Certainly! Here's a step-by-step explanation for the problem related to the lifting jack: --- ## Given Data: - Pump piston diameter = 15 mm - Load piston diameter = 75 mm - Load to be lifted = 100 kN --- ## Step 1: Describe how the lifting jack works based on fluid mechanics The lifting jack operates on **Pascal's Principle**, which states that **pressure applied to a confined fluid is transmitted equally in all directions**. - When the pump piston is pressed, it exerts pressure on the hydraulic fluid. - This pressure is transmitted to the load piston. - Because the pressure is the same in both pistons, the force is related to the cross-sectional area. --- ## Step 2: Role of each piston in force transmission - **Pump piston** (smaller piston): When you apply force here, it creates pressure in the fluid. - **Load piston** (larger piston): This receives the transmitted pressure and exerts a force to lift the heavy load. --- ## Step 3: Calculate the force needed on the pump piston to lift 100 kN Using Pascal's Principle: \[ \text{Force on pump piston} (F_p) = \text{Force on load piston} (F_L) \times \frac{A_p}{A_L} \] - \(A_p\): Cross-sectional area of pump piston - \(A_L\): Cross-sectional area of load piston Calculate the areas: \[ A = \pi \times \left(\frac{d}{2}\right)^2 \] ### Pump piston area: \[ A_p = \pi \times \left(\frac{15\, \text{mm}}{2}\right)^2 = \pi \times (7.5)^2 = \pi \times 56.25 \approx 176.71\, \text{mm}^2 \] ### Load piston area: \[ A_L = \pi \times \left(\frac{75\, \text{mm}}{2}\right)^2 = \pi \times (37.5)^2 = \pi \times 1406.25 \approx 442.72\, \text{mm}^2 \] ### Calculate the force on pump piston: \[ F_p = F_L \times \frac{A_p}{A_L} \] Convert 100 kN to Newtons: \[ F_L = 100\, \text{kN} = 100,000\, \text{N} \] Now, \[ F_p = 100,000 \times \frac{176.71}{442.72} \approx 100,000 \times .0399 \approx 399\, \text{N} \] **Final answer:** The force required on the pump piston is approximately **399 N**. --- ## Summary: - The jack operates via hydraulic pressure transmission. - The force needed on the small piston (pump piston) to lift 100 kN load is approximately **399 N**. - This demonstrates the force multiplication capability based on piston areas. --- Let me know if you'd like explanations for the remaining steps!