A hydraulic lift uses a lever to apply force. If an 80 N force is applied 28 cm from a pivot, and piston A is 4 cm from the pivot with an area of 20 cm², while piston B has an area of 100 cm², calcula

The moment about the pivot due to the applied force must equal the moment due to the force on piston A.

$$F_{applied} \times d_{applied} = F_A \times d_A$$

Given:
Applied force, $F_{applied} = 80 \text{ N}$
Distance from pivot to applied force, $d_{applied} = 28 \text{ cm}$
Distance from pivot to piston A, $d_A = 4 \text{ cm}$

$$80 \text{ N} \times 28 \text{ cm} = F_A \times 4 \text{ cm}$$ $$F_A = \frac{80 \text{ N} \times 28 \text{ cm}}{4 \text{ cm}}$$ $$F_A = 80 \text{ N} \times 7$$ $$F_A = \boxed{560 \text{ N}}$$

The pressure exerted by piston A on the hydraulic fluid is transmitted equally to piston B.

$$P_A = P_B$$ $$\frac{F_A}{A_A} = \frac{F_B}{A_B}$$

Given:
Force on piston A, $F_A = 560 \text{ N}$ (from part i)
Area of piston A, $A_A = 20 \text{ cm}^2$
Area of piston B, $A_B = 100 \text{ cm}^2$

$$\frac{560 \text{ N}}{20 \text{ cm}^2} = \frac{F_B}{100 \text{ cm}^2}$$ $$F_B = \frac{560 \text{ N}}{20 \text{ cm}^2} \times 100 \text{ cm}^2$$ $$F_B = 28 \text{ N/cm}^2 \times 100 \text{ cm}^2$$ $$F_B = \boxed{2800 \text{ N}}$$