How Do Levers Work? Explained Simply

  • Thread starter pete
  • Start date
  • Tags
    Work
In summary, the conversation revolves around understanding how levers work and how to use free body diagrams to analyze them. There is confusion about calculating the forces and moments on levers, especially when multiple levers are involved. The expert suggests drawing free body diagrams and making certain assumptions to understand the underlying physics of levers. They also clarify that the moment on a lever is equal on both ends, and the forces to consider are only the ones at the ends of the lever. The expert also gives an example of using a crowbar as a lever to illustrate how levers work.
  • #1
pete
72
5
So my last drawing was too confusing so I’ve tried to sketch it as simply as possible to explain. In the first picture a lever in red is prying the two blue parts apart. The Blue parts are fixed.
I’m not too good at all this but as I understand it the arm of the lever is twice as long on the side with the force being applied so the force is doubled, then equal applied to both half's of the blue part.
Given that I’ve understood that correctly then in the second picture the same is happening except that the lever instead of acting on the inside of the blue part is acting on another lever. The first lever is 3 to 1 so that’s 1.5 pushing on the pivot and 1.5 on the next lever and so on.
What I’m asking is do levers work in sequence like this or do you calculate it somehow as one long lever? Or is it all completely wrong?
levers_on_levers_3.jpg
 

Attachments

  • levers_on_levers_3.jpg
    levers_on_levers_3.jpg
    20 KB · Views: 1,051
Engineering news on Phys.org
  • #2
The best way to learn how these levers work is by drawing the free-body diagrams yourself and labeling the forces. Presumably you want to analyze these statically, so when determining the forces it's helpful to keep in mind that the sum of the forces in any component direction must be zero. Making other assumptions such as perfectly rigid bodies and prismatic beams is helpful as well to see the underlying physics of the problem.
 
  • #3
I'll give it a go. Any comment on whether I'm doing it right in the picture above?
 
  • #4
I watched a few tutorials on free body diagrams and they look very helpful but I do not really understand how to use them in the case of levers and I don't see how they are any use to understand the maths behind them.
For a lever its the mass x the distance to the fulcrum x the distance to find the force applied. What I don't understand is if you have a lever acting on another lever or a set of levers in sequence do you calculate each lever individually as I did in the picture above or is that wrong, am I doing it wrong?
If I draw free body diagrams without understanding the underlying maths how will I know how long my arrows should be?
Or have I misunderstood you, are you asking me to draw a free body diagram because you don't understand the picture I’ve done? I’ll give it a go but it’s going to look very similar I think.
 
  • #5
I tried to draw the problem and my solution out clearer here:
free_body_lever_fix.jpg

free_body_levers_2_Solution_fix.jpg
 

Attachments

  • free_body_lever_fix.jpg
    free_body_lever_fix.jpg
    17.7 KB · Views: 684
  • free_body_levers_2_Solution_fix.jpg
    free_body_levers_2_Solution_fix.jpg
    22.7 KB · Views: 985
  • #6
pete said:
For a lever its the mass x the distance to the fulcrum x the distance to find the force applied.
No, it's the force times the distance to the fulcrum.

pete said:
What I don't understand is if you have a lever acting on another lever or a set of levers in sequence do you calculate each lever individually as I did in the picture above or is that wrong, am I doing it wrong?
Your pictures in post #5 are confusing, since it's unclear whether either of the two bodies shown is fixed. The idea with a lever is that the fulcrum is fixed, but each arm of the lever is allowed to move. In your drawing is the L-shaped body allowed to move? Is the small rectangle allowed to move?

The basic idea of a lever is that the two moments are equal. Looking at the right-most lever in your drawing in post #5, let's assume for now that its fulcrum is fixed. If a force F is applied to the right end, at a distance of 3m from the fulcrum, the moment is 3 x F (probably in units of Newton-meters). The moment at the left end of this lever is also 3 x F, but since the lever arm is only 1 m., the force is 3F. Again, assuming the fulcrum is fixed, the only forces to consider are the forces at each end of the lever, not the force down on the fulcrum itself.

Both of your levers have forces applied at the ends of the levers, but levers also work with with fulcrum being at one end. For example, when you use a crowbar to pull a nail out of a piece of wood, you apply a relatively small force at the end of the crowbar (with a long lever arm), and the crowbar applies a much greater force with a shorter lever arm to the nail. For example, if the crowbar is 1 ft long, with the fulcrum at the far end of the crowbar, and you apply an upward force of 50 lb, the moment of your applied force is 50 ft-lb. If the force on the nail is applied 2" from the end of the crowbar, the moment is also 50 ft-lb, but the pulling force is 300 lb.

