The large piston in a hydraulic lift has a radius of 250 cm^2. What force must be applied to the small piston with a radius of 25 cm^2 in order to raise a car of mass 1500 kg?

According to Pascal's principle, pressure applied to a confined fluid is transmitted equally throughout the fluid in all directions. In the case of a hydraulic lift, this means that a force applied to a small piston will be transmitted through the hydraulic fluid and result in a larger force on a larger piston. The relationship between force, pressure, and area is given by the equation: {eq}F = Ptimes A {/eq} where F is the force applied, P is the pressure exerted, and A is the area over which the force is distributed. Since we know the radius of the large piston is {eq}sqrt{250 cm^2/ pi} \approx 8.9 cm {/eq}, its area is: {eq}A_1 = pi r_1^2 \approx 248.2 cm^2 {/eq} Similarly, the radius of the small piston is {eq}sqrt{25 cm^2/ pi} \approx 1.8 cm {/eq}, so its area is: {eq}A_2 = pi r_2^2 \approx 10.2 cm^2 {/eq} We are given that the car has a mass of 1500 kg. To lift the car, we need to exert an upward force equal to its weight, which is given by: {eq}F_g = mg = (1500 kg)(9.8 m/s^2) \approx 14,!700 N {/eq} Since the pressure must be the same on both pistons, we can set up the following equation: {eq}P_1 = P_2 Rightarrow \frac{F_1}{A_1} = \frac{F_2}{A_2} {/eq} where {eq}F_1 {/eq} and {eq}F_2 {/eq} are the forces applied to the large and small pistons, respectively. Solving for {eq}F_2 {/eq}, we get: {eq}F_2 = \frac{A_2}{A_1} times F_1 {/eq} Plugging in the values we know, we get: {eq}F_2 = \frac{10.2}{248.2} times 14,!700 \approx 603 N {/eq} Therefore, a force of \approx imately 603 N must be applied to the small piston to lift the car.