Chapter 3 · Section 3.2

3.2 Problems in Mechanics

Why does a raindrop not hit you at the speed of a bullet? A drop falling from a cloud 2 km high would reach about $200$ m/s in a vacuum. Real drops arrive at a few metres per second, because air resistance grows with speed until it balances the weight. This balance, the terminal velocity, falls straight out of a first-order differential equation.

In this section we turn Newton's second law into differential equations for the velocity of a moving body. We treat falling with and without air resistance, throwing upward, and sliding down a ramp. The equations are separable or linear, so the methods of Chapter 2 are all we need. The real work is in the modelling: choosing directions, listing forces and getting the signs right.

Newton's second law

For a body of constant mass $m$ moving along a straight line, let $x (t)$ be its position, $v = x^{'}$ its velocity and $a = v^{'} = x^{''}$ its acceleration. Newton's second law says that the mass times the acceleration equals the resultant (total) force $F$ acting on the body:

m \frac{d v}{d t} = F .

If $F$ depends only on $t$ and $v$ , this is a first-order equation for $v (t)$ . Once $v$ is known, one more integration gives the position: $x (t) = x (0) + \int_{0}^{t} v (s) d s$ .

We use SI units throughout:

mass in kilograms (kg), length in metres (m), time in seconds (s);
force in newtons, $1 N = 1 kg \cdotp m/s^{2}$ ;
near the Earth's surface, gravity pulls a mass $m$ with the force $m g$ (its weight), and we take $g = 9.8 m/s^{2}$ .

Mass is not weight

A 5 kg body has mass $m = 5$ kg and weight $m g = 49$ N. Put the mass on the left of $m v^{'} = F$ and forces (in newtons) on the right. Mixing the two is the most common source of wrong answers in this section.

Method in a nutshell

Choose an axis and a positive direction (for falling bodies, "down" is convenient). Put the origin at the starting point.
List all forces along the axis, each with its sign: positive if it points in the positive direction.
Write $m v^{'} = F_{1} + F_{2} + \dots$ and the initial condition $v (0) = v_{0}$ .
Solve for $v (t)$ (separable or linear equation). Integrate once more for $x (t)$ if positions are asked.
Check the answer: units, signs, and the behaviour as $t \to \infty$ .

Free fall without air resistance

If gravity is the only force and the positive direction is down, the equation is $m v^{'} = m g$ , i.e. $v^{'} = g$ . Integrating twice with $v (0) = v_{0}$ and $x (0) = 0$ :

v (t) = v_{0} + g t, x (t) = v_{0} t + \frac{1}{2} g t^{2} .

The mass has cancelled: without air, a hammer and a feather fall together. If "up" is positive instead, gravity is $- m g$ and the signs of the $g$ terms flip.

Example 3.2.1

A steel ball is thrown straight up at $19.6$ m/s from the edge of a roof $24.5$ m above the ground. Ignore air resistance. Find the greatest height, the time at which the ball hits the ground, and its speed at impact.

Solution

Axis. Let $y$ be the height above the ground, positive upward. Then $y (0) = 24.5$ , $v (0) = 19.6$ .
Equation. Gravity points down: $m v^{'} = - m g$ , so $v^{'} = - 9.8$ .
Solve. $v = 19.6 - 9.8 t, y = 24.5 + 19.6 t - 4.9 t^{2} .$
Top. At the top $v = 0$ , so $t = 2$ s and $y (2) = 24.5 + 39.2 - 19.6 = 44.1$ m.
Impact. $y = 0$ gives $4.9 t^{2} - 19.6 t - 24.5 = 0$ . Divide by $4.9$ : $t^{2} - 4 t - 5 = 0$ , so $(t - 5) (t + 1) = 0$ and $t = 5$ s (the root $t = - 1$ has no meaning here).
Speed at impact. $v (5) = 19.6 - 49 = - 29.4$ m/s. The minus sign says "downward"; the speed is $29.4$ m/s.