Your moment: 50 lb x 1 ft = 50 ft-lb
Nail moment: 300 lb x 1/6 ft = 50 ft-lb
 
  • #7
Thanks for helping me out with this. I remember from school that the force (not the mass) is multiplied by the distance from the fulcrum. This is clear, if F is 100N then it will apply 200Nm upwards on Body A.

What I’m trying to understand firstly is if A and B are fixed and with F you are trying to force them apart how do you calculate the force on the two body's?

lever_single.jpg
 

Attachments

  • lever_single.jpg
    lever_single.jpg
    15 KB · Views: 1,532
  • #8
pete said:
Thanks for helping me out with this. I remember from school that the force (not the mass) is multiplied by the distance from the fulcrum. This is clear, if F is 100N then it will apply 200Nm upwards on Body A.

What I’m trying to understand firstly is if A and B are fixed and with F you are trying to force them apart how do you calculate the force on the two body's?

View attachment 220549
If A and B are fixed, you can't pry them apart. "Fixed" means that they don't move. If the fulcrum (fixed point) is at A, and you apply a downward force of 100 Nt at the end, there will be a force of 200 Nt downward at B.
Similarly if the fulcrum is at B, a 100 Nt downward force at the end results in a 200 Nt force upward at A.
 
  • #9
pete said:
I remember from school that the force (not the mass) is multiplied by the distance from the fulcrum.
The defining equation says that the two moments will be equal.
If FA is one of the forces, applied at a distance of LA from the fulcrum, the moment is FALA. The other moment, defined similarly, is FBLB. The equation is FALA = FBLB.
 
  • #10
Mark44 said:
If A and B are fixed, you can't pry them apart. "Fixed" means that they don't move. If the fulcrum (fixed point) is at A, and you apply a downward force of 100 Nt at the end, there will be a force of 200 Nt downward at B.
Similarly if the fulcrum is at B, a 100 Nt downward force at the end results in a 200 Nt force upward at A.
A few people have said this to me that it’s fixed or it’s not and you can't talk about force on a fixed object. This question is from looking at a type of clamp and it can be very tightly done up or very loose without the two clamping surfaces moving about. What the object was fixed but had a breaking point of 500Nt, there must be a way to talk about the forces acting on a fixed object to understand when it’ll break?
 
  • #11
So for example if A and B are fixed unbreakable scales, so that when I push down on the lever I see the scales at A and B both move up. Now as in your example you apply 100Nt downward force at the end, what will the scales read? Will they both read 200Nm? Do they read the same? How do I work this out? What I’ve been doing in my attempt at a solution to the two lever problem was divide it in half, so 100Nt on A and 100Nt on B but I don't know if I’m doing it right?
 
  • #12
pete said:
What I’m trying to understand firstly is if A and B are fixed and with F you are trying to force them apart how do you calculate the force on the two body's?

Lets put some directions on the forces...

Lever.png


Lets say F = 10N.

If the lever isn't accelerating in rotation then the net torque must be zero. If we define clockwise as positive then taking moments about point B we can write..
-(A*1) + (2*F) = 0
or rearrange that to give..
A = 2F
and so A = 20N

If the lever isn't accelerating vertically then the net vertical force must also be zero. If we define up as positive then we can write..

(-A) + (+B) + (-F) = 0
or rearrange to give
B = F + A
= 10 + 20
= 30N
 

Attachments

  • Lever.png
    Lever.png
    932 bytes · Views: 629
  • #13
Those arrows really helped, Thanks, I appreciate the help. I've not disappeared I'm just trying to solve my original picture with the two levers in sequence.
 
  • #14
What I am confused about is that the clamp or anything you try to pry something apart with always applys equal pressure to both sides. If I push a lever, a crowbar between two posts and apply pressure it dose so to both sides equally, I thought? But in the examples like the one above you showed A and B are different? A=20N B=30N.
So in this picture: If A and B are the two surfaces and the lever a crowbar it should apply equal force to A and B. Whats the difference between my problem here and the example you gave?
lever_pry2.jpg
 

Attachments

  • lever_pry2.jpg
    lever_pry2.jpg
    14.4 KB · Views: 743
  • #15
So if F is 10Nt then A and B must both be 20Nt? I think the confusing thing is the idea of a kind of balance or see-saw with weights on ether side when I'm looking at a clamp between two surfaces. There is no up or down.
 
