Balanced forces

[Function]

Hello everyone

 

Let me just ask it:

 

If you look to the image below, which mass would need to be on the question mark?

(I never treated this kind of matter at school, yet I fear this is a very good question for the Med School Approval Exam)

 

 

I'd say:

 

[math]2(M\cdot g)=2(2m\cdot g)+6(5m\cdot g)=34 m\cdot g[/math]

[math]\Leftrightarrow (M\cdot g)=17m\cdot g[/math]

[math]M=17m[/math]

 

Yet I fear this is wrong. Don't know why.

 

Thanks.

 

Function

Edited February 14, 2014 by Function

[EdEarl]

Redraw so that all weights hang from the top balance bar, but do not move any weight right or left, just erase the horizontal lines, except the top one, and draw all vertical lines to connect the weights to the top bar (line). You will have to extend the top horizontal line to the right.

[Function]

Very well. Now, for our own ease, I drew it a bit more exact on paper

 

The 'new' balance is the following:

 

?-|-°-O-|-O-O-O-|-O-(OO)

 

With ? the mass to calculate, | an empty spot, O a mass 'm' and (OO) 2 masses hanging underneath one another and ° the hanging point

 

What now?

 

Once again, I bet this:

Mg = 1mg + 3mg + 4mg + 5mg + 7mg + 8*2mg = 36mg

 

M = 36m

Edited February 14, 2014 by Function

[swansont]

Your approach is correct: the torque needs to be zero, and T = r X F, so in this case it's simply rF. The image is misleading, though, because in the OP the bottom, right-most section itself does not look balanced.

[Implicate Order]

Hi Function.

 

I am also having difficulty with this problem. Hopefully someone here can put my brain to rest.

 

I am wondering whether the treatment of this problem may also be one of the scale of the experiment. In your ilustration for example I am not sure why your drawn example represents masses at different vertical locations. Is this for diagrammatic convenience or for some other purpose? This throws a red herring to me thinking in terms of the location of this experiment and it's relationship to a gravitational field. For example if this lever balance you have drawn was a theoretically huge lever balance located say between the earth and the moon, the vertical locations of the individual masses would have something to say in this experiment.

 

For example I could replace the vertical strings with a spring to measure true weight in a gravitational field and I would suspect the unknown mass calculation would be different.

 

If your diagram was merely representing a local experiment (eg. a medical experiment about levers) say here on earth where we can assume the scale if the experiment is inconsequential, then I agree with the horizontal treatments that have been applied here to cancel out any torque. If this is the case I therefore assume your different vertical locations of mass was drawn for ease as opposed for purpose. Can you please clarify this, and could someone here tell me if the treatment as we scale up in size would be different taking into account the varying weights of these masses from a varying gravitational field?

 

For example your experiment could be looking for a barycentre in a solar system as opposed to a simple centre of mass at a horizontal pivot point. I note in your working you are using the same value of g so I assume it is the latter.

Edited February 15, 2014 by Implicate Order

[Function]

Your approach is correct: the torque needs to be zero, and T = r X F, so in this case it's simply rF. The image is misleading, though, because in the OP the bottom, right-most section itself does not look balanced.

 

So the mass should indeed be 36 times a small mass?

 

 

Can you please clarify this?

 

It concerns just a simple experiment on earth. Let's say a penguin torturer and physics lover wants to put this to a test and hang 7 penguins on the scale as given in #1.

Edited February 15, 2014 by Function

[EdEarl]

A mass at length 2 will balance a 2xmass at length 1.

[swansont]

 

So the mass should indeed be 36 times a small mass?

 

 

No, the original treatment was valid. But the fact that you get 36 vs 17 at a distance of 2 with your two analyses indicates there's a problem with the drawing.

[Function]

 

No, the original treatment was valid. But the fact that you get 36 vs 17 at a distance of 2 with your two analyses indicates there's a problem with the drawing.

 

Then why EdEarl want me to draw it all on a horizontal line?

Never mind: realizing that the most lower, left mass should be 3m, I found that in both cases (the original balance and the horizontal line way of doing it), M should be 23m

[Implicate Order]

 

 

It concerns just a simple experiment on earth. Let's say a penguin torturer and physics lover wants to put this to a test and hang 7 penguins on the scale as given in #1.

 

Thanks Function. Then I support EdEarl and Swansonts approach unless of course you descriminate between Antarctic or Arctic penguins/ tic .

[swansont]

 

Thanks Function. Then I support EdEarl and Swansonts approach unless of course you descriminate between Antarctic or Arctic penguins/ tic .

 

Arctic penguins are massless.

[Function]

 

Arctic penguins are massless.

 

How about I take some Antarctic penguins, take them to the Arctic and make them breed there

Edited February 15, 2014 by Function

[Implicate Order]

LOL

[J.C.MacSwell]

 

Arctic penguins are massless.

Unless of course, they are actually an Antarctic Penguin...with reversed polarity.

[swansont]

Unless of course, they are actually an Antarctic Penguin...with reversed polarity.

 

Antiantarctic penguins?

[Implicate Order]

 

Antiantarctic penguins?

 

Hey guys cut it out and stop playing with my head. You know they are either one or the other when the are observed. pffft Antiantarctic penguins.....

[Function]

 

Antiantarctic penguins?

 

Aren't Antarctic penguins already 'anti-Arctic' penguins?

So.. anti-anti-Arctic penguins would be.. Arctic penguins?

[Implicate Order]

It's confusing to say the least. I wonder what the penguins think?

[Function]

It's confusing to say the least. I wonder what the penguins think?

 

It's as easy as (-(-(+1)) = +1

 

Do birds even think? O.O

[I-try]

Hello Function.

Your original drawing is supposedly an exact right and left side equal balance.

The small horizontals are added as a red herring and should be ignored because gravity is acting on all hanging weights according to their mass content and only leverage should be considered. There is a leverage of two on the left hand of the balance verses the differing positions of the masses on the right.

[I-try]

Hello again Function.

 

Judging from the drawing, each of the pair of larger circles contain twice the mass that each of the group of five contain; thereby and due to their position, providing 4 units balancing 4 units on the left side of the balance; the outer 5 provide the remainder mass that due to their position enables the even balance. Perhaps the O=M signifies that mass is involved without indication of quantity.

[nivek]

Hello everyone

 

Let me just ask it:

 

If you look to the image below, which mass would need to be on the question mark?

(I never treated this kind of matter at school, yet I fear this is a very good question for the Med School Approval Exam)

 

Untitled.png

 

I'd say:

 

[math]2(M\cdot g)=2(2m\cdot g)+6(5m\cdot g)=34 m\cdot g[/math]

[math]\Leftrightarrow (M\cdot g)=17m\cdot g[/math]

[math]M=17m[/math]

 

Yet I fear this is wrong. Don't know why.

 

Thanks.

 

Function

It's not wrong M=17

All the hanging weights can be calculated according to there mass as I-try states.There mass is centred around the hanging bar.

The positions of the weights can be simplified to half the distances ie:- The unknown mass on the left is at point 1 instead of 2 and the other masses on the right are at 1and 3 instead of 2and 6.

So a mass of 5 on the right at position 3 will lift 15 on the left at position 1

and a mass of 2 on the right at position 1 will lift 2 on the left at position 1

makeing a total of 17.