Efficence - Calculate the Force needed to life this load

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SUMMARY

The discussion focuses on calculating the force required to lift a load of 2000N using a mechanical car jack with 40% efficiency. Given that one rotation of the screw raises the load by 10mm and the lever arm is 0.5m long, the work done by the lever must equal the work done in raising the vehicle. Due to the jack's inefficiency, the applied force must account for an additional 40% work to achieve the desired lift.

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Homework Statement


Consider a mechanical car jack. To change a wheel it may need to lift a load of 2000N. Suppose that one rotation of the screw raises the car by 10mm. You apply a force to the end of the lever, 0.5m long, through a full circle. The jack is only 40% efficient. Calculate the force that you have to apply.

(This is a textbook question from OCR, AS Physics.)

Homework Equations



probably to do with moments:
anti-clockwise moments = clockwise moments

work done:
work done = force x perpendicular distance...

GPE:
gravitational pe = mass x g x height

I don't know where to begin, etc..

The Attempt at a Solution



There is answer at the back of the textbook, but I won't show it for now, so that it doesn't affect your way of workings, or anything... I'll show it if needed though.

Thank you for any help!
 
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The work done by the movement of the lever must equal the work done in raising the vehicle. With the jack not being 100% efficient due to friction, you'll have to supply 40% more work to raise it.

So begin thinking about work being done.
 

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