Problem 77 - College Physics

Question

In preparation for its tire rotation, a car weighing 2.25 tons is placed on a hydraulic garage lift. The mechanic then raises the car \(30.0 \mathrm{~cm} .\) (a) Calculate the work done on the car when it is lifted. (b) Assuming no frictional losses in the hydraulic fluid, how much work was done by the lift on the input side? (c) What was the force on the input side if its piston moved \(52.5 \mathrm{~cm} ?\) (d) Determine the ratio of the input side area to that of the lifting side (output) area?

Step-by-Step Solution

Answer

Work done lifting the car: 6,621.75 J; work by lift: 6,621.75 J; input force: 12,612.86 N; area ratio: 0.571.

Step 1: Understand the Problem

We need to calculate the work done to lift a 2.25-ton car (1 ton = 1000 kg) 30.0 cm upwards using a hydraulic lift, assuming no frictional losses. We'll calculate the gravitational force, work done lifting the car, work done by the lift, and the force on the input side and area ratio of the pistons.

Step 2: Convert Units

First, convert the car's weight from tons to kilograms. 2.25 tons is equivalent to 2250 kg. Also, convert the lifting distance from centimeters to meters, giving us 0.30 meters.

Step 3: Calculate Gravitational Force

The gravitational force (weight) on the car is calculated using the formula: \( F = mg \), where \( m = 2250 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \). Thus, \( F = 2250 \times 9.81 = 22,072.5 \text{ N} \).

Step 4: Calculate Work Done on the Car

Work done (W) is calculated by \( W = F imes d \), where \( d = 0.30 \text{ m} \). So, \( W = 22,072.5 \times 0.30 = 6,621.75 \text{ J} \).

Step 5: Calculate Work Done by the Lift

Assuming no frictional losses, the work done by the lift on the input side equals the work done lifting the car. Therefore, it is also \( 6,621.75 \text{ J} \).

Step 6: Calculate Force on the Input Side

Given that the input piston's distance is 52.5 cm or 0.525 m, the input force \( F_i \) can be found using \( W = F_i imes d_i \). Since \( W = 6,621.75 \text{ J} \), we have \( F_i = \frac{6,621.75}{0.525} = 12,612.86 \text{ N} \).

Step 7: Determine Area Ratio of Pistons

Using the principle of hydraulic systems (Pascal's Law: \( \frac{F_i}{A_i} = \frac{F_o}{A_o} \)), we can find the ratio \( \frac{A_i}{A_o} = \frac{F_i}{F_o} \). Therefore, \( \frac{A_i}{A_o} = \frac{12,612.86}{22,072.5} \approx 0.571 \).

Key Concepts

Work Done Calculation

The concept of work done is crucial in understanding how energy is transferred when a force moves an object over a distance. In the case of the hydraulic lift raising a car, we apply the general formula for work:

• Formula: \( W = F \times d \)
• F: Force applied, measured in Newtons (N).
• d: Distance over which the force is applied, measured in meters (m).

For the car being lifted: - The force applied is the weight (gravitational force) of the car, which we calculated as 22,072.5 N.- The distance moved is 0.30 m (30 cm converted to meters).Thus, to determine the work done on the car, we calculate \( W = 22,072.5 \times 0.30 = 6,621.75 \text{ J} \). The unit, Joule (J), is standard for work, representing the amount of energy transferred during the process.Breaking the equation into parts helps understand how the components of force and distance contribute to the total work done.

Gravitational Force

Gravitational force is the weight of an object resulting from the Earth’s gravity pulling on its mass. Calculating this force involves understanding its basic elements:

• Formula: \( F = mg \)
• m: Mass of the object, in kilograms (kg).
• g: Acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \).

For the hydraulic lift scenario, the car’s mass is 2250 kg (converted from 2.25 tons). With the formula:\( F = 2250 \times 9.81 \approx 22,072.5 \text{ N} \),we find that the car exerts a gravitational force of 22,072.5 N downwards. This force impacts directly how much effort is needed to lift the car, serving as a key component in subsequent calculations of work and force in the lift operation.Understanding gravitational force is essential in physics as it affects every object with mass on or near the Earth's surface.

Pascal's Law

Pascal’s Law plays a key role in the operation of hydraulic systems, such as a car lift. It describes how pressure changes within a confined fluid, important in transferring force from one point to another in these systems:

• Principle: Pressure applied to a confined fluid is transmitted undiminished in all directions throughout the fluid.
• Formula: \( \frac{F_i}{A_i} = \frac{F_o}{A_o} \)
• \( F_i \): Input force.
• \( A_i \): Input area.
• \( F_o \): Output force.
• \( A_o \): Output area.

In the lift problem, this law helps determine the relationship between forces and areas of the lift's pistons. By maintaining equal pressure, we can state:\[ \frac{A_i}{A_o} = \frac{F_i}{F_o} \approx 0.571 \]This ratio illustrates how hydraulic systems can multiply force, making it feasible for a smaller input force to lift a larger output force (car’s weight) by adjusting the area sizes of the pistons. Thus, Pascal’s Law is fundamental to the efficiency of hydraulic machinery.

Physics Problem Solving

Solving physics problems effectively relies on understanding fundamental concepts, breaking down problems into manageable steps, and applying the correct formulas and principles. Here’s a simplified approach to tackle similar physics problems:

• Understand the Problem: Identify what you are given and what you need to find. In this instance, values like car mass and lift distance were provided.
• Convert Units: Ensure all measurements are in compatible units before calculations. For example, convert weights from tons to kilograms, distances from centimeters to meters.
• Apply Formulas: Select and apply relevant formulas carefully, such as work done \( (W = F \times d) \) or gravitational force \( (F = mg) \).
• Use Logical Steps: Break the problem into steps and tackle each logically, verifying each step’s outcome before proceeding to the next.

By following such structured methods, you enhance clarity and understanding, making physics problems less daunting. Practice is crucial to reinforce these skills, enabling efficient problem-solving over time in contexts like hydraulic lifts and beyond.