that there's an angle of 30+90 = 120 degrees between [math]c[/math] and [math]T[/math].

 

Ofcourse, that's when you assume that [math]c[/math] and [math]mg[/math] are at 90 degrees to each other. What remains is the angle between [math]T[/math] and [math]mg[/math], and it's 360 - (120 + 90) = 150 degrees.

 

Now the only thing remaining for you to do is use Lami's theorem (for three forces in equilibrium)

 

It's going to be written out as:

 

[math] \frac{T}{Sin90} = \frac{mg}{Sin120} = \frac{c}{Sin150} [/math]

 

Use the values you know, and find all the unknowns. For more info on Lami's theorem, do see the wiki link I've given.

[ronthepon]

And, er, for general problems like these (eg: wit more or less than just three forces) You can perform a vector resolution of all the forces present in any two mutually perpendicular directions.

 

Have you studied vectors, taki?

[Jay-qu]

I am not so sure our approach will work in this case ron.. well not yet at any rate, we first need to find what C or T is