Problem 32 - Physics

Question

In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of \(45 \mathrm{~N}\) to the input piston, which has a radius \(r_{1}\). As a result, the output plunger, which has a radius \(r_{2}\), applies a force to the car. The ratio \(r_{2} / r_{1}\) has a value of 8.3. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.

Step-by-Step Solution

Answer

The force applied to the car by the output plunger is approximately 3102.85 N.

Step 1: Understand the Problem

We need to determine the force applied to the car by the output plunger of a hydraulic jack. We know the input force is 45 N and the ratio of the radii of the output plunger to the input piston is 8.3.

Step 2: Apply Pascal's Principle

According to Pascal's Principle, the pressure applied to a confined fluid is transmitted undiminished throughout the fluid. Therefore, the pressure on the input piston is equal to the pressure on the output plunger.

Step 3: Establish the Relationship Between Force and Area

The pressure applied by the input force is given by \( P = \frac{F_1}{A_1} \) where \( A_1 = \pi r_1^2 \). Similarly, the pressure on the output plunger is \( P = \frac{F_2}{A_2} \) where \( A_2 = \pi r_2^2 \). Given \( P_1 = P_2 \), \( \frac{F_1}{\pi r_1^2} = \frac{F_2}{\pi r_2^2} \).

Step 4: Use the Ratio of Radii

Since the ratio \( \frac{r_2}{r_1} = 8.3 \), substitute into the area relationship: \( \frac{F_1}{r_1^2} = \frac{F_2}{(8.3r_1)^2} \). Simplifying gives \( F_2 = F_1 \times 8.3^2 \).

Step 5: Calculate the Output Force

Substitute the known value to find \( F_2 \) : \[ F_2 = 45 \times 8.3^2 \]. Calculating this gives \( F_2 \approx 3102.85 \text{ N} \).

Key Concepts

Pascal's Principle

Pascal's Principle is a fundamental concept in fluid mechanics. It states that when pressure is applied to a confined fluid, the pressure change is transmitted equally throughout the entire fluid. This principle is vital for devices like hydraulic jacks, which are used to lift heavy loads with minimum effort. The pressure that we're mentioning is the same in all directions in a closed system, which is why a small force can exert a much larger force through the device allowing a motorist to lift a car easily. Understanding this principle helps us see how force is multiplied in hydraulic systems.
This principle can be applied in the context of the hydraulic jack, where the force exerted on the input piston is transmitted equally to the output plunger. Thus, the same pressure that the motorist applies on a small piston results in a significantly higher force output through the larger plunger, effectively lifting heavy objects.

Force Calculation

In hydraulic systems, calculating the force applied at different points involves understanding the relationship between force, pressure, and area. The formula used here is derived from Pascal's Principle and is given by:

• Pressure, \( P = \frac{F}{A} \)

Where \( F \) is the force applied and \( A \) is the cross-sectional area.

• Force on the input piston: \( F_1 \)
• Area of the input piston: \( A_1 = \pi r_1^2 \)
• Force on the output plunger: \( F_2 \)
• Area of the output plunger: \( A_2 = \pi r_2^2 \)

By using the relation that pressure is transmitted equally, \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \), you can find \( F_2 \) if \( F_1 \) is known. This relationship allows calculations to be simplified to solve for the unknown force. By rearranging and substituting known values, you can find that the output force \( F_2 \) is the input force times the square of the ratio of the radii of the plunger and piston. This effective control of force is what allows the jack to lift a car effortlessly.

Pressure and Area

The concept of pressure and its relation to area is pivotal in understanding how hydraulic jacks work. Pressure is defined as a force distributed over an area, mathematically expressed as \( P = \frac{F}{A} \). This relationship means that for a constant pressure, the force is directly proportional to the area.

In the hydraulic jack scenario, two pistons with different cross-sectional areas are involved. The input piston (or handle) applies a small force over a smaller area, creating a pressure in the fluid. The output piston, having a much larger area, converts this pressure into a larger force, due to the area multiplier.

• The smaller area of the input piston allows for ease of applying a force.
• The larger area of the output plunger allows it to exert significantly more force.

This directly illustrates how the multiplication effect in hydraulic systems allows for the lifting of heavy loads like a car, leveraging the principle of pressure distribution over larger areas.