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Asked by Aravind Last Modified
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To determine the maximum pressure the smaller piston would have to bear in a hydraulic automobile lift, we can utilize the principle of Pascal's Law, which states that pressure applied to a confined fluid is transmitted undiminished throughout the fluid in all directions.
Firstly, let's convert the area of the cross-section of the piston from square centimeters to square meters to maintain consistency in units.
Given: Area of cross-section of the piston (A) = 425 cm²
Converting cm² to m²: 1 cm² = 1 × 10^-4 m² So, 425 cm² = 425 × 10^-4 m² = 0.0425 m²
Now, let's use the formula for pressure:
Pressure (P) = Force (F) / Area (A)
We know that the force exerted by the car (weight) is the maximum load it can bear, which is the product of its mass (m) and the acceleration due to gravity (g).
Given: Maximum mass the lift can bear (m) = 3000 kg Acceleration due to gravity (g) = 9.8 m/s²
So, Force (F) = mass (m) × gravity (g) = 3000 kg × 9.8 m/s² = 29400 N
Now, plug in the values into the pressure formula:
Pressure (P) = Force (F) / Area (A) Pressure (P) = 29400 N / 0.0425 m² ≈ 690588.24 Pa
Therefore, the maximum pressure the smaller piston would have to bear is approximately 690588.24 Pascal (Pa).
As an experienced tutor on UrbanPro, I hope this explanation clarifies the concept for you. If you have any further questions or need clarification, feel free to ask!
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