  • #16
pete said:
What I am confused about is that the clamp or anything you try to pry something apart with always applys equal pressure to both sides. If I push a lever, a crowbar between two posts and apply pressure it dose so to both sides equally, I thought?
Yes, because both surfaces you're trying pry apart have the same lever arm.
In your drawing the two posts don't have the same lever arms. If A is the fulcrum, a 10 N force downward at the right end causes a 30 N force downward at B. If B is the fulcrum, the same 10 N force downward at the end causes a 20 N force upward at A.
 
  • #17
So if A and B are round steel posts seen from above and the black line is a bar and I pull at F which is the fulcrum? I don't understand how there can be two answer to the problem. The only variable in the drawing is how much force you apply to the lever is it not?
 
  • #18
OK I think I understand. By fulcrum you mean one is fixed, say A and then the lever turns around it against B or B is fixed and the lever turns around it against A. But this is back to the same problem. They are both fixed. Like the clamp applying pressure to the inside of the fitting. They are both the fulcrum. Again if A and B were unmovable unbreakable scales giving a reading and I applied 10Nt to the end of the lever then what reading should I get at A and B? 20Nt?
 
  • #19
pete said:
So if A and B are round steel posts seen from above and the black line is a bar and I pull at F which is the fulcrum? I don't understand how there can be two answer to the problem. The only variable in the drawing is how much force you apply to the lever is it not?
Here's my understanding, which others are free to refute (it's been many years since I took a physics class);
Since the force against B will be 50% greater than the force against A, the post at B will be the one that is most likely to give. That's assuming that both posts resist with the same force.

Do you understand where I got the 50% figure? If not, around post B, the force applied to A is twice the force F, because one lever arm is 2 m. and the shorter lever arm is 1 m. Around post A, the force applied to B is three times the force F, because one lever is 3 m. and the shorter lever arm is 1 m.
 
  • #20
So if I take the scales example the reading I will get at A will be 20Nt, this is clear, but the reading at B your saying will be 30Nt. That derived from the length of the lever times the distance between A and B? I must be confused because that dose not seem to make sense, the force at B would increase the closer B got to your hands which is the wrong way round. How did you get the 30Nt at B?
 
  • #21
I found it, it's a second class lever. So the force times the length of the lever = 20Nm, that's the "moment?" on one side to find the other it's 20Nm divided by distance from A to B. So 20Nm. They are equal but only by coincidence. Anyhow I've had another go at the problem with my new physics powers, what do you think? I put the scales into try and help with the confusion.
Lever_solution_3.jpg
 

Attachments

  • Lever_solution_3.jpg
    Lever_solution_3.jpg
    35.3 KB · Views: 589
  • #22
My response before related to your drawing in post #14.
pete said:
So if I take the scales example the reading I will get at A will be 20Nt, this is clear, but the reading at B your saying will be 30Nt. That derived from the length of the lever times the distance between A and B?
No.I explained my reasoning in post #19.
Assuming that we consider A as the fulcrum, we have two moments. One is F x 3, or 3F. The other is ? x 1. For these moments to be equal, 3F has to be equal to ? x 1, so the force acting at B is 3F, or three times the force that is applied.

pete said:
I must be confused because that dose not seem to make sense, the force at B would increase the closer B got to your hands which is the wrong way round. How did you get the 30Nt at B?
Why would B get closer to your hands. I'm not assuming that points A and B are moving laterally.
 
  • #23
No their not moving I just got it wrong, I see what you mean now I'll have another go at it.
 
  • #24
pete said:
I must be confused because that dose not seem to make sense, the force at B would increase the closer B got to your hands which is the wrong way round.

If you imagine the goal is to exert a 10 Newton force on A, then the closer B is to your hands (where force F is applied) the more force you would have to exert there to accomplish that goal. So it makes sense that the force on B would increase.

You are imagining a situation where you have picked a force to exert with your hands and you are willing to let the forces on A and B take whatever values are needed to balance the situation. In that situation, it is true that the closer B is to your hands, the less force is exerted on A (and thus on B).
 
  • #25
You can see the diagram in #21 must be wrong by looking at the external forces...you have 90+10=100N acting downwards and only 60N upwards so there is a net force of 40N downwards. It implies the whole assembly is accelerating downwards.
 
  • #26
CWatters said:
You can see the diagram in #21 must be wrong by looking at the external forces...you have 90+10=100N acting downwards and only 60N upwards so there is a net force of 40N downwards. It implies the whole assembly is accelerating downwards.
Correction: I made a mistake with the directions, that should be 60+10=70N downwards and 90N upwards. Still doesn't sum to zero so it's accelerating.
 
