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Lever with spread load: formula

  • Thread starter barvas11
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  • #1
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Hi Guys,
How much force will I need to lift 5000 kg block and how much force will be acting on the pivot point?
I know how to calculate a normal lever but couldn't find any formula for lever with load evenly spread as per the attached diagram.
see the attached diagram
Thanks
 

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Answers and Replies

  • #2
785
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Hi Guys,
How much force will I need to lift 5000 kg block and how much force will be acting on the pivot point?
I know how to calculate a normal lever but couldn't find any formula for lever with load evenly spread as per the attached diagram.
see the attached diagram
Thanks
Do you know about "the principle of lever " ? Minimum force you'd require will balance the load of 5000 kg. Hence lever will be in equilibrium. Clockwise and anticlockwise moments will add up to 0.
 
  • #3
haruspex
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Assuming the 5t load is uniform, you can take it as acting through its centre of mass. Can you post the equation you get?
 
  • #4
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Do you know about "the principle of lever " ? Minimum force you'd require will balance the load of 5000 kg. Hence lever will be in equilibrium. Clockwise and anticlockwise moments will add up to 0.
Hi,
Thanks I know this one..... I am not that stupid :D
 
  • #5
5,439
7
So what was your attempt?

Posting this is a requirement of the forum and Haruspex has already offered a hint.

By the way a simpler and smaller diagram would make things easier for you.
 
  • #6
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Assuming the 5t load is uniform, you can take it as acting through its centre of mass. Can you post the equation you get?
I kinda thought that. It would make sense but I need to be 100% sure.
The equation will be (5000 * 1250) / 800
 
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  • #7
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Is that an equation?
 
  • #8
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So what was your attempt?

Posting this is a requirement of the forum and Haruspex has already offered a hint.

By the way a simpler and smaller diagram would make things easier for you.
Sorry for the size of my diagram.... it's a printed PDF from Acad
 
  • #9
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Is that an equation?
yes if you check my diagram you will see. I would post the picture on here but it won't allow me.
 
  • #10
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Well, like I said, I think your diagram makes it more difficult for yourself.

I assume that the mechanism is turned through 90 degrees and is actually some sort of beam balance with force F applied by some sort of bent arm?

You have correctly (I hope) taken the appropriate perpendicular distance for the moment arm for F about the pivot.

However if you wrote a proper equation (an equation has an equals sign!!:wink:) it would (should) become apparent that you have omitted something on the other side.
Do we measure force in kg?
 
  • #11
haruspex
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I kinda thought that. It would make sense but I need to be 100% sure.
The equation will be (5000 * 1250) / 800
That looks right, but don't forget g.
 
  • #12
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Well, like I said, I think your diagram makes it more difficult for yourself.

I assume that the mechanism is turned through 90 degrees and is actually some sort of beam balance with force F applied by some sort of bent arm?

You have correctly (I hope) taken the appropriate perpendicular distance for the moment arm for F about the pivot.

However if you wrote a proper equation (an equation has an equals sign!!:wink:) it would (should) become apparent that you have omitted something on the other side.
Do we measure force in kg?
That's exactly what it is :)
Yes the force will be in KGs
 
  • #13
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That looks right, but don't forget g.
I was checking formulas for levers and they don't mention the gravity......
 
  • #14
5,439
7
What units do you use for force in your school?
 
  • #15
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What units do you use for force in your school?
newtons of course :)
 
  • #16
5,439
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So do you understand what haruspex was getting at?
 
  • #17
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to be honest; no I don't
 
  • #18
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OK I have sketched your arangement as I understand it.

5000kg is 5 tonnes ie quite a heavy block so I have drawn the pivot mounted near the edge of something solid and the restraint as the bent arm being tied to the vertical face of the support.

Now moment = force time distance from the line of action of the force to the pivot.

So the block exerts a moment equal due to the downward force of gravity on its mass.

This downward force is given by multiplying its mass (5000kg) by g to get a force in newtons.

So the block exerts an anticlockwise moment of 5000g newtons.

This is balanced by a clockwise moment of F times y newtons.

I mentioned that I hope you understood the correct distance since in your diagram x = y.
Would you have got it correct if x did not equal y?

An equation contains an equals sign so an equation is

5000 x g x 1.25 = F x 0.8

I am using consistent units here (length in metres) . If you work in millimetres you may get the wrong answer if asked for the force.

In the second diagram I have shown a different block that does not extend the whole length of the left hand side.

This is to show that the principle is the same.

Can you see this?
 

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  • #19
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OK I have sketched your arangement as I understand it.

5000kg is 5 tonnes ie quite a heavy block so I have drawn the pivot mounted near the edge of something solid and the restraint as the bent arm being tied to the vertical face of the support.

Now moment = force time distance from the line of action of the force to the pivot.

So the block exerts a moment equal due to the downward force of gravity on its mass.

This downward force is given by multiplying its mass (5000kg) by g to get a force in newtons.

So the block exerts an anticlockwise moment of 5000g newtons.

This is balanced by a clockwise moment of F times y newtons.

I mentioned that I hope you understood the correct distance since in your diagram x = y.
Would you have got it correct if x did not equal y?

An equation contains an equals sign so an equation is

5000 x g x 1.25 = F x 0.8

I am using consistent units here (length in metres) . If you work in millimetres you may get the wrong answer if asked for the force.

In the second diagram I have shown a different block that does not extend the whole length of the left hand side.

This is to show that the principle is the same.

Can you see this?
Thanks you are a star :)

Another question is what will be force acting on the pivot point. My guess is that the force on the lifting arm will counteract the weight of of the block so it will be my result from the lever calc? Is that correct?
Thanks
 
  • #20
5,439
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Vertical equilibrium tells you that the upward reaction at the pivot must equal the weight of the block.

Can you tell me what horizontal force must act to balance the tie force F?
I haven't shown it on my diagram
 
  • #21
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The horizontal force will be
F=(a*W1)/Y
 
  • #22
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It would help if you indicated your level of knowledge.

Do you know about the three conditions for static equilibrium?

That is

1)The algebraic sum of the vertical forces = 0 (sum of downward acting forces = sum of upward acting forces)

2)The algebraic sum of the horizontal forces = 0 (sum of leftward acting forces = sum of rightward acting forces)

3)The sum of the clockwise moments = The sum of the anticlockwise moments

We used (3) to calculate F from W and the geometry.

However geometry doesn't enter into (1) and (2).

You asked me what force acts on the pivot.
 
  • #23
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Hi Studiot,
I am not sure what you mean. The weight required to balance my 5t block is 7.8t (77.7 KN)
 
  • #24
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am not sure what you mean.
That is why I asked about your level of knowledge.

This is not a criticism, but there is no point describing things in terms you have not met.

You drew a horizontal force originally preventing the beam holding the block from tipping. A horizontal force is not a weight.
 
  • #25
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The reason I am using weight for this homework is that there will be a piston pulling that arm. So I need to choose the right piston that will be able to tilt the mechanism in 90 degrees. Pistons are always rated in tonnes rather than KN
Sorry for causing a confustion ;)
 
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