# Homework Help: A rod with two balls on each side - what do the scales on each side show?

1. Mar 16, 2010

### Ryker

1. The problem statement, all variables and given/known data
There's a rod, 1 m long and by itself weighing 1 kg, with two balls attached at each of the sides, one weighing 2 kg, and the other 4 kg. The balls are put upon two scales. What force is applied upon each of the scales (e.g. what do the scales show)?

2. Relevant equations
This is a problem that I'm trying to resolve not putting exact numbers to it, but just mentally to get to see how the forces work. So right now I don't have any equations to operate with.

3. The attempt at a solution
I was thinking each of the scales would show at least the weight x g of each of the balls. What I'm having problems with is how much of the weight of the rod would "show". Would I be correct in assuming that I need to find the center of gravity of the said rod (when the balls are attached to it) and then add the weight (force of gravity) of the rod, left and right of the said center, to the respective scales (weight of the rod's part left of the center to the left's ball weight and the rod's part right of the center to the right ball's weight)?

I'm sorry if I didn't put the question and the attempt at the solution in proper english terms, but I'm not yet familiar with them.

2. Mar 16, 2010

Unless the question states otherwise assume that the centre of gravity of the rod is at its geometrical centre.Your attempt seems to be okay but to see it more clearly try taking moments.Remember that weight=Mg.

3. Mar 16, 2010

### Jmf

A good place to start is to notice that this is a statics (nothing's moving) problem in (effectively) two dimensions.

You have two unknowns (the forces on the rod from the scales), and so if you can come up with two equations linking them then you have enough information to solve the system. You don't _need_ to find the centre of gravity, but it might help your understanding of the problem if you did. :)

Does this help? If not, then let me know and I'll give you an example of how to come up with the two equations.

4. Mar 16, 2010

### Ryker

I was thinking of something along the lines of:

2kg . y + (y/1m) . 1kg . y/2 = 4kg . (1m - y) + ((1m - y)/1m) . 1 kg . ((1m - y)/2),

where y is the distance from the ball that weighs 2kg to the centre of gravity of the whole system.

I tried equalizing the leverage of the left side to that of the right side, but don't know whether that's the right approach.

5. Mar 16, 2010

### Jmf

I would draw a diagram like the one I've attached (mspaint, sorry :) putting in the numerical values that we know (instead of the words I have). R1 and R2 in this are your unknown 'reaction' forces - they are the forces exerted on the bar + balls, which will be equal and opposite to the forces exerted on the scales by the balls (by Newton's 3rd Law).

We need two equations; we can get one of these by noting that if the whole thing is going to stay stationary, then the sum of the forces upwards must equal the sum of the forces downwards.

The other we can get by taking moments ('equalising the leverage', if you like).

EDIT: Kind of like the equation you wrote above, yeah. To make it easier on yourself though, you can consider the weights of the different components seperately, and take moments around any point you like. Then you don't have to worry about where the centre of mass of the whole assembly is.

Can you do that? :)

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6. Mar 16, 2010

### Ryker

Hmm, so would my equation then be correct? If I solved it and got y, calculate a side of the equation and derive force from it, would this equal the forced applied to one of the scales? And I figured that the weight of the whole systems equals the combined numbers on the scales, is that right?

7. Mar 16, 2010

### Jmf

Your equation for y (the position of the centre of gravity) is okay as far as I can tell, yes.

and you've got the right idea I think :) The weight of the whole system will equal the combined numbers on the scales. Be careful how you consider the forces on either side though (take moments).

are you going to work it through? I've got a slightly different method (a bit simpler; requires less thought) that I can show you afterwards if you like.

8. Mar 17, 2010

### Ryker

Yeah, I'd be grateful for that simpler method, if you're willing to share it with me :)

As for my equation, the one thing I wasn't sure is whether those (y/2) and ((1m - y)/2) were correct. I namely intended those to be half way from the centre of gravity to the end of the rod (because I assumed the centre of the part of rod left or right of the centre of gravity would be). So is that cool, too, or should I have sticked with y and 1m - y for that?

