Question

The effort measured in S.I. system in lifting a load through a simple machine is numerically equal to its mechanical advantage. If the mechanical advantage of the machine is increased by 20%, the same effort can lift a load of 12 kgwt. Find the magnitudes of the effort and the original load. (Take `g=10 ms^(-2)`)

Answer 1

I case: `MA_(1)=E=(L)/(E)rArrL=E^(2)`
II case ` = MA_(2)`
`=MA_(1)+20%` of `MA_(1)`
`=1.2 MA_(1)`
`=1.2E`
But `MA_(2)=(L^(1))/(E)`
Given `L^(1)=12` kgwt
120 N
Hence, `MA_(2)=(120)/(E)=1.2E`.
Or
` E^(2)=100 rArr E = 10N`
Hence, the original load `= E^(2)=100N`