Question about the moment/torque principle

[Theg]

Hi. We all know that example where they show you a balancing lever on a fulcrum, with 2 weights on each side. And you have the equation F1L1=F2L2. (F=force, L= distance)
But my question is different...
Does the total weight of the lever applied on the fulcrum changes with the distance of the weights from the center? Let's say we have 2 weights of 5kg each, and both of them are 1 meter away from the center. Let's say the lever weight is also 5kg, so the total amount of the weights and the lever on the fulcrum is 15 kg. Now we move both weights another meter away from the center... so we have 2 weights of 5 kg that are 2meter away from the center (4 meter distance between the 2 weights). So the question is, now that we moved the weights further apart, is the total weight of the lever and the weights on the fulcrum is still 15 kg or did it change (grow)?
Thank you.

 

[Theg]

 

[Theg]

Let me keep talking on this problem so you might see where I have a problem.
Now after thinking about it, my logic tells me that even after moving the weights further apart, the total weight of the system supposed to be still 15 kg. I mean they say that center of mass is a point at which all the forces are applied, and since after moving the weights, the center of mass didn't change and amount of forces didn't change, the total weight supposed to be still 15 kg.
(i will refer to the weights as "objects" from now on, so not to confuse with the "weight" as physical property).
But here I have another thought... let's say instead of moving both objects 1 meter further from the center, we only move the left object. So now in order for our lever to be in balance we have to move the right object also one meter to the right, or replace the right object with another object that weighs 10 kg (F1L1=F2L2; 5kg*2meter=10kg*1meter).
But in this new scenario will you still argue that the total weight of the system is 15kg? or since we changed the right object to a heavier one by 5 kg, so our total weight also supposed to grow by 5 kg (to 20kg instead of preaviously 15kg).
Got what I'm saying?

 

[Mark44]

Let me keep talking on this problem so you might see where I have a problem.
Now after thinking about it, my logic tells me that even after moving the weights further apart, the total weight of the system supposed to be still 15 kg.

The total mass of the system is 15 kg. Total weight (a force) would by 9.8 Nt/kg * 15 kg = 147.0 Nt.

I mean they say that center of mass is a point at which all the forces are applied, and since after moving the weights, the center of mass didn't change and amount of forces didn't change, the total weight supposed to be still 15 kg.

Total weight is 147.0 Nt.

(i will refer to the weights as "objects" from now on, so not to confuse with the "weight" as physical property).
But here I have another thought... let's say instead of moving both objects 1 meter further from the center, we only move the left object. So now in order for our lever to be in balance we have to move the right object also one meter to the right, or replace the right object with another object that weighs 10 kg (F1L1=F2L2; 5kg*2meter=10kg*1meter).
But in this new scenario will you still argue that the total weight of the system is 15kg? or since we changed the right object to a heavier one by 5 kg, so our total weight also supposed to grow by 5 kg (to 20kg instead of preaviously 15kg).
Got what I'm saying?

If you add more weight, the total weight increases. You seem to be confusing two different concepts: the total mass/total weight of the system on one hand, and the net torque of the system, on the other.

 

[Theg]

wait... let me get it clear.
1) so you say that the distance of the objects will not effect on the total weight of the system when meassured at the fulcrum point (center)?
2) so there is a difference of total weight between scenarios (5kg*2meter=5kg*2meter) and (5kg*2meter=10kg*1meter)? Even though both of those scenarios have balance, they weigh differently at fulcrum point?

 

[Mark44]

wait... let me get it clear.
1) so you say that the distance of the objects will not effect on the total weight of the system when meassured at the fulcrum point (center)?

Right. The distance of the objects from the fulcrum (also called lever arm) doesn't affect the mass of the system. However, the lever arm is used in the calculation of torque

2) so there is a difference of total weight between scenarios (5kg*2meter=5kg*2meter) and (5kg*2meter=10kg*1meter)?

Of course, the system went from a mass of 15 kg to 20 kg.

Even though both of those scenarios have balance, they weigh differently at fulcrum point?

"Weigh differently at fulcrum point" doesn't mean anything. The torque in the clockwise direction due to the 10 kq mass is 9.8 Nt/kg * 10 kg * 1 m = 98 Nt-m. The counterclockwise torque due to the 5 kg mass is 9.8 Nt/kg * 5 kg * 2m = 98 Nt-m. Since the two torques are equal in magnitude but oppositely directed, the net torque is 0. I'm ignoring the contributions to each torque by the board, since it makes the calculations slightly more complicated. In any case, the unloaded board is evenly balanced, so its net torque is also 0.

