How Do Levers Work? Explained Simply

[pete]

When you say it will accelerate in a direction what do you mean? Do you mean it will actually fly off somewhere or is it some kind of rule that your referring too? When you looked at the drawing in post #14 you pointed out that there was a 30N force on the body B and a 20N force on the body A but did not talk about it accelerating anywhere. If the crowbar can apply more force to one post that the other why can the clamp not do it?
If I stand between two walls and push against one wall with my back against the other I’m applying all my force in one direction but I feel it equally on my back. If I put a scale behind me and one in front of my hands they would read the same despite me pushing all my force in one direction. This is why I took the total force, unbalance, 90Nt to the right and 100Nt to the left combined them then dived them by 2 to give the readings on the scales, what is it I am doing wrong?u say it will acc

 

[Mark44]

If I stand between two walls and push against one wall with my back against the other I’m applying all my force in one direction but I feel it equally on my back.

Yes, because your arms are applying a force against one wall and your back is applying an oppositely directed force to the other wall. The magnitudes of the forces are equal, but their directions are opposite. Note also, that each wall is also applying a force against you. If you can apply a greater force than either wall can match, there will be motion of one or both walls.

If I put a scale behind me and one in front of my hands they would read the same despite me pushing all my force in one direction.

You're actually pushing in two directions. Even if there weren't a wall behind you, to push against the wall with your arms, you have to push against something with your feet (the ground). If you were on slippery ice, you wouldn't be able to push against the wall.

 

[pete]

OK I understand this and it seems to make sense to me that both should be equal. To do the calculations though I have followed the rules we have been talking about and this drawing is just a more complex example of the drawing in post #14. This is why I drew #14 and tried to solve it first to understand the rules so if it unbalanced in #14 it’s going to be unbalanced in this one. What am I missing?
Here is the shape between the two walls and the lever and seems no matter how I do it it will always be more force on one than the other?

 

[CWatters]

That is correct, there will be more force on the lower "wall". The force on the lower wall will be equal to the sum of the force on the upper wall and the applied force.

Consider the lever on its own. Newton's law F=ma applies. If a =0 then the net force f=0. In other words the vertical forces sum to zero as I explained earlier.

 

[CWatters]

Your drawing in #27 looks correct.

 

[pete]

I don’t really understand some of the terminology you use, the sum of the force on the upper wall, that’s the numbers I get from doing the first class lever calculation? And the applied force, that’s the number I get from doing the second class lever calculation I guess?
And everything summing to zero. The walls will always be stronger than the clamp so won't the force always sum to zero? I don’t know how to apply an equation with acceleration to a static object. You talk about acceleration a lot and I really don’t understand what you mean when everything if fixed in place.
It’s good that the drawing looks OK, was it not a problem that the numbers for the two walls are different, I though they had to be equal? That’s the summing to zero bit?

 

[Tom.G]

Consider this.

1. We consider the bottom wall to be fixed (such as attached to the floor)
2. There is some upward force being supplied (via the lever) to lift the upper wall
3. Consequentially, since the upper wall is immovable, there is an equal downward reaction force
4. This reaction force is supported by the lower wall (if you assume equilibrium)
5. To supply the lifting force to the upper wall, there is a (smaller) downward force at the right end of the lever
   (perhaps you standing there and pushing the lever down)
6. Again for equilibrium, this downward force at the right end of the lever must be supported by something (again, the lower wall), otherwise things are going to move

As you can probably see, the lower wall is supporting all the downward forces; which is the sum of the reaction force at the left of the lever plus your downward force at right of the lever. This holds whether the top wall is fixed, is a free weight that is not being lifted, or is attached to the bottom wall.

In general, for a fulcrum to be considered 'fixed' it must somehow be referenced to some applied force. In this scenario the reference is the floor you are standing on, and the fulcrum (the lower wall), being attached to it.

An alternate configuration could be the upper wall being a free weight and the lower wall being spring mounted to the floor. For this situation, you might want to consider the free weight at the left end to be the fulcrum (i.e. maintain its position), while you apply a force to the right end of the lever. Here the reference could be the Earth thru Gravitational coupling and eventually to the floor.

