UY1: Resistive Forces

Interaction between moving object and medium (liquid, gas) sometimes cannot be neglected.

The medium exerts a resistive force $f$ on the object opposite to its direction of motion. (Common example: Air resistance)

The magnitude of the resistive force, $f$, generally increases with increasing speed. The actual dependence of the magnitude of the resistive force with the speed is complicated. Hence, the common approximations used are as follows:

  • For small objects at low speeds. (E.g. Dust in air) $$f = kv$$
  • For large objects at high speeds (E.g. Skydiver) $$f = D v^{2}$$

Resistive Forces At Low Speed

Neglecting upthrust or buoyancy force, the vertical forces on the object is:

$$\begin{aligned} \sum F_{y} &= ma \\ &= m \frac{dv}{dt} \end{aligned}$$

Since $f = kv$,

$$\begin{aligned} mg-kv &= m \frac{dv}{dt} \\ \frac{dv}{dt} &= g-\frac{k}{m} v \end{aligned}$$

, where k is a constant. The value of k depends on the medium and the object.

When $v = 0$, the resistive force $-kv$ is also zero and the acceleration is simply $g$.

As t increases, the resistive force increases and the acceleration decreases.

Eventually, the acceleration becomes zero and object moves with a constant speed $v_{t}$, which is the terminal speed. The terminal speed is:

$$\begin{aligned} mg-kv_{t} &= 0 \\ v_{t} &= \frac{mg}{k} \end{aligned}$$

Let’s find v as a function of t, solving $\frac{dv}{dt} = g-\frac{k}{m} v$, we have:

$$v = \frac{mg}{k} \left( 1-e^{-\frac{kt}{m}} \right)$$

Defining the time constant, $\tau = \frac{m}{k}$, we have:

$$v = v_{t} \left( 1-e^{\frac{-t}{\tau}} \right)$$

A graph showing the effect of resistive force on the speed of the object:

Terminal speed

Air Drag At High Speeds

For objects moving at high speeds through the air, the resistive force is proportional to the square of the speed,

$$f = Dv^{2}$$

A more detailed formula is:

$$f = \frac{1}{2} D \rho A v^{2}$$

, where

  • $\rho$ is the density of air
  • $A$ is the cross-sectional area of the falling object; measured in a plane perpendicular to its motion
  • $D$ is the drag coefficient

Let’s find the terminal velocity for the more detailed formula. We have:

$$\begin{aligned} mg-\frac{1}{2} D \rho A v^{2} &= ma \\ a &= g-\left( \frac{D \rho A}{2m} \right) v^{2} \\ v_{t} &= \sqrt{\frac{2mg}{D \rho A}} \end{aligned}$$

Next: Uniform Circular Motion & Non-uniform Circular Motion

Previous: Friction

Back To Mechanics (UY1)

Mini Physics

As the Administrator of Mini Physics, I possess a BSc. (Hons) in Physics. I am committed to ensuring the accuracy and quality of the content on this site. If you encounter any inaccuracies or have suggestions for enhancements, I encourage you to contact us. Your support and feedback are invaluable to us. If you appreciate the resources available on this site, kindly consider recommending Mini Physics to your friends. Together, we can foster a community passionate about Physics and continuous learning.

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.