 

[Theg]

"Weigh differently at fulcrum point" doesn't mean anything.

what do you mean "doesn't mean anything"? I want to know if I to try to messure the weight of the board at the center, will it be different in the two scenarios ? (as you said it will be different, 15 and 20).

 

[jbriggs444]

what do you mean "doesn't mean anything"? I want to know if I to try to messure the weight of the board at the center, will it be different in the two scenarios ? (as you said it will be different, 15 and 20).

For most of us and I think for @Mark44, "weight" is a synonym for the total gravitational downforce on the system in question. For this notion of "weight", there is no notion of "weight at a point". There is just "weight".

I am guessing here that the notion of "weight" that you have in mind is the measured vertical force that it would take to support a rigid object at a particular point so the point remains motionless. i.e. "Weight is what a scale measures".

As long as the object is resting at equilibrium, supported only at the point of measurement, both notions of weight are identical.

Of course, in order for the object to be at equilibrium, the point of support/measurement needs to align vertically with the object's center of gravity. If one tried to weigh it at any other point, the measured support force would be low because the object would be tilting and would have a non-zero downward component of acceleration as a result.

 

[Theg]

ok... but you didn't answer. yes I mean by "weight" as something that you messure with scales, like in vegetables shop. So will there be a difference in what the scales show between (5*2=5*2) and (5*2=10*1) scenarios?

 

[Theg]

 

[A.T.]

So will there be a difference in what the scales show between (5*2=5*2) and (5*2=10*1) scenarios?

This type of scale shows whether the torques are equal, and they are in both cases.

 

[Theg]

This type of scale shows whether the torques are equal, and they are in both cases.

the weight... is the weight will be the same in both cases?

 

[A.T.]

So will there be a difference in what the scales show between (5*2=5*2) and (5*2=10*1) scenarios?

View attachment 241524

The weight measured with scales under the pivots is not the same between these two scenarios.

 

[Mark44]

So will there be a difference in what the scales show between (5*2=5*2) and (5*2=10*1) scenarios?

The weight measured with scales under the pivots is not the same between these two scenarios.

Which is pretty much what I said in posts #4 and #6.
It should be crystal clear that the 5*2, 5*2 scenario has a total mass of 15 kg (147.0 Nt) and the 5*2, 10*1 scenario has a total mass of 20 kg (196 kg).

@Theg, what part of this is confusing to you?

 

[A.T.]

@Theg, what part of this is confusing to you?

@Theg More importantly, why don't you just try it out? It's not like you need hi-tech equipment to make the experiment.

 

[Theg]

the thing is... i guess I'm looking at this wrong. i guess i have a trouble separating the two notions of torque and weight.

 

[A.T.]

the thing is... i guess I'm looking at this wrong. i guess i have a trouble separating the two notions of torque and weight.

For the bar to be static, you need both: force balance and torque balance.

 

[Mark44]

the thing is... i guess I'm looking at this wrong. i guess i have a trouble separating the two notions of torque and weight.

Well, you need to separate them, because they are different concepts, with different units. In the SI system, the unit of force is Newtons (Nt.) and the unit of torque is Newton-meters (Nt-m).

Have you ever had a flat tire, and needed to taken the wheel off a car? If you have a short lug wrench, you might not be able to apply enough force to break loose the wheel nut, but if you extend the wrench with a length of pipe, the same force but with extra leverage (more torque) can often make the difference.

The force applied in both cases is the same, but with a longer lever arm on the wrench, the torque (a twisting force) is greater.

 

[A.T.]

Good one. But regarding the last one: You don't need to melt it - expanding the screw and hole by the same amount will also expand the gap between them.

 

[nasu]

What you should ask about is the force on the fulcrum rather than "weight at the fulcrum" (which is meaningless, as already pointed). The force on the fulcrum may be equal to the weight of some objects but not necessarily. If you replace one of the object with the force applied by a hand holding the lever in equilibrium, the force on the fulcrum won't depend on the weight of your hand but on the force you apply. The force on the fulcrum can be found easily from the equilibrium conditions (for forces and torques).