Well, this ended up a rather esoteric post. I hope it helps rather than confuses.

Cheers,
Tom

 

[CWatters]

I don’t really understand some of the terminology you use, the sum of the force on the upper wall, that’s the numbers I get from doing the first class lever calculation? And the applied force, that’s the number I get from doing the second class lever calculation I guess?

Imagine you had 1000+ levers all arranged in some complicated way between points A,B and F. Let's draw a box around those levels and call it a block. If the block is stationary then it's not accelerating so the external forces at A, B and F must sum to zero.

And everything summing to zero. The walls will always be stronger than the clamp so won't the force always sum to zero? I don’t know how to apply an equation with acceleration to a static object. You talk about acceleration a lot and I really don’t understand what you mean when everything if fixed in place.

It doesn't really matter _why_ it's stationary. If it is stationary it's not accelerating so the net force acting on it (in any direction) must be zero.

In this case we are assuming all your forces are vertical so we can write...

(-A) + (-F) + (+B) = 0

The signs in brackets are needed because this is vector addition and we need to take into account the direction of the forces acting on the block of levers.

It’s good that the drawing looks OK, was it not a problem that the numbers for the two walls are different, I though they had to be equal? That’s the summing to zero bit?

Indeed. Not only is the whole block of levers stationary (eg not accelerating) but each individual lever in the block is stationary (not accelerating) so the vertical forces on each individual lever will sum to zero. The force at points A can't be equal to B because of F.

Note I'm assuming that F acts vertically. If it acts at an angle then you have to use the vertical component of F in the sum above. The horizontal component of F would then cause there to be a horizontal component at A or B. This is because the horizontal forces must also sum to zero if the block of lever is stationary (eg not accelerating) horizontally. But perhaps best we assume F is vertical until you fully understand this.

 

[CWatters]

This issue of forces summing to zero is important... Consider an aircraft flying along straight and level at some constant speed.

It's not accelerating vertically so we can write...

Lift + (-weight) = 0

or rearrange that to...

Lift = weight

Likewise it's not accelerating horizontally so we can write...

Thrust + (-drag) = 0

or rearrange it to give

Thrust = drag

So even though the airplane is moving (but not accelerating) you can apply Newton's f=ma to the situation.

 

[pete]

I had to get some sleep, we are in different time zones.

I understand that an object must have equal forces operating on it or it will start moving in the direction of the weakest force proportional to the discrepancy, this makes sense to me. As with the aeroplane.

When you talk about the drawings I find the up down thing very confusing. All of these drawings are of me trying to refer to a clamp and are top down, looking down on the drawings from above. So everything is equal and talking about lifting things throws me off a bit.

I’m very pleased that the calculations I did are correct, you say the whole block of levers are stationary and nothing is accelerating and all the maths looks good to me as far as I understand the two equations I have to solve this.

But the total force one way is 90 and the other totals 100 so they are not equal. As the example of standing in an ally way pushing on the wall, if I apply pressure with my hands I also apply pressure with my back and it must be equal, a zero sum or I would accelerate towards one wall.

This leads me to exactly what I’m confused about. I think I understand the zero sum thing, if it’s not then it’s moving. That makes sense to me. So I’ll explain very specifically what’s not adding up for me. In this example. Let's say we are looking down on a gap in a wall. The wall is to heavy to move and to hard to break. I’m going to apply force on the lever and at 30n the surface of the lever will deform leaving marks and this is what I want. I want to apply enough force to do the job but no more.

First I look at B, using A as the fulcrum. The difference in the distance of B to A and B to The force is 3 to 1! So I times the Force by 3, 30n at A. That’s my answer, 30n acting to the right on the wall at B and I get my mark on the lever.

Now A, I take the distance from the new fulcrum B to the Force, that’s 4, and times it by the force, that’s 40n, then I divide this by the distance from A to B to find the force at A and it’s 40n acting on the wall to the left at A. it’s 10 too much and I make a bad mark in my lever.