Air resistance

A body moving through air feels a drag force that always points against the velocity. Its exact size depends on the shape, the speed and the fluid, but two simple laws cover most textbook and many engineering situations:

Linear drag, magnitude $k ∣ v ∣$ with $k > 0$ in N·s/m. It describes slow motion of small bodies in viscous fluids (fine dust, droplets in mist, a particle settling in oil).
Quadratic drag, magnitude $c v^{2}$ with $c > 0$ in kg/m. It describes faster motion of larger bodies (balls, skydivers, vehicles). In fluid mechanics, $c = \frac{1}{2} ρ C_{D} A$ , where $ρ$ is the air density, $A$ the frontal area and $C_{D}$ the drag coefficient.

Both are models. In practice $k$ or $c$ is measured, and the model is trusted only in the speed range where it was checked.

Linear drag: the falling body

Take "down" as positive. While the body falls ( $v > 0$ ), the forces are (Figure 3.2.1):

the weight $m g$ , pointing down, so it enters with a plus sign;
the drag $k v$ , pointing up, so it enters as $- k v$ .

Figure 3.2.1. Free-body diagram of a falling body with linear air resistance. With

x

measured downward, the weight is positive and the drag is negative.

Newton's law gives the initial-value problem

m \frac{d v}{d t} = m g - k v, v (0) = v_{0} .

Notice that $- k v$ has the correct sign even if the body moves up ( $v < 0$ ): then $- k v > 0$ points down, against the motion. So this single equation covers both directions.

Let us solve it as a linear equation, in small steps.

Standard form. Divide by $m$ : $v^{'} + \frac{k}{m} v = g$ .
Integrating factor. $μ = e^{k t / m}$ , so $(e^{k t / m} v)^{'} = g e^{k t / m}$ .
Integrate. $e^{k t / m} v = \frac{m g}{k} e^{k t / m} + C$ .
Divide. $v = \frac{m g}{k} + C e^{- k t / m}$ .
Initial condition. $v (0) = v_{0}$ gives $C = v_{0} - \frac{m g}{k}$ .

Two combinations of the data appear again and again, so we give them names.

Definition 3.2.1 (Terminal velocity and time constant)

For the falling body with linear drag, the terminal velocity (limiting velocity) and the time constant are

v_{T} = \frac{m g}{k}, τ = \frac{m}{k} .

Theorem 3.2.1 (Falling body with linear drag)

The solution of $m v^{'} = m g - k v$ , $v (0) = v_{0}$ , $x (0) = 0$ , is

v (t) = v_{T} + (v_{0} - v_{T}) e^{- t / τ}, x (t) = v_{T} t + (v_{0} - v_{T}) τ (1 - e^{- t / τ}) .

Whatever $v_{0}$ is, $v (t) \to v_{T}$ as $t \to \infty$ .

Why the terminal velocity is

m g / k

You can find $v_{T}$ without solving anything. The velocity stops changing when $v^{'} = 0$ , i.e. when $m g - k v = 0$ : drag balances weight. That gives $v = m g / k$ . It is the equilibrium solution of the differential equation, and the exponential term shows that every other solution approaches it. After one time constant $τ$ , the gap $∣ v - v_{T} ∣$ has shrunk by the factor $e^{- 1} \approx 0.37$ ; after $3 τ$ , to about $5%$ .

Example 3.2.2

A $5$ kg package is dropped from rest from a hovering helicopter. Air resistance is $2.5 v$ newtons, where $v$ is the speed in m/s. Find $v (t)$ and $x (t)$ , the speed after $2$ s, and the time and distance needed to reach $95%$ of the terminal velocity.

Solution

Equation. Down is positive: $5 v^{'} = 5 (9.8) - 2.5 v = 49 - 2.5 v$ , with $v (0) = 0$ .
Constants. $v_{T} = \frac{m g}{k} = \frac{49}{2.5} = 19.6$ m/s and $τ = \frac{m}{k} = \frac{5}{2.5} = 2$ s.
Velocity and position. By Theorem 3.2.1 with $v_{0} = 0$ : $v = 19.6 (1 - e^{- t /2}), x = 19.6 t - 39.2 (1 - e^{- t /2}) .$
After 2 s. $v (2) = 19.6 (1 - e^{- 1}) \approx 12.39$ m/s.
95% of $v_{T}$ . We need $1 - e^{- t /2} = 0.95$ , i.e. $e^{- t /2} = 0.05$ , so $t = 2 ln 20 \approx 5.99$ s.
Distance. At that time $e^{- t /2} = 0.05$ , so $x = 19.6 (2 ln 20) - 39.2 (0.95) \approx 117.4 - 37.2 = 80.2$ m.

