Problem 41 - College Physics

Question

A hypodermic syringe has a plunger of area \(2.5 \mathrm{~cm}^{2}\) and a \(5.0 \times 10^{-3}-\mathrm{cm}^{2}\) needle. (a) If a \(1.0-\mathrm{N}\) force is applied to the plunger, what is the gauge pressure in the syringe's chamber? (b) If a small obstruction is present at the end of the needle, what force does the fluid exert on it? (c) If the blood pressure in a vein is \(50 \mathrm{~mm} \mathrm{Hg},\) what force must be applied on the plunger so that fluid can be injected into the vein?

Step-by-Step Solution

Answer

(a) 4000 Pa; (b) 2.0 mN; (c) 1.67 N

Step 1: Calculate the pressure applied by the plunger

The pressure exerted by the plunger is given by the formula:\[P = \frac{F}{A_1}\]Here, \(F = 1.0 \; \text{N}\) is the force applied and \(A_1 = 2.5 \; \text{cm}^2 = 2.5 \times 10^{-4} \; \text{m}^2\) is the area of the plunger. Substitute these values into the formula:\[P = \frac{1.0 \; \text{N}}{2.5 \times 10^{-4} \; \text{m}^2} = 4000 \; \text{Pa}\]

Step 2: Finding the force exerted on the obstruction

Using the pressure found in Step 1 and knowing that pressure is the same throughout the fluid, we can find the force exerted by the fluid on the obstruction using the formula:\[F_{needle} = P \times A_2\]where \(A_2 = 5.0 \times 10^{-3} \; \text{cm}^2 = 5.0 \times 10^{-7} \; \text{m}^2\).Substitute the known values:\[F_{needle} = 4000 \; \text{Pa} \times 5.0 \times 10^{-7} \; \text{m}^2 = 2.0 \times 10^{-3} \; \text{N}\]

Step 3: Calculate minimum force required to inject fluid into the vein

First we need to calculate the pressure needed to overcome the blood pressure of the vein. The pressure of the vein is given as \(50 \; \text{mmHg}\) which can be converted to pascals using the conversion factor: \(1 \; \text{mmHg} = 133.322 \; \text{Pa}\).\[P_{vein} = 50 \; \text{mmHg} \times 133.322 \; \text{Pa/mmHg} = 6666.1 \; \text{Pa}\]To inject the fluid, the pressure we apply with the plunger must be at least the pressure of the vein. Thus,\[P_{needed} = P_{vein} = \frac{F_{needed}}{A_1}\]Solving for \(F_{needed}\), we get:\[F_{needed} = 6666.1 \; \text{Pa} \times 2.5 \times 10^{-4} \; \text{m}^2 = 1.6665 \; \text{N}\]

Key Concepts

Pressure Calculation

Understanding how to calculate pressure is key in fluid dynamics. Pressure is essentially the force applied per unit area on a surface. It is defined by the formula:

• \[P = \frac{F}{A}\]

where \(P\) is the pressure, \(F\) is the force, and \(A\) is the area. In our syringe exercise, the plunger applies a force of 1.0 Newtons on an area of \(2.5 \text{ cm}^2\). To calculate pressure, it's important to convert the area from square centimeters to square meters:

• \(A_1 = 2.5 \text{ cm}^2 = 2.5 \times 10^{-4} \text{ m}^2\)

Substitute these into the pressure equation, and the pressure exerted by the plunger becomes:

• \[P = \frac{1.0 \, \text{N}}{2.5 \times 10^{-4} \, \text{m}^2} = 4000 \, \text{Pa}\]

This shows that pressure calculation is about dividing the force by the area over which it acts.

Force and Area Relationship

The relationship between force and area is pivotal to understanding pressure dynamics. When you apply a force on a fluid via a surface, the distribution of that force across the area determines the pressure. For instance, in our problem, a syringe nozzle, or needle, has a small area compared to the larger plunger area. When pressure is constant throughout the fluid, the force exerted on smaller areas changes. The force exerted at the needle's obstruction is calculated with:

• \(F_{needle} = P \times A_2\)

where \(A_2 = 5.0 \times 10^{-7} \, \text{m}^2\) is the needle's area. Using the pressure derived, 4000 Pa, the force becomes:

• \[F_{needle} = 4000 \, \text{Pa} \times 5.0 \times 10^{-7} \, \text{m}^2 = 2.0 \times 10^{-3} \, \text{N}\]

This demonstrates that, as area decreases, force needed to maintain the same pressure also decreases, critical in machinery like syringes.

Conversion of Pressure Units

Converting pressure units is often necessary in fluid dynamics tasks because different fields may use varying units. In our exercise, the vein's pressure given in millimeters of mercury (mmHg) needs converting to Pascals (Pa), the SI unit of pressure. The conversion factor used is:

• 1 mmHg = 133.322 Pa

The blood pressure in the vein is 50 mmHg, which converts as:

• \[P_{vein} = 50 \, \text{mmHg} \times 133.322 \, \text{Pa/mmHg} = 6666.1 \, \text{Pa}\]

Understanding and applying unit conversions is essential for accurate calculations in engineering and medicine. The force required to inject fluid into the vein can now be calculated using:

• \(F_{needed} = 6666.1 \, \text{Pa} \times 2.5 \times 10^{-4} \, \text{m}^2\)
• \[F_{needed} = 1.6665 \, \text{N}\]

Converting units ensures precision and adaptability across diverse scientific environments.