And this is where I’m confused. I’ve done the maths correctly as far as I can tell and the force to the right, 30n, is different to the force to the left 40n.

Lets look at the clamp, here it is from above in a crack in a wall. As the lever expands the clamp the two sides bite into the wall but as in the example of me pushing my hands against the wall the force to the right and the left will always be the same because as you said you are always pushing in two directions. This is an absolute, there will always be the same force on ether side of an expanding object between two fixed planes, I just can imagine being able to push more with my hands than my back!

So I don’t understand how to resolve this. In my solution with the 2 levers I added the total force together and divided it by 2. It’s the only thing I could think of to do to resolve the numbers.

So in this case I’d take the 30n on B and the 40n on A and add them together to get 70n, divide them in 2 to get 35n. Both A and B make a mark in the lever but nether make a hole. But I did this because the answer did not fit and I’m not sure if I can?

I noticed some other folk joined in while I was asleep, trying to explain this to me and I appreciate the help. I am getting there and I’m trying do my own research too to fill in the blanks but I just seem to be stuck on this point.

 

[CWatters]

First I look at B, using A as the fulcrum. The difference in the distance of B to A and B to The force is 3 to 1! So I times the Force by 3, 30n at A. That’s my answer, 30n acting to the right on the wall at B and I get my mark on the lever.

Now A, I take the distance from the new fulcrum B to the Force, that’s 4, and times it by the force, that’s 40n, then I divide this by the distance from A to B to find the force at A and it’s 40n acting on the wall to the left at A. it’s 10 too much and I make a bad mark in my lever.

And this is where I’m confused. I’ve done the maths correctly as far as I can tell and the force to the right, 30n, is different to the force to the left 40n.

It will be different. You have done the maths correctly, you just don't like what the maths is telling you :-)

The Applied force is 10N.
The force at B is 30N.
The force at A is 40N.

Here is the Free Body Diagram for your lever showing the forces on the lever...

You can see how the horizontal forces sum to zero which is exactly what you expect if the lever isn't accelerating.

Your drawing with the clamp should look more like this...

Now your "back" and "hands" are pushing with same force. Your "back" pushes with 40N (at A), one hand pushes with 30N (at B) and the other hand 10N (at F). Newton is happy, the force on both walls is the same.

Adding more levers doesn't change anything.

 

[CWatters]

Perhaps this achieves what you are trying to do? The force on both walls would be the same...

 

[pete]

That’s it! It’s a threaded bolt not a spring but I think this is what has been driving me mad. So If the bolt is screwed tight applying 20n force, that will be 10 in each direction and the whole thing will add up. Is this right?

If it was a lever that someone was pulling on, like a crow bar jammed in a gap then picture could be unbalanced, not add to 0, but only because we don’t see the whole system. If I drew the guy doing the pulling with his feet on the floor then I could balance the whole system?

 

[CWatters]

That’s it! It’s a threaded bolt not a spring but I think this is what has been driving me mad. So If the bolt is screwed tight applying 20n force, that will be 10 in each direction and the whole thing will add up. Is this right?

Yes but the screw would only need to apply 10N on the lever. The 10N it applies to the wall comes free due to Newton's third law.

 

[CWatters]

That’s it! It’s a threaded bolt not a spring but I think this is what has been driving me mad. So If the bolt is screwed tight applying 20n force, that will be 10 in each direction and the whole thing will add up. Is this right?

Yes but the screw would only need to apply 10N on the lever. The 10N it applies to the wall comes free due to Newton's third law.

 

[pete]

Yes I think I got you, I’m not too up on these laws but at a guess its the one about an equal and opposite force. The screw can only be tightened in one direction I just figured that if it pushed 10n onto the lever then it had to push back 10n too.
After 2 days of staring at this I think I’ll remember it too my grave. Thanks to everyone who posted trying to explain this to me especially Mark who really stuck in there when I couldn't get my head around it. It’s been a real help.

 

[Mark44]

especially Mark who really stuck in there when I couldn't get my head around it

You're welcome!