Absolutely! Let's break down each part of this hydraulic lift problem step by step: --- ### **Given Data:** - **Vehicle weight (W):** 3500 lb - **Large piston diameter (D₂):** 14 in - **Small piston force (F₁):** 10 lb - **Automobile lift height (h₂):** 6 ft = 72 in --- ## **a) What diameter is required for the small input piston?** ### **Key principle: Pascal’s Law** The pressure is the same throughout a confined fluid: \[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \] So, \[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \] Where: - \(F_1\) = 10 lb (input force) - \(F_2\) = 3500 lb (output force) - \(A_1\) = Area of small piston = \(\frac{\pi d_1^2}{4}\) - \(A_2\) = Area of large piston = \(\frac{\pi d_2^2}{4}\), \(d_2 = 14\) in Plug in the values: \[ \frac{10}{A_1} = \frac{3500}{A_2} \implies A_1 = A_2 \cdot \frac{10}{3500} = A_2 \cdot \frac{1}{350} \] Calculate \(A_2\): \[ A_2 = \frac{\pi (14)^2}{4} = \frac{\pi \cdot 196}{4} = 49\pi \text{ in}^2 \] So, \[ A_1 = 49\pi \cdot \frac{1}{350} = \frac{49\pi}{350} \text{ in}^2 \] Now, area of a circle: \[ A_1 = \frac{\pi d_1^2}{4} \implies \frac{\pi d_1^2}{4} = \frac{49\pi}{350} \implies d_1^2 = 4 \cdot \frac{49}{350} = \frac{196}{350} = 0.56 \] \[ d_1 = \sqrt{0.56} \approx 0.75 \text{ in} \] **Answer:** **The required diameter for the small input piston is approximately 0.75 inches.** --- ## **b) How far must the mechanic move this piston to raise the automobile 6 ft?** The volume of fluid moved by both pistons must be equal: \[ A_1 \cdot d_1 = A_2 \cdot d_2 \] Here, \(d_1\) is the distance the input piston moves, \(d_2\) is the distance the car is raised (6 ft = 72 in). \[ A_1 \cdot x = A_2 \cdot 72 \implies x = \frac{A_2 \cdot 72}{A_1} \] From above: - \(A_2 = 49\pi\) - \(A_1 = \frac{49\pi}{350}\) \[ x = \frac{49\pi \cdot 72}{\frac{49\pi}{350}} = 72 \cdot 350 = 25,200 \text{ in} \] Convert to feet: \[ \frac{25,200}{12} = 2,100 \text{ ft} \] **Answer:** **The mechanic must move the piston 2,100 feet!** --- ## **c) What is the operating pressure of the system?** \[ \text{Pressure} = \frac{F_1}{A_1} = \frac{10}{A_1} \] From before, \(A_1 = \frac{49\pi}{350} \approx 0.44 \text{ in}^2\): \[ \text{Pressure} = \frac{10}{0.44} \approx 22.7 \text{ psi} \] **Answer:** **The operating pressure is approximately 22.7 psi.** --- ## **d) Is this a practical system?** - The input piston is **very tiny** (less than 1 inch in diameter). - The mechanic would have to **move it 2,100 feet** (over 1/3 mile) to lift the car 6 feet! **Conclusion:** **No, this is not a practical system.** While the pressure is reasonable, the distance the mechanic must move the piston is absurdly large, making the design impractical. --- ### **Summary Table** | Part | Answer | |------|--------| | a | Small piston diameter ≈ **0.75 in** | | b | Mechanic must move piston **2,100 ft** | | c | Operating pressure ≈ **22.7 psi** | | d | **Not practical** (unrealistic piston travel distance) | If you have follow-up questions or want to see the steps in more detail, just ask!