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?
Question:
The large piston in a hydraulic lift has a radius of {eq}rm 250 cm^2 {/eq}. What force must be applied to the small piston with a radius of {eq}rm 25 cm^2 {/eq} in order to raise a car of mass 1500 kg?
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Answers (1)
HopeApril 19, 2023 в 13:29
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.
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