Balancing Moments: A Ruler's Center of Mass

[Rumplestiltskin]

Homework Statement

A metre stick has its centre of mass at the 50cm mark, and weighs 0.92N . A 2.00N weight is stuck to the 10cm mark with massless glue. About which point will the ruler balance?

Homework Equations

Σclockwise = Σanticlockwise

The Attempt at a Solution

2 N * 0.4 m = 0.8 Nm
There needs to be a 0.8 Nm moment on the other side.

 

[faradayscat]

For the stick to be in equilibrium the total torque must be zero:

$Στ = 0$

To solve this, you should pick an arbitrary position "a" at which the meter stick is balanced, this position is your axis of rotation. Then you can find the distance between "a" and the center of mass as well as the 2.00N weight. For instance, the distance "r" between "a" and the center of mass at x = 50 cm is:

$r = a - 0.5$

Now you can sum up the torques (assume clockwise torque is positive) and find "a".

 

[Jose Confraria]

It will rotate on the new center of mass( 2*0.1+0.92*0.5) / 2.92

(which is btw the definition that the other users are explaining to you)

 

[billy_joule]

Homework Statement

A metre stick has its centre of mass at the 50cm mark, and weighs 0.92N . A 2.00N weight is stuck to the 10cm mark with massless glue. About which point will the ruler balance?

Homework Equations

Σclockwise = Σanticlockwise

The Attempt at a Solution

2 N * 0.4 m = 0.8 Nm
There needs to be a 0.8 Nm moment on the other side.

True if you want to balance it at the midpoint but that's not the question.
You need to find the distance from an end to the pivot point when the sum of moments is zero. Let x be that distance and express all other distances in terms of x, then solve for x.

 

[Rumplestiltskin]

It will rotate on the new center of mass( 2*0.1+0.92*0.5) / 2.92

(which is btw the definition that the other users are explaining to you)

Makes sense, but turned up 0.36m which was incorrect.

I'm not really sure about the other explanations, you're going to have to spoonfeed this to me.

 

[billy_joule]

Let the balance point be 'x' metres from the end of the ruler closest to the 2N weight (draw a picture with dimensions).
Let anticlockwise torques about that point be positive.
For balance we know: ΣT=0
So
0= 2N(x-0.1m) - 0.92N(0.5m-x)
Solve for x

 

[Rumplestiltskin]

Let the balance point be 'x' metres from the end of the ruler closest to the 2N weight (draw a picture with dimensions).
Let anticlockwise torques about that point be positive.
For balance we know: ΣT=0
So
0= 2N(x-0.1m) - 0.92N(0.5m-x)
Solve for x

Sorry, I was just getting to your post.

"Here was my attempt. About x,
Total anticlockwise moment: 2N(x - 0.1m) + moment of the ruler to the left of x, how would I find this?
Total clockwise moment: 0.92N(1m - x)"

So you did this a little differently, not finding the anticlockwise moment of the ruler (why not?) and using 0.5m - x instead of 1m - x (but wouldn't that just give you the clockwise moment exerted by a small portion of the ruler right from x?)

 

[billy_joule]

Did you draw a picture? It seems like you haven't.
Drawing a free body diagram should be the first step for many physics problems, including this one. Draw one, if your error doesn't become clear post your drawing and we can see where you went wrong.

 

[Rumplestiltskin]

Did you draw a picture? It seems like you haven't.
Drawing a free body diagram should be the first step for many physics problems, including this one. Draw one, if your error doesn't become clear post your drawing and we can see where you went wrong.

Did something rudimentary, but here's V2.

What I'm seeing is that I'm considering left hand/right hand COMs whereas you've done away with that altogether and said that the weight of the ruler would act down it's own COM at 0.5m. Are you sure? I've tried to learn from the other moments question I posted to consider two COMs. Wouldn't the weight produce two moments about a given pivot?

 

[billy_joule]

What I'm seeing is that I'm considering left hand/right hand COMs whereas you've done away with that altogether and said that the weight of the ruler would act down it's own COM at 0.5m. Are you sure? I've tried to learn from the other moments question I posted to consider two COMs. Wouldn't the weight produce two moments about a given pivot?

Either approach will work but in this case considering the COM of each half unnecessarily complicates the math (and the FBD).
If you do want to consider each half then you need to recognise that each half weighs half the total weight.

 

[Rumplestiltskin]

Either approach will work but in this case considering the COM of each half unnecessarily complicates the math (and the FBD).
If you do want to consider each half then you need to recognise that each half weighs half the total weight.

Oh, right. Could you take a look at the other moments thread I posted and tell me how you'd do that (if it isn't too much)?
FBD?
I'm not considering each half, but the splits that x make. So x/2 (left hand COM) and 1 - x (right hand COM).

But going by the lone COM:
2(x - 0.1) = 0.92(0.5 - x)
2x - 0.2 = 0.46 - 0.92x
2.92x = 0.66
x = 0.23

 

[billy_joule]

Oh, right. Could you take a look at the other moments thread I posted and tell me how you'd do that (if it isn't too much)?

I had a look. I'd take the same approach you did in post #6. Be careful with units, at one point you add a force to a moment which is like trying to add litres to kilometres - it doesn't make sense. But you got to the right answer in the end.

FBD?
I'm not considering each half, but the splits that x make. So x/2 (left hand COM) and 1 - x (right hand COM).

FBD = free body diagram.
That approach could also work but is even more complicated than considering each half as not only the lever arm length for the self weight depends on x but also the magnitude of the self weight force.

But going by the lone COM:
2(x - 0.1) = 0.92(0.5 - x)
2x - 0.2 = 0.46 - 0.92x
2.92x = 0.66
x = 0.23

That's the answer I got. Good work.

 

[Rumplestiltskin]

I had a look. I'd take the same approach you did in post #6. Be careful with units, at one point you add a force to a moment which is like trying to add litres to kilometres - it doesn't make sense. But you got to the right answer in the end.

But with one COM instead of two, right? I redid the question that way and arrived at the same answer.
Not strictly, but I did choose to represent a moment with F, which would cause that confusion.

Thanks for all the help!

 

[billy_joule]

But with one COM instead of two, right?

Yes, right, missed that.
There may be situations where splitting up a beam into multiple sections is needed each with it's own COM is required/simpler but neither of your cases need it IMO

Thanks for all the help!

You're welcome.

 

[ehild]

Makes sense, but turned up 0.36m which was incorrect.

How did you make ( 2*0.1+0.92*0.5) / 2.92 equal to 0.36 m? Forgot parentheses? :)

 

[Rumplestiltskin]

How did you make ( 2*0.1+0.92*0.5) / 2.92 equal to 0.36 m? Forgot parentheses? :)

Using Google for a calculator. :P