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werji · Mar 29, 2009, 02:54 PM

It takes a 60 lb force on a jack to lift a quarter (a corner) of a 2000 lb car. The force is applied over a distance of 10 inches and the car's corner moves up one half inch.
Calculate IMA, AMA and Efficiency of the jack?

stingRay18 · Apr 5, 2009, 06:03 AM

Originally Posted by werji

It takes a 60 lb force on a jack to lift up a quarter (a corner) of a 2000 lb car. The force is applied over a distance of 10 inches and the car's corner moves up one half inch.
Calculate IMA, AMA and Efficiency of the jack?

$IMA=\frac {D_e}{D_r}\\<br /> AMA=\frac {F_r}{F_e}\\<br /> \frac {AMA}{IMA}= eff$

The effort distance is 10 inches and a resistance distance of 0.5 inches you can find the IMA.

The resistance force is one quarter of the 2000 pound car which is lifted and the effort force is 60 pounds calculate the AMA and then the efficiency.