What Calculations Are Needed to Solve This Third Order Lever Problem?

[junkie_ball]

Homework Statement

The 200Kg Crates are placed on a hydraulic lifting beam: (Diagram attached)

A. State the order of lever used here
B. Calculate the effort required to lift the load
C. Calculate mechanical advantage for this system
D. Calculate the velocity ratio for this system. If the work put in is 3000J.Calculate the useful energy output.
E. System has efficiency of 65%. Calculate the velocity ratio for this system.

Homework Equations

Effort Required = large distance x small effort = small distance x large load

Mechanical advantage = load/effort

Velocity ratio = distance moved by effort/distance moved by load OR
Velocity ratio = Mechanical Advantage/Efficiency

Efficiency = output/input

Efficiency = mechanical advantage/velocity ratio

The Attempt at a Solution

A. Third Order as the effort lies between to the fulcrum and load

B. Force Effort (Fe) x Distance Effort (De) = Force Load (Fl) x Distance Load (Dl)

Fe = (Fl x Dl)/De
Fe = (200 x 6)/1
Fe = 1200N

C. Mechanical Advantage (Ma) = Load / Effort

Ma = 200/1200
Ma = 0.16 Recurring

D. This is where i have become slightly confused do i assume the system is 100% efficient for this part of the question? In which case i could use the following to calculate the valocity ratio

Velocity ratio (Vr) = Mechanical Advantage (Ma) /Efficiency (E)
Vr = 0.16/100
Vr = 0.01666 (THIS SEEMS INCORRECT TO ME)

As such the equation Velocity ratio = distance moved by effort/distance moved by load would be the one to use. The problem is i have not been given the distances moved. I know i need to take the work put in somehow to get my answer but not sure how to use this. Could someone please point me in the right direction?

E. I will calculate this once I've solved for part D

 

[junkie_ball]

Hi,

I guess i worded this question wrong. Anyway have spend some more time on it and now have a slightly different question for this system. The force load is 200KG but as there is gravity acting on the crate will the force load be (200 x 9.81) making it 1962N rather than 200N?

I have now figured out how to use the work put into calculate the displacement. One last question as the effort displacement is 5m from the load effort will the displacement be 5x larger or is it 6x as this is the distance from the fulcrum. I believe it will be 5x?