Find the weight of a metre stick – using the law of the lever

[Vivara]

Hi Guys.

 

I'm newly joined, this forum looks great!

 

I was reading through some exam papers from the state exams here in Ireland, and one of the questions was:

 

Create a experiment to find the weight of a metre stick using the law of the lever and without using a weighing scales/balance or Newton scales.

 

I have to admit I'm curious and have been told there are numerous ways - can't think of any after two or three weeks. I know, that's pathetic!

 

Any ideas?

 

Warm Regards,

Ed.

[J.C.MacSwell]

Hi Guys.

 

I'm newly joined, this forum looks great!

 

I was reading through some exam papers from the state exams here in Ireland, and one of the questions was:

 

 

 

I have to admit I'm curious and have been told there are numerous ways - can't think of any after two or three weeks. I know, that's pathetic!

 

Any ideas?

 

Warm Regards,

Ed.

 

Can you use a known weight and a fulcrum point?

[Martin]

nice problem. If it were homework you should take it to the homework section. But it sounds to me like you are considering problems for fun, so this is the right place and we are free to help.

 

I think.

 

I believe you must assume you are given some standard weight, like a 100 gram weight.

 

If you have nothing in the room to compare with, I don't think it's possible.

 

So assume you have a standard steel cylinder 100 grams.

 

One way is hang it on one end of the meter stick and see where the combined object balances. Perch it on your finger at the place it exactly balances and read off where that place is.

 

Hopefully other people will have other suggestions, maybe more clever. That's one anyway

[Vivara]

Can you use a known weight and a fulcrum point?

 

That's the question, there are no other notes, and there's no-one to ask in the exam, so I presume you can use the fulcrum to conduct the experiment. And yes, you can use a known weight.

 

nice problem. If it were homework you should take it to the homework section. But it sounds to me like you are considering problems for fun, so this is the right place and we are free to help.

 

I think.

 

I believe you must assume you are given some standard weight, like a 100 gram weight.

 

If you have nothing in the room to compare with, I don't think it's possible.

 

So assume you have a standard steel cylinder 100 grams.

 

One way is hang it on one end of the meter stick and see where the combined object balances. Perch it on your finger at the place it exactly balances and read off where that place is.

 

Hopefully other people will have other suggestions, maybe more clever. That's one anyway

 

Thanks Martin. I know I'm probably not getting something really simple, but I don't understand. To make what you said easier in my mind, instead of using my finger, I'd prefer to use a retort stand and string (silly I know). Ok, so let's say hang a 2 Newton weight from one end – and then move the string the the place where it balances. How do I find the weight of the metre stick from then on? Something like FORCE X DISTANCE TO THE FULCRUM comes to mind, but that was a long time ago so I don't see how it could help me.

 

 

Yeah, I have to admit, I was looking for something a bit more elaborate! Thanks again Martin (and J.C.).

 

Ed.

[SkepticLance]

The principle of moments say that, when a lever is balanced, the centre of mass on one side of the fulcrum multiplied by the distance from fulcrum equals the centre of mass on the other side multiplied by the distance from fulcrum on that side.

 

In this case, I would hang the metre rule by a piece of string one quarter of the way from the end, and add known weights to the short end until it balanced horizontally.

Since one quarter of the ruler to the left weighs the same as one quarter to the right, we can temporarily ignore those weights. This means that the weight of half the ruler, multiplied by the distance of its centre of mass (0.375 metres) equals the weights you have hung times the distance to the fulcrum (0.25 m). When you have this, just double the result to get the mass of the whole ruler.

 

Suppose you got the ruler to balance with the fulcrum point at 0.25 metres, and 100 grams added at the short end. Call the weight of the ruler x. Then 0.5x times 0.375 = 100 times 0.25.

Solve this equation for x. The result will be in grams. Simple.

Edited March 3, 2009 by SkepticLance

[Vivara]

Thanks a million SkepticLance!

 

So, if I read that correctly, the equation to find the mass of the total ruler is:

 

x = Weight of half the ruler

Let's just say that the force of the Newton weights is 5N.

 

x(0.75) = 5(0.25)(2)

 

I'm really excited now!

 

Thanks,

Ed.

[SkepticLance]

Actually x was the weight of the whole ruler.

 

Then 0.5x times 0.325 = 100 times 0.25

 

Or x = (100 times 0.25) divided by (0.5 times 0.325)

 

Answer in grams.

[John Cuthber]

Would you get more marks or fewer for pointing out that, while it's perfectly possible to measure the mass of the ruler this way, it gives you no information about its weight. The balance point would be the same on the moon, but the ruler would weigh about 6 times less.

[Vivara]

Instead of using grams, I used the Newton – the unit of weight.

 

Also, thanks Skeptic Lance, the clockwise moment is equal to the anti-clockwise moment trick worked great.

 

I didn't have any 0.5x though and I did it slightly differently, but it worked out perfectly. You're obviously allowed to use a weighing scales as a control.

 

Ed.