Problem 10 - Applied Physics

Question

The mechanical advantage of a hydraulic jack is \(450 .\) Find the weight of the heaviest automobile that can be lifted by an applied force of \(60.0 \mathrm{~N}\).

Step-by-Step Solution

Answer

The heaviest automobile that can be lifted is 27,000 N.

Step 1: Understanding the Problem

A hydraulic jack uses a small force to lift a heavy load by a mechanical advantage. Mechanical advantage ( MA) is defined as the ratio of the load force (weight lifted) to the applied force. We need to find the load force given the mechanical advantage and the applied force.

Step 2: Calculate Load Force

The formula for mechanical advantage is: \( MA = \frac{F_{load}}{F_{applied}} \). Given that \( MA = 450 \) and \( F_{applied} = 60.0 \, N \), we can rearrange this formula to find \( F_{load} \) as follows: \( F_{load} = MA \times F_{applied} = 450 \times 60.0 \, N \).

Step 3: Perform the Calculation

Perform the multiplication: \( F_{load} = 450 \times 60.0 = 27000 \, N \). So, the load force, or the weight of the heaviest automobile that can be lifted, is \( 27000 \, N \).

Key Concepts

Mechanical Advantage

Mechanical advantage is a crucial concept in physics, especially when dealing with machines like hydraulic jacks. It allows a small input force to produce a much larger output force. This is achieved by the mechanical system itself, which in this case, is the hydraulic jack.

• Mechanical advantage (MA) is defined as the ratio of the load force (the force exerted by the machine) to the applied force (the force you exert).
• The formula is: \( MA = \frac{F_{load}}{F_{applied}} \)

This means that if you apply a small force, like with your arm, the machine multiplies this force. In the given exercise, a hydraulic jack with a mechanical advantage of 450 allows the user to lift very heavy objects using minimal effort. Understanding this principle helps in calculating how much force (or how little) is needed to operate various mechanical systems.

Load Force Calculation

Load force calculation involves using the concept of mechanical advantage to determine the maximum weight that can be lifted. In our original exercise, we've calculated this using the given mechanical advantage and the applied force.To calculate the load force \( F_{load} \), you can rearrange the mechanical advantage formula:\[ F_{load} = MA \times F_{applied} \]

• Given \( MA = 450 \), you multiply this by the \( F_{applied} = 60.0 \, N \).

Performing this calculation gives us \( F_{load} = 450 \times 60.0 = 27000 \, N \). This result tells us that the heaviest automobile that can be lifted with the given applied force is quite substantial, demonstrating the power of the hydraulic jack.

Applied Force

The applied force is the initial effort you put into a system. Understanding it is essential when working with mechanical systems, such as hydraulic jacks.

• In this problem, the applied force is relatively small: \( 60.0 \, N \).
• This force is manageable and can be exerted by a person using limited physical strength.

The beauty of machines like hydraulic jacks lies in their ability to transform this small applied force into something much larger, making heavy lifting tasks feasible. This transformation process is the essence of mechanical advantage, where your small input becomes a large output, allowing even a single person to lift a heavy automobile.

Problem Solving in Physics

Problem-solving in physics often requires breaking down a scenario into understandable parts, much like we did in this exercise with the hydraulic jack. Here's how to approach such problems:

• Begin by understanding the problem setup and identifying known values and equations involved.
• Use relevant physics formulas—in this case, the mechanical advantage equation.
• Substitute the known values into your formula to find what you need—like the maximum load force.

Each step in problem solving is crucial, as missing out on a minor detail can lead to incorrect conclusions. In physics, precision and methodical calculations are key to solving complex problems, such as determining the capacity of a hydraulic jack.