Velocity ratio of a third class lever

[rad10k]

Homework Statement

3rd class lever 5m distance from load to effort and 1m distance from effort to pivot.

Load = 1962

Effort required to lift load = 11772N

MA = 0.16 or 1/6 ( so there really is no MA)

Calculate the velocity ratio of the system ?

Homework Equations

I have been told to use vertical distance to solve but I don't know how to work ouyt the vertical distance

VR = Distance moved by effort / distance moved by load

The Attempt at a Solution

My attempt was

1 / 6 = 0.167777 or 0.17

I have been told this is wrong and I must use the vertical distance but I can't find anywhere that will help me work this out can someone offer me guidance please.

 

[PhanthomJay]

That looks correct to me. You can find the ratio using the geometry of similar triangles ( draw a sketch)...if the load end moves up 1 m, the effort point moves up by 1/6 m, during that same time period. You can assume any distance you want...the ratio will be the same for vertical distance or velocity.

 

[rad10k]

thanks

 

[SammyS]

Homework Statement

3rd class lever 5m distance from load to effort and 1m distance from effort to pivot.

Load = 1962

Effort required to lift load = 11772N

MA = 0.16 or 1/6 ( so there really is no MA)

Calculate the velocity ratio of the system ?

Homework Equations

I have been told to use vertical distance to solve but I don't know how to work ouyt the vertical distance

VR = Distance moved by effort / distance moved by load

The Attempt at a Solution

My attempt was

1 / 6 = 0.167777 or 0.17

I have been told this is wrong and I must use the vertical distance but I can't find anywhere that will help me work this out can someone offer me guidance please.

Homework Statement

Homework Equations

The Attempt at a Solution

That looks correct to me. You can find the ratio using the geometry of similar triangles ( draw a sketch)...if the load end moves up 1 m, the effort point moves up by 1/6 m, during that same time period. You can assume any distance you want...the ratio will be the same for vertical distance or velocity.

Since it's the load which moves at a greater velocity (It moves a greater distance in a given time.), the velocity ratio is:

$${{v_{Load}}\over{v_{Effort}}}={{6}\over{1}}\ .$$

 

[PhanthomJay]

Since it's the load which moves at a greater velocity (It moves a greater distance in a given time.), the velocity ratio is:

$${{v_{Load}}\over{v_{Effort}}}={{6}\over{1}}\ .$$

But the VR is V_effort/V_load = 1/6.

 

[SammyS]

But the VR is V_effort/V_load = 1/6.

If I did that wrong, I apologize.

I was thinking that since Mechanical Advantage is:

$$\displaystyle \text{MA}={{F_{Load}}\over{F_{effort}}}\,,$$

then velocity ratio would be similar.

 

[PhanthomJay]

I wasn't sure of the ratio either, which was the numerator and which was the denominator? I gather for a third class lever, where you have to apply a large force to lift a small load, the MA is less than 1, sacrificed to get a higher load velocity.

 

[rad10k]

yes the MA is 1/6 which I am dividing 1 by 6 to get 0.17(0.167777) to us in the VR formula



Assuming 100% efficiency , if the energy put in was 3000j would I be correct in think that
the output energy would also be 3000J since ouput = input/efficiency ie. 3000J/100*100 = 3000J ?

Then the next question relating is : If the efficiency of the system is 65% , calculate the Velocity ratio?

The formula I am using for this is : VR = MA/efficency

so, VR 0.17/65*100 = 0.26

Although that is more than I started with so I think how can that be correct?

Thanks for any help

 

[PhanthomJay]

Efficiency is output/input.

 

[Dr Seuss]

Hi, I'm stuck on the same question.

The velocity ratio seems at first glance to be 1/6 as the load is 6 times further from the fulcrum than the effort though in the next question it states the efficiency is 65% giving a vr of 0.256?