Problem 13 - Applied Physics

Question

The large piston on a hydraulic lift has radius \(40.0 \mathrm{~cm} .\) The small piston has radius \(5.00 \mathrm{~cm}\) to which a force of \(75.0 \mathrm{~N}\) is applied. (a) Find the force exerted by the large piston. (b) Find the pressure on the large piston. (c) Find the pressure on the small piston. (d) Find the mechanical advantage of the lift. (e) What happens when the area of the small piston is half as large? (f) What happens when the radius of the small piston is half as large?

Step-by-Step Solution

Answer

(a) 4800 N (b) 9552 Pa (c) 9552 Pa (d) 64 (e) Input force needs doubling. (f) Input force needs quadrupling.

Step 1: Finding Force Exerted by Large Piston (a)

First, calculate the area of the small piston using the formula for the area of a circle, \( A = \pi r^2 \). The radius of the small piston is 5.00 cm, which is 0.050 m. Thus, the area \( A_s = \pi (0.050)^2 = 0.00785 \text{ m}^2 \). Using the force applied to the small piston, we want to find the force exerted by the large piston. According to Pascal's principle, the pressure is the same in the two pistons: \( \frac{F_s}{A_s} = \frac{F_L}{A_L} \), where \( A_L \) is the area of the large piston. Now, find \( A_L = \pi (0.40)^2 = 0.5024 \text{ m}^2 \). Therefore, \( F_L = F_s \times \frac{A_L}{A_s} = 75.0 \times \frac{0.5024}{0.00785} = 4800 \text{ N} \).

Step 2: Calculating Pressure on Large Piston (b)

Pressure on any piston is given by the formula \( P = \frac{F}{A} \). For the large piston, the force is 4800 N and the area \( A_L = 0.5024 \text{ m}^2 \). Therefore, \( P_L = \frac{4800}{0.5024} = 9552 \, \text{Pa} \).

Step 3: Calculating Pressure on Small Piston (c)

The pressure on the small piston is calculated as \( P_s = \frac{F_s}{A_s} \). Given \( F_s = 75.0 \text{ N} \) and \( A_s = 0.00785 \text{ m}^2 \), the pressure \( P_s = \frac{75.0}{0.00785} = 9552 \, \text{Pa} \). Notice that the pressure exerted by the large piston and the small piston is the same, as expected from Pascal’s Principle.

Step 4: Finding Mechanical Advantage of Lift (d)

The mechanical advantage (MA) of the lift is the ratio of the output force to the input force, \( \text{MA} = \frac{F_L}{F_s} = \frac{4800}{75.0} = 64 \). This means that the lift amplifies the input force 64 times.

Step 5: Analyzing Effect of Halving Small Piston Area (e)

If the area of the small piston is halved, because pressure remains constant (as per Pascal's Principle), the force on the large piston would also be reduced. Since \( F_L = P \times A_L \) and \( P \) remains constant, halving \( A_s \) leads to the input force being multiplied by twice to maintain the force output from the large piston.

Step 6: Analyzing Effect of Halving Small Piston Radius (f)

If the radius of the small piston is halved, its area becomes one quarter of the original (since \( A = \pi r^2 \)). This decrease reduces the original applied force by a factor of four for the same pressure output, unless the applied pressure is increased accordingly.

Key Concepts

Pascal's Principle

Pascal's Principle is a fundamental law in fluid mechanics. It states that pressure applied to a confined fluid is transmitted undiminished in all directions throughout the fluid. This principle is the key behind the functioning of hydraulic lifts. When you apply a force on the smaller piston, it creates pressure in the fluid. This pressure is then transmitted equally to a larger piston, allowing you to lift heavy loads with minimal effort.

In our exercise, the small and large pistons form a hydraulic lift. The pressure created by applying force to the small piston is transferred unchanged to the large piston. This relationship can be expressed through the equation: \[\frac{F_s}{A_s} = \frac{F_L}{A_L}\]where \(F_s\) and \(F_L\) are the forces on the small and large pistons, respectively, and \(A_s\) and \(A_L\) are their corresponding areas.

Mechanical Advantage

Mechanical Advantage (MA) is a measure of the force amplification achieved by a tool or system. In the context of hydraulic lifts, it describes how much the lift system helps increase the input force applied.

The formula for mechanical advantage in hydraulic systems is:\[\text{MA} = \frac{F_L}{F_s}\]where \(F_L\) is the force exerted by the large piston and \(F_s\) is the force applied to the small piston. A higher MA indicates a greater force amplification. In our situation, the lift has an MA of 64. This means the input force applied to the small piston is multiplied 64 times by the hydraulic system, allowing heavy loads to be easily lifted.

• High MA: Allows lifting heavier objects with less input force.
• Low MA: Reduces the benefit of the lift, requiring more input force.

Pressure Calculation

Pressure is the force per unit area applied on a surface in a direction perpendicular to that surface. The formula is given by:\[P = \frac{F}{A}\]where \(P\) is the pressure, \(F\) is the force, and \(A\) is the area.

Let's break it down with the pistons:

• The small piston has a force of 75 N and an area calculated using the radius 5 cm, giving an area \( A_s = \pi (0.05)^2 \).
• The larger piston has a calculated area based on its 40 cm radius, resulting in \( A_L = \pi (0.40)^2 \).

For both pistons, the pressures are found to be the same due to Pascal's Principle. This means that, for hydraulic systems, the applied pressure is consistent across all pistons regardless of their sizes.

Force Exerted by Pistons

Understanding the force exerted by pistons in a hydraulic system requires considering both the area of the pistons and the applied pressure. For a hydraulic lift, the force exerted by one piston can be significant when compared to the force applied to another due to differences in piston area.

For our two pistons:

• Small Piston: With an applied force of 75 N and an area of 0.00785 m\(^2\), this creates a specific pressure that affects the entire hydraulic fluid.
• Large Piston: Utilizing the same pressure due to Pascal's principle and a larger area of 0.5024 m\(^2\), the force exerted is found to be 4800 N. This significant difference is why hydraulic systems can lift heavy objects: they magnify the input force by exploiting larger areas.