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Steer me in the right direction. What class of lever are we talking about?

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Steer me in the right direction. What class of lever are we talking about?

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Fermat

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quasar987

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I believe that the mention of Mechanical Advantage imediately tells you what kind of lever we're talking about...

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Chi Meson

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HallsofIvy

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Chi Meson said:

Not

The basic concept for a lever is "conservation of energy" together with "work = force times distance". To have a mechanical advantage of 4, the load must be 4 times as great as the force applied. Set up your equation with "force applied*applied lever arm= (4*force applied)*load lever arm" to determine the relation between the two lever arms.

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--------------------------------------------------------------------------Fermat said:

A force of P newtons equals

1000 Newtons

------------

4

= 250 Newtons

So what we have so far is a load of 1000 Newtons on one end and a force of 250 Newtons on the other.

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HallsofIvy

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Chi Meson

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I knew that. Don't you think I knew that? I knew that. Of course I knew that. I did. :grumpy:HallsofIvy said:Notquitetrue. With a class 3 lever, the mechanical advantage is always less than 1.

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-----------------------------------------------------HallsofIvy said:

The length of the entire lever is 4m.

The ma=4 x 1

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Doc Al

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Then why does the problem state "A lever 2 meters long..."scientist said:The length of the entire lever is 4m.

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scientist said:The length of the entire lever is 4m.

The ma=4 x 1

i'd say, by logics, that the fulcrum is at 1/5th of the length, starting at the end which lifts, am i right?

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-------------------------------------------------------------Fermat said:

Is there a formula I use to calculate the reaction at the fulcrum?

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Fermat

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------------------------------------------------------------Fermat said:

Ok then,

the load = 1000 Newtons + force of 250 Newtons = 1250 Newtons, reaction force at the fulcrum.

the effort = load

------

MA

= 1000N

-----

4

= 250 Newtons

---------------------------------------------

Taking moments about the fulcrum:

Would clockwise moments = anticlockwise moments apply in this question?

This is what my book says.

Please post your reply and correct my answer.

scientist

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Fermat

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You haven't actually said that the force at the fulcrum = 1250N but, if so, then that is correct. Your downward forces (load plus effort) = 1250N, so your single upward force (fulcrum reaction) is also 1250N.scientist said:------------------------------------------------------------

Ok then,

the load = 1000 Newtons + force of 250 Newtons = 1250 Newtons, reaction force at the fulcrum.

Code:`the effort = load ------ MA = 1000N ----- 4 = 250 Newtons`

---------------------------------------------

Taking moments about the fulcrum:

Would clockwise moments = anticlockwise moments apply in this question?

This is what my book says.

Please post your reply and correct my answer.

scientist

Since the lever is assumed to be in static equilibrium, then

Σ Fv = 0

Σ Fh = 0

Σ T = 0

where Σ Fv is the sum of Vertical forces, Σ Fh is the sum of Horizontal forces, and Σ T is the sum of the Torques.

There are no horizontal forces and you have dealt with the vertical forces when finding the reaction at the fulcrum.

They could, yes.scientist said:------------------------------------------------------------

...

Taking moments about the fulcrum:

Would clockwise moments = anticlockwise moments apply in this question?

...

Now you can do Σ T = 0, as

Take moments about the fulcrum. Call the distance from the fulcrum to the load as x. So what is the distance fornt the fulcrum to the effort?

Then the moment of the Load about the fulcrum is equal to the moment of the Effort about the fulcrum. Solve for x.

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----------------------------------------------------------------Fermat said:This is what you mean to post, yes?

You haven't actually said that the force at the fulcrum = 1250N but, if so, then that is correct. Your downward forces (load plus effort) = 1250N, so your single upward force (fulcrum reaction) is also 1250N.

Since the lever is assumed to be in static equilibrium, then

Σ Fv = 0

Σ Fh = 0

Σ T = 0

where Σ Fv is the sum of Vertical forces, Σ Fh is the sum of Horizontal forces, and Σ T is the sum of the Torques.

There are no horizontal forces and you have dealt with the vertical forces when finding the reaction at the fulcrum.

They could, yes.

Now you can do Σ T = 0, asoneway of evaluating the position of the fulcrum

Take moments about the fulcrum. Call the distance from the fulcrum to the load as x. So what is the distance fornt the fulcrum to the effort?

Then the moment of the Load about the fulcrum is equal to the moment of the Effort about the fulcrum. Solve for x.

Yes, I mentioned at the beginning of my post that the reaction force at the fulcrum is 1250 Newtons.

The length between the application and the fulcrum is 4 meters. The length between the fulcrum and the load is 1 meter.

This is taken from the MA =4:1.

I added the 4 + 1 = 5 meters

The total lever length is 2 meters as stated in the question.

I divided the lever length of 2 meters by 5 meters.

2/5=0.4 meters.

X = 0.4 meters as the final answer.

So the pivot point is 0.4 meters from the load arm of the lever.

This is a first class lever.

The lever is said to be in a static equilibrium because the lever has no movement and the lever is balanced. So in other words the lever is perfectly straight across and there are no outside forces acting upon this lever.

Please post a reply.

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Doc Al

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Careful. Thescientist said:The length between the application and the fulcrum is 4 meters. The length between the fulcrum and the load is 1 meter.

This is taken from the MA =4:1.

If you call the load-fulcrum distance X, then the two distances are X and 4X. Their sum must equal 2 meters, so X + 4X = 5X = 2 meters. Thus X = 2/5 meters.I added the 4 + 1 = 5 meters

The total lever length is 2 meters as stated in the question.

I divided the lever length of 2 meters by 5 meters.

2/5=0.4 meters.

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HallsofIvy

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Yes, and 2/5= 0.4 meters as he said.

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-------------------------------------------Neohaven said:i'd say, by logics, that the fulcrum is at 1/5th of the length, starting at the end which lifts, am i right?

You are correct! Because the total length of 2 meters divide by

the MA of 4:1 equals 0.4 meters.

2 meters

-------

4+1=5

=0.4 meters

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I would like to thank all my physics tutors for the outstanding information I received during this lengthly post! I have learned a lot! I will study and learn these principles.

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