Fictitious forces and resistive forces

This week there are two separate topics, and both carry Newton’s second law into a new situation. The first is what happens when the observer is accelerating: fictitious forces. The second is liquid and air resistance (resistive force), a force that grows with speed, and the terminal speed it produces. On the exam, the fictitious force appears more as a concept question and a “at what angle does the pendulum hang” question, and the resistive force as terminal speed and the exponential speed formula.

Golden rules:

1. The fictitious force is in the direction opposite to the acceleration, $-m\vec a$. For an observer on the ground it does not exist at all.
2. Both viewpoints give the same answer. On the exam choose whichever is easier, but do not mix the two.
3. At terminal speed the acceleration is zero: resistance = weight.
4. If $R = bv$ then $v_T = mg/b$; if $R \propto v^2$ then $v_T \propto \sqrt m$.

Non-inertial systems

Newton’s laws hold only in inertial frames. An observer in an accelerating car, a braking bus or a rotating platform is not inertial (noninertial frame). If this observer wants to write $\sum \vec F = m\vec a$, they have to add a fictitious force to the real forces:

$\vec F_{fictitious} = -m\vec a_{system}$

The source of the fictitious force is not an object, so it has no action-reaction pair. It is just a correction term that makes the calculation easier.

Conceptual example. When a bus brakes suddenly, you are thrown forward. The observer on the ground: “The bus slowed down, and according to the first law the passenger kept going at the same speed.” The observer in the bus: “There is a force pushing me forward.” Both make the right prediction; only the second uses a fictitious force.

A hanging object in an accelerating vehicle

A small mass hanging from the ceiling hangs backward at an angle $\theta$ in a truck accelerating forward with $a$.

View from the truck (equilibrium): T sinθ = m a   (fictitious force backward, horizontal component of tension forward)
                                   T cosθ = m g
Divide: tanθ = a / g
View from the ground (accelerating): T sinθ = m a, T cosθ = m g → the same result

Example 1. $a = 2.5\ \text{m/s}^2$.

tanθ = 2.5 / 9.80 = 0.255  →  θ = 14.3°
T = m √(a² + g²) = m · 10.1 m/s²   (the string is tensioned slightly more than the weight)

This setup is an accelerometer: if you measure the angle, you find the acceleration.

The elevator, through the fictitious force

You can solve the elevator question of week 6 from inside the elevator too. A 60 kg person is in an elevator accelerating upward at 2.0 m/s². Inside the elevator the person is in equilibrium; the real forces are $N$ (upward) and $mg$ (downward), and the fictitious force $ma$ is opposite to the acceleration, that is, downward.

Equilibrium: N - mg - m a = 0
N = m (g + a) = 60 (9.80 + 2.0) = 708 N
Magnitude of the fictitious force: m a = 60 · 2.0 = 120 N

The result is the same as the view from the ground. The person’s “feeling heavy” is the feeling of this extra 120 N fictitious force. In a freely falling elevator, $a = g$ downward, and the fictitious force $mg$ upward exactly cancels the weight: $N = 0$, weightlessness.

Rotating systems: centrifugal and Coriolis

An observer on a platform rotating with angular speed $\omega$, standing at a distance $r$ from the center, feels an outward force: the centrifugal force.

$F_{cf} = m\omega^2 r = \frac{mv^2}{r}\quad(\text{outward})$

Seen from the ground the situation is different: the person is going around in a circle, and there is a real force (friction, the bar they hold) pulling them toward the center. The centrifugal force does not exist in an inertial frame. Since in week 8 we solved the curve from the ground, we did not use the centrifugal force there.

Example 2. A 50 kg child, 3.0 m from the center of a platform rotating at 1.2 rad/s.

From the platform: F_cf = m ω² r = 50 · 1.44 · 3.0 = 216 N, outward
To avoid sliding, static friction 216 N, inward. μs ≥ 216 / (50 · 9.80) = 0.44

On an object that is moving in a rotating system, the Coriolis force also acts; it is perpendicular to the velocity and deflects sideways. This is how the curving of the path of a ball thrown across a rotating platform is explained. The rotation of large air masses on Earth is this effect too. In this course it is usually asked only as a concept.

