Why would you expect the lever shown in the top image...

Why would you expect the lever shown in the top image (below) to have a greater efficiency than the inclined plane shown in the bottom image (below)? IMA = Le / Lr IMA = L / h IMA = L / t The effort arm is shorter in case of the inclined plane. The resistance arm is shorter in case of the inclined plane. The area of contact is greater in case of the inclined plane.

As seen in Fig. $7-5$, a jackscrew has a lever arm of $40 \mathrm{~cm}$ and a pitch of $5.0 \mathrm{~mm}$. If the efficiency is 30 percent, what horizontal force $F$ applied perpendicularly at the end of the lever arm is required to lift a load $F_{W}$ of 270 kg? When the jack handle is moved around one complete circle, the input force moves a distance $$ 2 \pi r=2 \pi(0.40 \mathrm{~m}) $$ while the load is lifted a distance of $\mathrm{m}$. The IMA is therefore $$ \mathrm{IMA}=\text { Distance ratio }=\frac{2 \pi(0.40 \mathrm{~m})}{0.0050 \mathrm{~m}}=0.50 \times 10^{3} $$Since efficiency = AMA / IMA, we have $$ \text { AMA }=(\text { Efficiency })(\mathrm{IMA})=(0.30)(502)=0.15 \times 10^{3} $$ But AMA = (Load lifted)/(Input force) and therefore $$ F=\frac{\text { Load lifted }}{\mathrm{AMA}}=\frac{(270 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{151}=18 \mathrm{~N} $$

Question

Please give Ace some feedback