Velocity Ratio of a 3rd Class Lever

[aaronh25]

Homework Statement

1962N Load 5 m from effort (hydraulic ram) and fulcrum 1m from effort.

Calculate the Velocity ratio, if work input is 3000j calculate work output.

Homework Equations

VR = distance moved by effort / distance moved by load

effort = 11772N

MA = load / effort = 0.17

The Attempt at a Solution

Now i thought this would be 1 / 6 = 0.16 However this doesn't really truly reflect the distances moved by either effort or load. So would it be correct to assume that I need to use trigonometry to work this problem out and give me correct hights etc.

any help would be very much appreciated

 

[eczeno]

why do you think it should be 1/6? you need to just think about the definition of velocity. velocity is just how far the thing goes divided by how long it took to get there (well, that is average velocity, but that's all we need here). you can use trig to solve the problem, but that is the long way around, a little geometry will do the trick.cheers

 

[HallsofIvy]

I don't see how "work input is 3000J" is relevant here. As eczeno says, this really just depends on the geometry.

 

[aaronh25]

I used the 1 / 6 because of the assumption that vr = ma in frictionless systems, so i assumed that the 1 being distance from the fulcrum and 6 being the 5m plus the 1. So i take it I am completley barking up the wrong tree here?

In what way would I be able to use geometry to solve this problem, if I am quite honest with you I don't really know where to start.

Your help is much appreciated

 

[eczeno]

so, for velocity, we only need to worry about how far each end of the lever moves when we perform our task, and how long it takes to move that distance. we can ignore all the real physics stuff like force and energy for now. so picture it in your head, the long end of the lever is pulled down and the sort end goes up. the nice thing is we know each of these motions took the same amount of time, so the ratio of the velocities is just the ratio of the distances.

hope this helps

 

[aaronh25]

Thank you I think that makes perfect sense to me, so in other words if the short end being 1m and the long end being 5m, we can use these distances to then give us the VR. for example 1/5.

Thats the way I have interpreted your description. Your wise words and patience are very helpful.

 

[aaronh25]

The 3000j input relates to a subquestion in this asking what would then be the useful work output

 

[eczeno]

you have interpreted correctly . though i feel the need to say that the distances traveled will not be 1m and 5m, but due to similar triangles, the ratio will be 1/5, and that is all we care about.

cheers

 

[aaronh25]

Thank you eczeno, will i need to convert the ratio into a decimal, and i take it that this will be like a dimensionless result, i.e not a measured unit?

Thanks for your help

 

[eczeno]

i typically leave ratios in rational form, but that is aesthetics more than anything. and yes, ratios are dimensionless.

 

[aaronh25]

Thanks again eczeno, i have one more question, how do i calculate the useful energy output of this system if the work input is 3000j, woulkd this question be related to ma?

 

[eczeno]

oops, now that i read the problem more carefully, i see that the MA (and so VR also) is 1/6. i read it as saying the load was 5m from the fulcrum, but it is 5m from the effort, so 6m from the fulcrum. ok, then the load and effort force make sense.

now, as for the work, that's easy, energy is conserved so work in equals work out.