PITFALL · mixing the two viewpoints. When solving from the ground, do not add the centrifugal force to the diagram. If you do, the forces balance and you write $0$ instead of $\sum F = mv^2/r$. The rule: either look from the ground and use only real forces, or look from the rotating system and add the fictitious force and write equilibrium.

Resistive forces

A resistive force opposite to the velocity acts on an object moving in a liquid or gas. Its magnitude depends on speed. Two models are used.

Model 1: resistance proportional to speed, $R = bv$

Small objects, slow motion, dense liquids (a bead in oil, a dust particle). $b$ is a constant that depends on the size of the object and the liquid, with the unit kg/s. For an object released from rest, take down as +:

$mg - bv = m\frac{dv}{dt}$

At the start $v = 0$ and $a = g$. As the speed increases, the resistance increases and the acceleration decreases. When $a = 0$, the speed becomes constant: the terminal speed.

$v_T = \frac{mg}{b},\qquad v(t) = v_T\left(1 - e^{-t/\tau}\right),\qquad \tau = \frac{m}{b}$

The acceleration also falls to zero as $a(t) = g\,e^{-t/\tau}$. $\tau$ is the time constant.

Example 3. $m = 10$ g, $b = 0.050$ kg/s, released from rest. What are $v_T$, $\tau$ and the speed at $t = 0.30$ s?

m = 0.010 kg  (gram → kilogram!)
v_T = m g / b = 0.010 · 9.80 / 0.050 = 1.96 m/s
τ = m / b = 0.010 / 0.050 = 0.20 s
v(0.30) = 1.96 (1 - e^(-0.30/0.20)) = 1.96 (1 - e^(-1.5)) = 1.52 m/s
a(0.30) = 9.80 e^(-1.5) = 2.19 m/s²

PITFALL · a unit left in the exponent. If you write $e^{-t}$, seconds is left in the exponent; the exponent must always be dimensionless. The right one is $e^{-t/\tau}$. If you write $\tau$ as $b/m$, its unit becomes s⁻¹, which is wrong again.

Model 2: resistance proportional to the square of the speed

For objects moving at everyday sizes and speeds (a skydiver, a ball, a car), air resistance:

$R = \tfrac12 D\rho A v^2$

$D$ is the dimensionless drag coefficient, $\rho$ is the air density (about 1.20 kg/m³), and $A$ is the cross-sectional area perpendicular to the motion. At terminal speed $R = mg$:

$v_T = \sqrt{\frac{2mg}{D\rho A}}$

At any moment: $a = g - \dfrac{D\rho A v^2}{2m} = g\left(1 - \dfrac{v^2}{v_T^2}\right)$.

Example 4. A 0.45 kg ball, $D = 0.50$, $A = 0.038\ \text{m}^2$.

v_T = √(2 · 0.45 · 9.80 / (0.50 · 1.20 · 0.038))
    = √(8.82 / 0.0228) = √386.8 = 19.7 m/s

When a parachute opens, $A$ and $D$ become much larger and $v_T$ drops to a few m/s. If the mass is quadrupled, $v_T$ doubles ($\sqrt m$); in the $R = bv$ model it would have quadrupled.

Example 5. The terminal speed of an 80 kg skydiver is 43.2 m/s (with the values in the questions). What is the acceleration at the moment the speed reaches 30 m/s?

a = g (1 - v² / v_T²) = 9.80 (1 - 900 / 1867) = 9.80 · 0.518 = 5.08 m/s²

The acceleration starts from $g$ and keeps decreasing, becoming zero at terminal speed. That is why the $v$–$t$ graph is a curve that first rises with slope $g$ and then flattens at $v_T$. The $a$–$t$ graph is a curve that falls from $g$ to zero. Recognize these two shapes in graph questions.

TIP · which model? If the question gives $b$, it is $R = bv$; if it gives $D$, $\rho$ and $A$, it is the $v^2$ model. The exponential $e^{-t/\tau}$ formula is valid only for $R = bv$.

Common mistakes