Check of units: $k$ has units N·s/m $=$ kg/s, so $m / k$ is in seconds and $m g / k$ in m/s, as it should be.

Figure 3.2.2. Velocity for

m = 5

kg,

k = 2.5

N·s/m and three initial velocities. A body thrown down faster than

v_{T}

slows down to

v_{T}

. The grey tangent to the lowest curve at

t = 0

(slope

g

) meets the line

v = v_{T}

exactly at

t = τ

Try other values in the simulator below. Change the mass, the drag constant or the initial velocity and watch how $v_{T}$ and $τ$ change. (A negative $v_{0}$ means the body is thrown upward, since the positive direction is down.)

Example 3.2.3

A $0.2$ kg ball is thrown straight up at $30$ m/s. Air resistance is $0.04 v$ newtons. How long does the ball rise, and how high does it go? Compare with the case without air resistance.

Solution

Axis. Here "up" is more natural, so let $y$ be the height, positive upward, $y (0) = 0$ , $v (0) = 30$ .
Forces. Gravity $- m g = - 1.96$ N. Drag $- 0.04 v$ (while rising, $v > 0$ , so it points down; while falling it points up; the same formula works for both).
Equation. $0.2 v^{'} = - 1.96 - 0.04 v$ , i.e. $v^{'} + 0.2 v = - 9.8$ .
Solve. With $μ = e^{0.2 t}$ : $(e^{0.2 t} v)^{'} = - 9.8 e^{0.2 t}$ , so $e^{0.2 t} v = - 49 e^{0.2 t} + C$ . From $v (0) = 30$ , $C = 79$ : $v = - 49 + 79 e^{- t /5} .$
Position. Integrate with $y (0) = 0$ : $y = - 49 t + 395 (1 - e^{- t /5}) .$
Top. $v = 0$ when $e^{- t /5} = \frac{49}{79}$ , i.e. $t_{top} = 5 ln \frac{79}{49} \approx 2.39$ s.
Maximum height. At $t_{top}$ , $1 - e^{- t /5} = \frac{30}{79}$ , so the second term is $395 \cdot \frac{30}{79} = 150$ and $y_{m a x} = 150 - 245 ln \frac{79}{49} \approx 33.0 m .$
Without drag. $v = 30 - 9.8 t$ vanishes at $t \approx 3.06$ s, and $y_{m a x} = \frac{3 0 ^{2}}{2 ( 9.8 )} \approx 45.9$ m.

Drag cuts the height by about a quarter. Figure 3.2.3 shows a second effect: with drag the flight is no longer symmetric, and the ball takes longer to come down than to go up.

Figure 3.2.3. Height of the ball in Example 3.2.3. Without air resistance (dashed) the curve is a symmetric parabola. With linear drag (solid) it is lower and leans to the right: the descent is slower than the climb.

Quadratic drag

For a body falling fast through air, the drag is closer to $c v^{2}$ . With "down" positive and $v \geq 0$ :

m \frac{d v}{d t} = m g - c v^{2}, v (0) = 0.

This equation is not linear, but it is separable. Its equilibrium is where $m g = c v^{2}$ , which suggests the notation

v_{T} = \frac{m g}{c}, so that m g - c v^{2} = c (v_{T}^{2} - v^{2}) .

Let us solve it step by step.

