Balance a meter stick

[rasen58]

A uniform meter stick of mass 0.24 kg can be balance by a mass of 2.16 kg at the 100 cm mark if the fulcrum is placed at the point in cm ___.

 

So the mass on one side is just the mass of the stick (0.24 kg) and the other side is (2.16 kg + .24 kg = 2.4 kg)

So the torques have to be equal

So on one side, it's 0.24 kg * 10 m/s^2 * d

And the other is 2.4 kg * 10 m/s^2 * (100-d)

So I set them equal to each other and got

26.4 d = 2400

d = 90 cm.

 

But the answer is apparently 95 cm.

why?

[studiot]

You have made it to simple, trying to combine too much.

 

You have three moments to balance. Each has a different lever arm.

 

Here I will start you off.

 

I strongly suggest you draw a diagram.

 

Working with zero art the left hand end

 

Let d cm be the distance from the zero mark to the fulcrum.

 

So the distance from the fulcrum to the 100cm mark at the right hand end is (100-d) cm

 

The mass of the ruler is 0.24kg ie 0.024 kg per centimetre.

So the mass on the left hand side producing counterclockwise moments is 0.024d kg

The counterclockwise moment is therefore (0.024gd) x(d/2) N-cm.

 

Can you now complete by working out the right hand side clockwise moments properly?

Edited February 8, 2015 by studiot

[rasen58]

Oh, I see, it makes sense that it would be 0.024 kg per centimetre.

But why is the counterclockwise moment (0.024gd) x(d/2) N-cm?

Where did the d/2 come from?

 

I got the mass on the right side is (100-d)(.024 kg), but how do I calculate the moment?

[studiot]

Have you drawn a diagram?

 

 

Where did the d/2 come from?

 

Where is the centre of gravity of the chunk of metre stick to the left of the fulcrum?

[rasen58]

Yes, I drew a diagram.

Why do you need the centre of gravity though?

[Fuzzwood]

Where else would you like to pivot it on?

[studiot]

Was your diagram something like this?

 

 

What is the definition of cente of gravity in relation to the weight or mass of an object?

 

I am not going to do all the work for you so I have left some blanks on the diagram for you to fill in.

 

Ask if you need more help, but you should be able to post an equation balancing the moments of the three forces and leading to a quadratic equation in d.

Edited February 8, 2015 by studiot

[rasen58]

Where else would you like to pivot it on?

I thought it would pivot on the fulcrum.

 

 

Was your diagram something like this?

 

It was like that, but I don't understand why there are three forces.

[studiot]

Rasen, you managed to solve the chemistry and maths questions you posted with a little help and can obviously cope with science.

However it is clear from your questions that you have missed some physics so try this explanation.

 

I asked what was the definition of the centre of gravity (for your purposes it is also the centre of mass if you prefer to call it that).

 

Since you did not answer

 

The centre of gravity (mass) is the point through or at which all the weight (mass) of a body can be considered to act.

 

So that is the exact centre of a rectangular bock such as the one in my Fig1.

I have drawn an arrow to show the weight (W) acting vertically downwards through this point.

 

Now any force has a moment about any point not on its line of action.

So I have shown that W has a moment about the corner (A) of the block and also about some point B outside the block.

 

You should understand that you can take a moment about any point you like. Obviously some points are more convenient than others and the trick (experience) is choosing the best point.

 

Ok so if we imagine a light frame of negligable mass, shown dashed in Fig2, that just fits out blocks, and slide a block of weight W into one end as shown.

This arrangement is not in balance but don't worry we will build up to that.

 

The frame is mounted on a fulcrum at the centre so our standard block exerts a moment of WL₁ about the fulcrum.

 

 

If we now slide the block along the frame as in Fig3 the moment changes to WL₂.

In fact the moment varies as we slide the block along the frame, but we are still not in balance.

 

If we now add a second matching block to the frame but on the other side of the fulcrum we can achieve balance if the distances from the fulcrum are equal.

 

That is balance requires a mass on each side of the fulcrum.