  • #27
I don’t understand the accelerating downward bit, it’s a clamp restricted between two scales it can't go anywhere and I don’t see why it has to be balanced. If I had a fitting with a bolt that you tightened so it came out of one side then all of the force would come from that point but as it did so it would push it’s self against the opposite side so the force would be equal on both sides, in a 2D representation. Though different amounts of force are applied on both sides I then add them together and divide it by 2 to give the results on the scales. Here’s another go at:
Lever_correction_2.jpg

dose this one look right?
 

Attachments

  • Lever_correction_2.jpg
    Lever_correction_2.jpg
    35.7 KB · Views: 433
  • #28
Or do you mean both side have to add up to the same no matter what?
 
  • #29
pete said:
Though different amounts of force are applied on both sides I then add them together and divide it by 2 to give the results on the scales.
Which is not how it works. In this drawing, there's a force of 90 N on the upper scale, and a force of 100 N on the lower scale. If the upper scale reads 80 N, then that's the upward force, and similarly for the lower scale.
 
  • #30
pete said:
Or do you mean both side have to add up to the same no matter what?
If the system has a net force in some direction, then it will accelerate in that direction -- F = ma is the basic equation here.
 
  • Like
Likes CWatters
  • #31
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
 
  • #32
pete said:
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.

pete said:
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.
 
  • #33
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?
wedge.jpg
 

Attachments

  • wedge.jpg
    wedge.jpg
    11.1 KB · Views: 273
  • #34
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.
 
  • #35
Your drawing in #27 looks correct.
 
<h2>1. How do levers work?</h2><p>Levers work by using a pivot point, or fulcrum, to lift or move objects with less force than would be required without the lever. The lever is made up of a rigid bar, a fulcrum, and two points of force known as the effort and the load.</p><h2>2. What is the principle behind levers?</h2><p>The principle behind levers is the law of the lever, which states that the effort force multiplied by its distance from the fulcrum is equal to the load force multiplied by its distance from the fulcrum.</p><h2>3. What are the three types of levers?</h2><p>The three types of levers are first-class, second-class, and third-class. In a first-class lever, the fulcrum is located between the effort and the load. In a second-class lever, the load is between the fulcrum and the effort. In a third-class lever, the effort is between the fulcrum and the load.</p><h2>4. How do you calculate the mechanical advantage of a lever?</h2><p>The mechanical advantage of a lever can be calculated by dividing the distance from the fulcrum to the effort by the distance from the fulcrum to the load. This ratio represents the amount of force amplification that the lever provides.</p><h2>5. What are some real-life examples of levers?</h2><p>Some common examples of levers in everyday life include seesaws, scissors, crowbars, and wheelbarrows. Other examples include the human arm, which acts as a third-class lever, and the claw of a hammer, which acts as a first-class lever.</p>

1. How do levers work?

Levers work by using a pivot point, or fulcrum, to lift or move objects with less force than would be required without the lever. The lever is made up of a rigid bar, a fulcrum, and two points of force known as the effort and the load.

2. What is the principle behind levers?

The principle behind levers is the law of the lever, which states that the effort force multiplied by its distance from the fulcrum is equal to the load force multiplied by its distance from the fulcrum.

3. What are the three types of levers?

The three types of levers are first-class, second-class, and third-class. In a first-class lever, the fulcrum is located between the effort and the load. In a second-class lever, the load is between the fulcrum and the effort. In a third-class lever, the effort is between the fulcrum and the load.

4. How do you calculate the mechanical advantage of a lever?

The mechanical advantage of a lever can be calculated by dividing the distance from the fulcrum to the effort by the distance from the fulcrum to the load. This ratio represents the amount of force amplification that the lever provides.

5. What are some real-life examples of levers?

Some common examples of levers in everyday life include seesaws, scissors, crowbars, and wheelbarrows. Other examples include the human arm, which acts as a third-class lever, and the claw of a hammer, which acts as a first-class lever.

Similar threads

Replies
21
Views
1K
Replies
8
Views
1K
Replies
28
Views
763
  • Mechanical Engineering
Replies
2
Views
872
Replies
5
Views
1K
  • Mechanical Engineering
Replies
30
Views
6K
Replies
4
Views
2K
  • Mechanical Engineering
Replies
1
Views
5K
Replies
16
Views
10K
Replies
2
Views
2K
Back
Top