An automobile sack is lifted by a hydraulic jack that consists of two pistons. The large piston is `1 m` in diameter and the small piston is `10 cm` in diameter. If W be the weight of the car, how much smaller a force is needed on the small piston to lift the car?

Answer

To solve the problem of how much smaller a force is needed on the small piston to lift the car using a hydraulic jack, we can follow these steps:

### Step 1: Understand the relationship between pressure and force
In a hydraulic system, the pressure exerted on both pistons is the same. The pressure (P) is defined as the force (F) applied per unit area (A):
\[ P = \frac{F}{A} \]

### Step 2: Calculate the area of both pistons
The area of a piston can be calculated using the formula for the area of a circle:
\[ A = \frac{\pi D^2}{4} \]
where D is the diameter of the piston.

- For the large piston (diameter = 1 m):
\[ A_{large} = \frac{\pi (1)^2}{4} = \frac{\pi}{4} \, m^2 \]

- For the small piston (diameter = 10 cm = 0.1 m):
\[ A_{small} = \frac{\pi (0.1)^2}{4} = \frac{\pi (0.01)}{4} = \frac{\pi}{400} \, m^2 \]

### Step 3: Set up the equation based on equal pressures
Since the pressure is the same on both pistons, we can set up the equation:
\[ \frac{W}{A_{large}} = \frac{F}{A_{small}} \]
where W is the weight of the car and F is the force applied on the small piston.

### Step 4: Substitute the areas into the equation
Substituting the areas we calculated:
\[ \frac{W}{\frac{\pi}{4}} = \frac{F}{\frac{\pi}{400}} \]

### Step 5: Simplify the equation
Cross-multiplying gives:
\[ W \cdot \frac{\pi}{400} = F \cdot \frac{\pi}{4} \]
Cancelling \(\pi\) from both sides:
\[ \frac{W}{400} = \frac{F}{4} \]

### Step 6: Solve for F
Now, we can solve for F:
\[ F = \frac{W}{400} \cdot 4 \]
\[ F = \frac{W}{100} \]

### Conclusion
Thus, the smaller force needed on the small piston to lift the car is:
\[ F = \frac{W}{100} \]