Consider a hydraulic lift whose cross-sectional areas A1 and A2 are both circles. If a force F1=125N is required to lift a 1520-kg car, what must be the ratio d2/d1 where d1 and d2 are the diameters of the cross-sectional areas A1 and A2?
F2/F1 = 1520/125 = Area2/Area1 = pi d2^2/4 / pi d1^2/4
= d2^2/d1^2
d2/d1 = sqrt (F2/F1) = sqrt (1520/125)
To solve this problem, we need to apply Pascal's law, which states that pressure in a fluid is transmitted equally in all directions. In this case, we can consider the hydraulic lift as a simple Pascal's law application.
Let's break down the problem step-by-step:
1. Convert the given force F1 from newtons (N) to pressure (P):
Pressure (P) = Force (F) / Area (A)
Pressure (P1) = F1 / A1
2. Use Pascal's law to establish the pressure in the second area (P2):
P1 = P2
3. Determine the relationship between areas A1 and A2:
P1 = P2
F1 / A1 = F2 / A2
F1 / A1 = F2 / (π * (d2/2)^2) [Since the areas are circles, A2 = π * r^2]
F1 / A1 = F2 / (π * (d2^2 / 4)) [Expanded the formula for the area of a circle]
4. Rearrange the equation to solve for the ratio d2/d1:
F1 / F2 = A1 / (π * (d2^2 / 4))
F2 / F1 = (π * (d2^2 / 4)) / A1
F2 / F1 = (π * (d2^2 / 4)) / (π * (d1/2)^2) [Substituting A1 with π * (d1/2)^2]
F2 / F1 = (d2^2 / 4) / (d1^2 / 4)
F2 / F1 = (d2^2 / d1^2)
Therefore, the ratio d2/d1 is given by:
d2 / d1 = sqrt(F2 / F1) = sqrt(125N/1520kg)
5. Calculate the numerical value of the ratio d2/d1:
d2 / d1 ≈ sqrt(0.0822368) ≈ 0.2864
So, the ratio d2/d1 is approximately 0.2864.