Separate. For $0 \leq v < v_{T}$ : $\frac{d v}{v _{T}^{2} - v ^{2}} = \frac{c}{m} d t .$
Partial fractions. $\frac{1}{v _{T}^{2} - v ^{2}} = \frac{1}{2 v _{T}} (\frac{1}{v _{T} - v} + \frac{1}{v _{T} + v}) .$
Integrate. $\frac{1}{2 v _{T}} ln \frac{v _{T} + v}{v _{T} - v} = \frac{c}{m} t + C .$ With $v (0) = 0$ the left side is $0$ at $t = 0$ , so $C = 0$ .
Simplify the constant. $\frac{2 v _{T} c}{m} = \frac{2 c}{m} \frac{m g}{c} = 2 \frac{c g}{m} = \frac{2 g}{v _{T}}$ . Hence $ln \frac{v _{T} + v}{v _{T} - v} = \frac{2 g}{v _{T}} t .$
Solve for $v$ . Write $s = g t / v_{T}$ . Then $\frac{v _{T} + v}{v _{T} - v} = e^{2 s}$ , which gives $v = v_{T} \frac{e ^{2 s} - 1}{e ^{2 s} + 1} = v_{T} \frac{e ^{s} - e ^{- s}}{e ^{s} + e ^{- s}} = v_{T} tanh s .$

Theorem 3.2.2 (Falling from rest with quadratic drag)

The solution of $m v^{'} = m g - c v^{2}$ , $v (0) = 0$ , $x (0) = 0$ , is

v (t) = v_{T} tanh (\frac{g t}{v _{T}}), x (t) = \frac{v _{T}^{2}}{g} ln cosh (\frac{g t}{v _{T}}), v_{T} = \frac{m g}{c} .

Since $tanh s \to 1$ as $s \to \infty$ , again $v \to v_{T}$ . For small $t$ , $tanh s \approx s$ , so $v \approx g t$ : at first the body falls as if there were no air.

Velocity as a function of distance

Sometimes we want the speed after a given distance, not a given time. By the chain rule, $\frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x}$ . The quadratic-drag equation becomes

m v \frac{d v}{d x} = m g - c v^{2},

which is linear in the new unknown $w = v^{2}$ (because $v d v / d x = \frac{1}{2} d w / d x$ ): $\frac{m}{2} w^{'} + c w = m g$ . With $w (0) = 0$ the solution is

v^{2} = v_{T}^{2} (1 - e^{- 2 gx / v_{T}^{2}}) .

Quadratic drag going up

The formula $- c v^{2}$ is always negative, so it points in the negative direction whatever the sign of $v$ . It is correct only while the body moves in the positive direction. The physically correct force is $- c v ∣ v ∣$ . For a ball thrown upward, you must solve the rising phase ( $m v^{'} = - m g - c v^{2}$ , up positive) and the falling phase ( $m v^{'} = m g - c v^{2}$ , down positive) separately and join them at the top. Linear drag $- k v$ does not have this problem.

Example 3.2.4

A skydiver of mass $80$ kg (with equipment) jumps from a hovering helicopter and falls in a spread position with drag $0.25 v^{2}$ newtons. Find the terminal velocity, $v (t)$ , the speed after $5$ s, and the time and distance needed to reach $50$ m/s.

Solution

Terminal velocity. $v_{T} = \frac{80 \cdot 9.8}{0.25} = 3136 = 56$ m/s (about $200$ km/h, a realistic value).
Velocity. $\frac{g}{v _{T}} = \frac{9.8}{56} = 0.175 s^{- 1}$ , so $v = 56 tanh (0.175 t) .$
After 5 s. $v (5) = 56 tanh (0.875) \approx 39.4$ m/s. Without air it would be $49$ m/s.
Time to 50 m/s. $tanh (0.175 t) = \frac{50}{56}$ , so $t = \frac{1}{0.175} tanh^{- 1} \frac{50}{56} = \frac{1}{0.35} ln \frac{56 + 50}{56 - 50} \approx 8.20 s .$
Distance. Use the velocity–distance formula with $v_{T}^{2} / (2 g) = 3136/19.6 = 160$ m: $x = - 160 ln (1 - \frac{5 0 ^{2}}{5 6 ^{2}}) = 160 ln \frac{3136}{636} \approx 255 m .$ (The same number comes from $x = 320 ln cosh (0.175 \cdot 8.20)$ .)

