Calculating Mass to Balance Electrostatic Field on See-Saw

[cerium]

b]1. Homework Statement [/b]
A see-saw with a central pivot is made of insulating material. the left hand side supports a conducting sphere of charge Q= 5.0x 10-6 which experiances an electrostatic force from an idetical sphere 10cm below with +Q
the right end has anouther conducting sphere of charge +3Q which is 10cm above a sphere with charge +4Q. A block is placed on the right hand side of the see-saw to balance the system and keep the rod horizontal.

Homework Equations

The Attempt at a Solution

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My question is calculate the mass of the block required to balance the see-saw. I have no idea were to start
Thanks in advance

 

[tiny-tim]

Hi cerium! Welcome to PF!

Show us how far you've got, and where you're stuck, and then we'll know how to help!

Start with the forces on the two spheres.

 

[cerium]

Thank you for that hopefully I can get going now

 

[cerium]

I have worked out the electrostatic force for both sides of the see-saw but I am unsure of how to proceed to get the mass of the block
Thanks in advance

 

[tiny-tim]

ok, now you have the three forces (two electrostatic and one gravitational), find the moment (the torque ) of each force about the pivot, and decide what the mass needs to be for them to balance.

 

[cerium]

Im having a mental block how can I work out gravitatinal force if I haven't been given any masses,

 

[tiny-tim]

uhh? the mass is what you have to find.

call it m.

 

[coz]

ok, now you have the three forces (two electrostatic and one gravitational)

Hi tiny-tim,
Can I just check, to find the electrostatic force on either side of the see saw would you use the equation:
F_el = 1/4pi$$\epsilon$$_0 x q_1xq_2/r_2

 

[tiny-tim]

Hi coz!

(are you the same person as cerium?)

Hi tiny-tim,
Can I just check, to find the electrostatic force on either side of the see saw would you use the equation:
F_el = 1/4pi$$\epsilon$$_0 x q_1xq_2/r_2

A little difficult to read (try using the X² and X₂ tags just above the Reply box ), but yes, that looks like Coulomb's law !

 

[coz]

Hi, no I am not the same person as cerium, just have a similar question to work out. I am new to this, so thanks for the tips!

For the LHS I have
[F][/el] = [8.988x10][/9] N m2 C-2 x ([5.0x10][/-6] x [5.0x10][/-6])/(0.1m)2
= 22.47N

For the RHS I have
[F][/el] = [8.988x10][/9] N m2 C-2 x ([1.5x10][/-5] x [2.0x10][/-5])/(0.1m)2
= 269.64 N

I must have done something wrong though, as this shows that electrostatic force is greater on the RHS than the LHS, so why would the block be placed on the RHS to balance the see-saw?

(i hope all my calculations come out correctly)

 

[coz]

Obviously I didn't use the tags correctly! How do they work then? Thanks again!

 

[cerium]

Hi Coz I worked out the same as you and I am still not sure how to proceed Have you had any joy yet

 

[coz]

Not yet, but I won't let it beat me! :)

 

[tiny-tim]

Welcome to PF!

I must have done something wrong though, as this shows that electrostatic force is greater on the RHS than the LHS, so why would the block be placed on the RHS to balance the see-saw?

hmm … something to do with the direction of the electrostatic force?

Obviously I didn't use the tags correctly! How do they work then? Thanks again!

Press the "QUOTE" button under this post, and you'll see lots of exciting tags!

 

[coz]

hmm … something to do with the direction of the electrostatic force?

Ok, so the charges are like and so they repel, and the RHS repels with greater force than the LHS, because the charge is higher there.

 

[coz]

ok, now you have the three forces (two electrostatic and one gravitational

The only equations I know for gravitational force are;

_Fgrav = -G_m1_m2/^r2
_Fgrav (on m at r) = mg(r)

But both of these need the mass to work out the gravitational force. I could use F=ma to work out the mass, but this seems too easy! Also you mentioned torque earlier, so I don't think this is right!


Could I make the electrostatic force on th RHS equal the gravitational force and then rearrange _Fgrav (on m at r) = mg(r)? Again this doesn't use torque, so I am assuming it is not right!

 

[tiny-tim]

Hi coz!

(erm … you're using the X² tag on the wrong bits! )

The only equations I know for gravitational force are;

_Fgrav = -G_m1_m2/^r2
_Fgrav (on m at r) = mg(r)

But both of these need the mass to work out the gravitational force.

As I said to cerium … the mass is what you have to find.

call it m.

And your formula is for a general radius r from the Earth's centre.

When you're here (on the Earth's surface), just use F = mg.

I could use F=ma to work out the mass, but this seems too easy! Also you mentioned torque earlier, so I don't think this is right!

Could I make the electrostatic force on th RHS equal the gravitational force and then rearrange _Fgrav (on m at r) = mg(r)? Again this doesn't use torque, so I am assuming it is not right!

?? you're just rambling … get some sleep! :zzz:

 

[coz]

Sorry for rambling!

m=F/g = 27.486kg

This seems too simple.

 

[tiny-tim]

m=F/g
…
This seems too simple.

Well, in this case all the distances happen to be the same, so it is that simple!

 

[coz]

Thank you so much for all your help :)

 

[sand]

Hi. I'm new but I wanted to clarify something in Coz's working. It looks like they used the electrostatic force of the LHS for the force in F=mg. But is that right? or should it be the difference between the LHS and the RHS? But that leads to me to wonder if you can just subtract forces like that?
Thanks

 

[tiny-tim]

Welcome to PF!

Hi. I'm new but I wanted to clarify something in Coz's working. It looks like they used the electrostatic force of the LHS for the force in F=mg. But is that right? or should it be the difference between the LHS and the RHS? But that leads to me to wonder if you can just subtract forces like that?
Thanks

Hi sand! Welcome to PF!

(i'm not sure what you're asking, but …)

technically, we weren't subtracting forces, we were subtracting moments of forces …

but the distances are all the same, so the moments are all proportional to the forces in this question.

 

[sand]

Thanks. That was helpful.
My understanding of this side of Physics isn't very good so I'm sorry if my questions aren't well worded.