Homework Help: Lever with spread load: formula

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.

 

Assuming the 5t load is uniform, you can take it as acting through its centre of mass. Can you post the equation you get?

 

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.

 

Is that an equation?

 

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!!) it would (should) become apparent that you have omitted something on the other side.
Do we measure force in kg?

 

That looks right, but don't forget g.

 

What units do you use for force in your school?

 

So do you understand what haruspex was getting at?

 

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?

 

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

 

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.

 

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.