Figure 3.2.4. Skydiver of Example 3.2.4. For the first two seconds the curve follows free fall; after about

15

s it is practically at

v_{T} = 56

m/s.

Motion on an inclined plane

A block slides down a straight ramp inclined at angle $θ$ to the horizontal. Measure $x$ along the ramp, positive downhill. Resolve every force into a component along the ramp and a component perpendicular to it (Figure 3.2.5).

Weight. $m g$ vertically down. Along the ramp: $m g sin θ$ (downhill, positive). Perpendicular: $m g cos θ$ into the ramp.
Normal force. The ramp pushes back with $N$ . There is no motion perpendicular to the ramp, so $N = m g cos θ$ .
Friction. For a sliding block, kinetic friction has magnitude $μ N = μ m g cos θ$ and points uphill, against the motion ( $μ$ is the coefficient of kinetic friction).
Air resistance. Say $k v$ , uphill.

Figure 3.2.5. Forces on a block sliding down a ramp. Only the components along the ramp change the speed.

Adding the components along the ramp,

m \frac{d v}{d t} = m g sin θ - μ m g cos θ - k v .

This is the falling-body equation again, with $m g$ replaced by the constant net driving force $F_{0} = m g (sin θ - μ cos θ)$ . So Theorem 3.2.1 applies with $v_{T}$ replaced by the limiting speed $F_{0} / k$ .

Will it slide at all?

The friction formula $μ N$ holds only while the block is moving. A block at rest stays at rest unless the downhill pull beats the maximum static friction $μ_{s} N$ , that is, unless $tan θ > μ_{s}$ . If you get $F_{0} < 0$ , the model is telling you that a block released from rest does not move.

Example 3.2.5

In a warehouse, crates slide down a straight loading ramp $10$ m long that makes an angle of $3 0^{\circ}$ with the floor. A $10$ kg crate starts from rest at the upper end. The kinetic friction coefficient between crate and ramp is $0.2$ , and air resistance is $2 v$ newtons. How long does the trip down the ramp take, and what is the crate's speed at the lower end?

Solution

Net driving force. $F_{0} = m g sin 3 0^{\circ} - μ m g cos 3 0^{\circ} = 98 (0.5) - 0.2 (98) (0.8660) \approx 49 - 16.97 = 32.03 N .$
Equation. $10 v^{'} = 32.03 - 2 v$ , $v (0) = 0$ .
Constants. Limiting speed $F_{0} / k \approx 16.01$ m/s, time constant $τ = m / k = 5$ s.
Velocity and position. $v \approx 16.01 (1 - e^{- t /5}), x \approx 16.01 t - 80.06 (1 - e^{- t /5}) .$
Time to the bottom. We need $x (T) = 10$ . This equation mixes $t$ and $e^{- t /5}$ and has no closed-form solution, so we solve it numerically (for instance by Newton's method or a calculator): $T \approx 2.73$ s.
Final speed. $v (T) \approx 16.01 (1 - e^{- 0.545}) \approx 6.73$ m/s.

For comparison, without air resistance the acceleration is the constant $F_{0} / m \approx 3.20$ m/s $^{2}$ , giving $T = 2 (10) /3.20 \approx 2.50$ s and a final speed of about $8.00$ m/s. Even on a short ramp the air makes a visible difference.

Exercises

Use $g = 9.8$ m/s $^{2}$ . Unless stated otherwise, "down" is the positive direction for falling bodies.