These masses generate opposing moments that balance each other out.

 

So how is all that related to our solid metre ruler?

 

Well if we now fill up the frame with blocks we will again achieve balance, since the fulcrum is at the middle point.

In just the same way we can balance our ruler on a fulcrum placed at its middle point.

 

So another way to look at this is to say that we can consider the ruler made up of an assemblage of blocks.

 

I have done just this in Fig5.

 

The left hand side is made up of three equal blocks.

The right hand side is made up of a single continuous block, like the ruler.

 

And these two sides balance!

 

So that shows how two of the three forces are generated.

From the left hand side and right hand side of the ruler look at my first diagram in post#4

 

The third force is of course the additional weight that is placed on the end of the ruler.

 

I have not done so but it is an interesting exercise to prove in my Fig5 that the anticlockwise moments do indeed equal the clockwise ones.

 

To do this work out the distances in terms of L₃.

You should do this to convince yourself that you can divide up a continuous block into pieces in this way or assemble the contributions from individual pieces to one whole.

 

You will meet this technique again and again in more advanced work.

[rasen58]

Okay so I finally understand how to actually go about doing these problems now.

 

So I made the fulcrum the pivot point and found the torques from there.

So d is the length from the fulcrum to the left end

100-d is the length from the fulcrum to the right end

 

So counter-clockwise moment is (.0024 kg)(10 m/s^2)(d/2)

Clockwise moments are (.0024 kg)(10 m/s^2) ( (100-d)/2 ) & (2.16 kg)(100 -d)

So the net torque = (.012)(d) - [ 1.2 - (.012)(d) ] - [ 216 - 2.16(d) ] = 0

Then I solved for d and got 99.45, which is off.

 

What did I do wrong this time?

 

Edit:

So I did it again with the pivot point being the right end.

So net torque = (Fbeam)(.5 m) - (Ffulcrum)(x) = 0

where x is the distance from the 100 cm to the fulcrum.

So Ffulcrum has to balance the weight of the beam and the mass so it turns out to be 24 N.

And then I solved for x and got .05 m, which means it's at the 95 cm mark.

Edited February 10, 2015 by rasen58

[studiot]

First let me congratulate you on thinking about the question and finding the easy way,

Exactly as I suggested

 

You should understand that you can take a moment about any point you like. Obviously some points are more convenient than others and the trick (experience) is choosing the best point.

 

It is usually convenient to take moments about one of the forces so you do not have to include that force in the calculation.

 

They do not normally teach this, so once again well done +1

 

 

 

As to what went wrong with the beam balance method:

 

After correcting my arithmetic error (0.024 kg per centimetre should have been 0.0024), you made one of your own.

 

Moments are force x distance and you forgot to multiply by the distance.

Compare my lines 4 and 6 with yours.

 

A flag should have been that I said a quadratic in d and you did not get one

 

It is good to spot short cuts, but you should become comfortable with my method for two reasons

Firstly because it is the general method, you will not always be able to take moments about an end.

Secondly because the problem is very similar to the workings of a Chemist's analytical beam balance.

The additional 2.6 kg weight is like the rider (although rather larger than most riders it is similar to riders used in medical and vetinary practice).

 

So I have reproduced the solution in some detail, correctly arriving at 95 cm.

 

Some points to note are that the value you use for g does not matter.

g cancels throught the equation.

This is important since it shows that the Chemist's beam balance is independent of g and does not need to be corrected for it.

 

Secondly the equation looks fearsome at first, but have faith.

College questions are designed to work out with easy numbers.

The quadratic terms cancel, reducing the equation to a simple one.

And the akward fractions, decimals and large numbers fall out to any easy non calculator solution.

 

This is also a good point to demonstrate how to pick up exam marks.

The key equation is line 8, as set out it shows you understand force x distance and balancing clockwise and anticlockwise moments.

You would get 3/4 marks out of say 5 available for this line.

 

There would probably ony be 1 more mark for the rest of the arithmetic down to line 15.

 

 

 

[rasen58]

Thank you so much for all the help once again!