Certainly! Let's go through each part of the problem step-by-step. --- ## Given Data: - Pump piston diameter, \( d_p = 15 \text{ mm} \) - Load piston diameter, \( d_l = 75 \text{ mm} \) - Load to be lifted, \( W = 100 \text{ kN} = 100,000 \text{ N} \) --- ## 1. How does the lifting jack operate based on fundamental fluid mechanics principles? ### Explanation: A hydraulic lifting jack works on Pascal's Principle, which states that **pressure applied to a confined fluid is transmitted equally in all directions**. - The pump piston applies a force \( F_p \) on the fluid, creating a pressure \( P \). - This pressure acts on the load piston, producing a force \( F_l \) that lifts the load. - The pressure in both pistons is the same because they are connected via the fluid. ### Mechanical relationship: \[ P = \frac{F_p}{A_p} = \frac{F_l}{A_l} \] where: - \( A_p \) = cross-sectional area of the pump piston - \( A_l \) = cross-sectional area of the load piston --- ## 2. Role of each piston in force transmission ### Pump piston: - **Role:** Creates hydraulic pressure by applying a force. - **Function:** When the operator applies force on the pump piston, it displaces fluid, generating pressure. ### Load piston: - **Role:** Transmits the hydraulic pressure to lift the load. - **Function:** The pressure acts on the load piston, producing a force that lifts the load. --- ## 3. Calculate the force needed on the pump piston to lift 100 kN ### Step 1: Find cross-sectional areas. The area of a piston: \[ A = \frac{\pi}{4} d^2 \] - Pump piston area: \[ A_p = \frac{\pi}{4} (15 \text{ mm})^2 = \frac{\pi}{4} \times (15)^2 = \frac{\pi}{4} \times 225 \approx 176.7 \text{ mm}^2 \] - Load piston area: \[ A_l = \frac{\pi}{4} (75 \text{ mm})^2 = \frac{\pi}{4} \times (75)^2 = \frac{\pi}{4} \times 5625 \approx 4417.9 \text{ mm}^2 \] ### Step 2: Convert areas to \( \text{m}^2 \): \[ A_p = 176.7 \times 10^{-6} \text{ m}^2 \] \[ A_l = 4417.9 \times 10^{-6} \text{ m}^2 \] ### Step 3: Calculate the pressure needed to lift the load: \[ P = \frac{W}{A_l} = \frac{100,000 \text{ N}}{4417.9 \times 10^{-6} \text{ m}^2} \] \[ P \approx 22,649,071 \text{ Pa} \approx 22.65 \text{ MPa} \] ### Step 4: Calculate the force on the pump piston: \[ F_p = P \times A_p = 22.65 \times 10^6 \times 176.7 \times 10^{-6} \] \[ F_p \approx 22.65 \times 176.7 \approx 4,005 \text{ N} \] **Final answer:** \[ \boxed{ \text{Force needed on pump piston} \approx \mathbf{4,005 \text{ N}} } \] --- ## 4. Calculate the pressure of the oil inside the jack From the previous calculations: \[ P \approx 22.65 \text{ MPa} \] --- ## 5. Effect of increasing the pump piston diameter to 25 mm ### Step 1: New pump piston area: \[ A_{p,new} = \frac{\pi}{4} (25)^2 = \frac{\pi}{4} \times 625 \approx 490.9 \text{ mm}^2 \] ### Step 2: Recalculate the force needed: Since the pressure remains the same, the force on the larger pump piston: \[ F_{p,new} = P \times A_{p,new} \] \[ F_{p,new} \approx 22.65 \times 10^6 \times 490.9 \times 10^{-6} \] \[ F_{p,new} \approx 22.65 \times 490.9 \approx 11,112 \text{ N} \] **Conclusion:** - Increasing the pump piston diameter to 25 mm **increases** the force required to generate the same pressure, approximately **11,112 N**. - This means **more effort** is needed on the pump handle to generate the same lifting force. --- # Summary: | Part | Result | |---|---| | Force on pump piston for 100 kN lift | **~4,005 N** | | Oil pressure inside the jack | **~22.65 MPa** | | Effect of increasing pump piston diameter to 25 mm | **Increased force (~11,112 N)** required on the pump piston | If you need further clarifications or calculations, feel free to ask!