A stone is dropped from rest from a height of $78.4$ m. Ignoring air resistance, when does it hit the ground and how fast is it moving?
Answer
$t = 4$ s; $39.2$ m/s.
A ball is thrown straight down at $4.9$ m/s from a window $29.4$ m above the ground. Ignoring air resistance, find the time of impact and the impact speed.
Answer
$4.9 t^{2} + 4.9 t - 29.4 = 0$ , so $t = 2$ s; $24.5$ m/s.
A $3$ kg body falls from rest with air resistance $0.6 v$ N. Find $v (t)$ , the speed after $5$ s, and the distance fallen in $10$ s.
Answer
$v = 49 (1 - e^{- t /5})$ ; $v (5) \approx 30.97$ m/s; $x (10) = 490 - 245 (1 - e^{- 2}) \approx 278.2$ m.
A $0.1$ kg object is thrown straight down at $60$ m/s. Air resistance is $0.05 v$ N. Show that the object slows down, and find when its speed is $25$ m/s.
Answer
$v = 19.6 + 40.4 e^{- t /2}$ decreases towards $v_{T} = 19.6$ m/s; $t = 2 ln (40.4/5.4) \approx 4.02$ s.
A $0.5$ kg ball is thrown straight up at $20$ m/s, with air resistance $0.1 v$ N. Find the time to the top and the maximum height. Compare with no air resistance.
Answer
Up positive: $v = - 49 + 69 e^{- t /5}$ ; $t_{top} = 5 ln (69/49) \approx 1.71$ s; $y_{m a x} = 100 - 245 ln (69/49) \approx 16.14$ m (versus $20.41$ m without drag).
A $2$ kg parcel falls from rest with air resistance $0.1 v^{2}$ N. Find the terminal velocity, the speed after $2$ s, and the distance fallen in $2$ s.
Answer
$v_{T} = 14$ m/s; $v = 14 tanh (0.7 t)$ , $v (2) \approx 12.39$ m/s; $x (2) = 20 ln cosh (1.4) \approx 15.32$ m.
A $4$ kg box slides from rest down a ramp inclined at $3 0^{\circ}$ . The coefficient of kinetic friction is $3 /6 \approx 0.289$ and air resistance is $0.8 v$ N. Show that the net driving force is $9.8$ N, and find $v (t)$ and $v (5)$ .
Answer
$F_{0} = 4 (9.8) (\frac{1}{2} - \frac{1}{4}) = 9.8$ N; $v = 12.25 (1 - e^{- t /5})$ ; $v (5) \approx 7.74$ m/s.
The coefficient of static friction between a box and a plank is $0.35$ . What is the smallest angle of inclination at which a box placed at rest on the plank starts to slide?
Answer
$tan θ = 0.35$ , $θ \approx 19. 3^{\circ}$ .
A $1200$ kg car coasts in neutral on a level road from $30$ m/s. Neglect rolling friction and take the air drag as $0.6 v^{2}$ N. Use $v d v / d x$ to find the distance over which the speed halves, and then find the time this takes.
Answer
$v = 30 e^{- x /2000}$ , distance $2000 ln 2 \approx 1386$ m; $1/ v = 1/30 + t /2000$ , time $\approx 66.7$ s.
A skydiver of total mass $75$ kg falls from rest with air resistance $15 v$ N. After $10$ s the parachute opens, and the air resistance becomes $147 v$ N. Find the speed at the moment the parachute opens, the new terminal velocity, and the speed $2$ s after opening.
Answer
$v (10) = 49 (1 - e^{- 2}) \approx 42.37$ m/s; new $v_{T} = 5$ m/s; for $t \geq 10$ , $v = 5 + 37.37 e^{- 1.96 (t - 10)}$ , so $v (12) \approx 5.74$ m/s.
Show that as $k \to 0^{+}$ the velocity and position in Theorem 3.2.1 tend to the free-fall formulas $v_{0} + g t$ and $v_{0} t + \frac{1}{2} g t^{2}$ .
Answer
Write $v_{T} = gτ$ and expand $e^{- t / τ} = 1 - t / τ + t^{2} / (2 τ^{2}) - \dots$ with $τ = m / k \to \infty$ : $v \to v_{0} + g t$ and $x \to v_{0} t + \frac{1}{2} g t^{2}$ .
A $10$ kg object is released from rest just below the surface of a lake. The buoyant force is $60$ N (up) and the water resistance is $20 v$ N. Find $v (t)$ and the limiting velocity.
Answer
$10 v^{'} = 98 - 60 - 20 v$ ; $v = 1.9 (1 - e^{- 2 t})$ ; limit $1.9$ m/s (sinking).