9. Mar 19, 2010

### Jmf

Of course. :) Sorry for not replying to this yesterday.

One general idea to keep in mind in statics problems such as this is that you can always relate forces to one another by one of two methods:

- You can say that the sum of all the forces acting on a body, resolved in a certain direction, is equal to zero. For n forces this gives an equation something like:

$$F_1 + F_2 + ... + F_n = 0$$

possibly with some sines or cosines in the expression for each force, if there's angles involved.

- We can also say that the sum of all the moments acting on the body around a point are zero. This gives an equation something like:

$$F_1d_1 + F_1d_2 + ... + F_nd_n = 0$$

Where the forces F are as above and the d's are the perpendicular distances (with either a positive or negative sign - decide in advance which direction will be positive, and stick to it) from where each force acts to the point we're taking moments around. Note that we can take moments around any point on or off the object, and they should sum to zero as long as the object is static.

These two ideas are actually enough to solve almost any statics problem of this type, as long as you're willing to do some algebra. I'll use your problem of the ball on the two scales to illustrate.

Oh, and refer back to my rubbish mspaint sketch in my last post :)

I'll call our 'reaction' forces from each scale $$R_1$$ and $$R_2$$. Since if we take moments about a point that a force passes through, we get an equation without that particular force in it, I'm going to make two equations by taking moments about the points at which each reaction force acts on the ball. So firstly, for the left (bigger) ball, we're going to get an equation that says:

moment from weight of rod + moment from weight of small ball - moment from R2 = 0

Notice the minus sign, since the force R2 acts in the opposite direction to the others. Using the numbers you gave originally, noting that force due to gravity is 9.81 ('g') newtons per kilogram of mass, this gives:

$$9.81 \times 1kg \times 0.5m + 9.81 \times 2kg \times 1m - R_2 \times 1m = 0$$

Rearranging and solving for $$R_2$$ gives:

$$R_2 = 9.81 \times 2.5 N$$

Doing exactly the same thing on the other side (moments around small ball) gives:

moment from weight of rod + moment from weight of large ball - moment from R1 = 0

hence:

$$9.81 \times 1kg \times 0.5m + 9.81 \times 4kg \times 1m - R_1 \times 1m = 0$$

so:

$$R_1 = 9.81 \times 4.5 N$$

Now consider our scales. The forces acting upon them are equal (by newton's 3rd law) to R1 and R2. If they give a reading in kilograms then we are going to lose the factors of 9.81 above, so the scale under the heavy ball will read 4.5kg, and that under the light ball will read 2.5kg.

Notice that these add to 7kg, which is the mass of the whole system, as we would hope. :)

Alternatively we could have used the fact that the forces must all add to zero to get a different equation above, and we could have taken our moments about the centre of gravity, but we would have to solve simultaneous equations then (since both equations would have had both R1 and R2 in them). We would get the same answer.

I hope this is clear. I've deliberately made this a lot more long-winded than it needs to be, so my apologies if it was a bit dull, but I was just trying to explain everything as fully as possible. :)

10. Mar 20, 2010

### Ryker

Hey, thanks for the "long-winded" post, it was really helpful. Though I still can't quite get my head around as to why the weight of the rod would "show" equally, half of it on each of the scales. Your equations make it really clear and sensible, but if the centre of gravity isn't in the centre of the rod, why does then the left scale show 0,5kg of the rod and not less than that (since less than half of the rod's weight would, due to the centre of gravity being more to the left, pull down on the left side, and more than half of the rod's weight would pull down on the right side)?

edit: Could you perhaps help me set up that "alternate" equation with forces you were talking about?

edit2: Is it possible to explain the system in whole WITHOUT using momentum or is the latter essential to the solution? I'm guessing you can't do it without momentum, but if you can, could you explain how?

Last edited: Mar 20, 2010