Lesson 5 · 35 min

Forces That Vary

So far every force has been constant, so the acceleration was too, and the constant-acceleration formulas did the rest. Real forces change. A spring pushes harder the more it is compressed, air drag grows with speed, and a motor's thrust changes with time. Newton's second law still gives the acceleration at every instant; what changes is how you turn that acceleration into the motion.

Learning objectives

Recognize whether a force depends on time, position or velocity, and pick the kinematic relation that separates the variables.
Integrate \(a = F(t)/m\) for the velocity, and \(v\,dv = a\,ds\) when the force depends on position.
Solve for the time or the distance when the force depends on speed, as with fluid drag.
Find a terminal velocity, and explain why a falling body approaches it but never exceeds it from rest.

Which relation to use

Newton's second law gives the acceleration, \(a = \sum F/m\), as a function of whatever the force depends on. Pair it with the kinematic relation that keeps only that variable and the one you want, separate, and integrate.

Separating variables in \(m\,a = F\)
The force depends on	Write \(a\) as	Separate and integrate
Time, \(F(t)\)	\(\dfrac{dv}{dt}\)	\(m(v_2 - v_1) = \displaystyle\int_{t_1}^{t_2} F(t)\,dt\)
Position, \(F(s)\)	\(v\dfrac{dv}{ds}\)	\(\tfrac12 m(v_2^2 - v_1^2) = \displaystyle\int_{s_1}^{s_2} F(s)\,ds\)
Velocity, \(F(v)\), for the time	\(\dfrac{dv}{dt}\)	\(t_2 - t_1 = \displaystyle\int_{v_1}^{v_2} \frac{m\,dv}{F(v)}\)
Velocity, \(F(v)\), for the distance	\(v\dfrac{dv}{ds}\)	\(s_2 - s_1 = \displaystyle\int_{v_1}^{v_2} \frac{m\,v\,dv}{F(v)}\)

The first two rows are previews of the modules that follow. The integral of a force over time is its impulse, and over distance its work; the Work and Energy and the Impulse and Momentum modules build whole methods on them.

A force that depends on time

A worker leans on a crate harder and harder, so the push grows steadily with time, \(P = ct\). At first the crate does not move: static friction matches the push exactly, whatever its size. Only when \(P\) reaches \(\mu_s N\) does the crate break free. Then friction drops to \(\mu_k N\) and the crate accelerates, faster and faster as the push keeps growing.

Figure 5.1 A push that grows with time, \(\colP{P = ct}\), on a crate on a rough floor. The chart plots the push and the size of the friction force against time. While the crate is at rest, friction follows the push up the same line. When it reaches \(\mu_s N\), the crate slips, friction drops to \(\mu_k N\) and stays there, and the gap between the two lines is \(\colMA{ma}\), growing as the push grows.

Example 5.1 — The push that grows

A \(10\ \text{kg}\) crate rests on a floor with \(\mu_s = 0.5\) and \(\mu_k = 0.4\). Starting at \(t = 0\), a horizontal push \(P = 20t\ \text{N}\) (\(t\) in seconds) acts on it. When does it start to slide, and how fast is it moving at \(t = 4\ \text{s}\)?

Show solution

\(N = mg = 98.1\ \text{N}\). The crate slips when the push reaches \(\mu_s N = 49.05\ \text{N}\):

\[ 20t_1 = 49.05 \quad\Rightarrow\quad t_1 = 2.45\ \text{s} \]

After that, friction is \(\mu_k N = 39.24\ \text{N}\) and

\[ m\frac{dv}{dt} = 20t - 39.24 \quad\Rightarrow\quad 10\,v = \int_{2.45}^{4} (20t - 39.24)\,dt = \Big[10t^2 - 39.24\,t\Big]_{2.45}^{4} \] \[ 10\,v = (160 - 156.96) - (60.15 - 96.24) = 39.13 \quad\Rightarrow\quad v = 3.91\ \text{m/s} \]

The acceleration jumps from zero to \(0.98\ \text{m/s}^2\) at the instant of slipping, because friction drops from \(\mu_s N\) to \(\mu_k N\). Check the jump in Figure 5.1.

A force that depends on position

Springs and gravity at large distances give forces that depend on where the particle is. Write the acceleration as \(a = v\,dv/ds\), so time drops out:

\[ m\,v\,dv = F(s)\,ds \quad\Rightarrow\quad \tfrac12 m v_2^2 - \tfrac12 m v_1^2 = \int_{s_1}^{s_2} F(s)\,ds \]

Example 5.2 — A spring launcher

A \(2\ \text{kg}\) block is held against a spring of stiffness \(k = 800\ \text{N/m}\), compressed \(\delta = 0.1\ \text{m}\), on a smooth floor. It is released. How fast does it leave the spring?

Show solution

Measure \(s\) from the released position. The spring pushes with \(F_s = k(\delta - s)\) until it reaches its free length at \(s = \delta\):

\[ \int_0^{v} 2\,v\,dv = \int_0^{0.1} 800(0.1 - s)\,ds \quad\Rightarrow\quad v^2 = 800\left[0.1s - \tfrac12 s^2\right]_0^{0.1} = 800(0.005) = 4 \]

so \(v = 2\ \text{m/s}\). In general \(v = \delta\sqrt{k/m}\). The block's acceleration falls from \(40\ \text{m/s}^2\) to zero during the launch, so the constant-acceleration formulas would give the wrong answer.

A force that depends on velocity

A body moving through a fluid feels a drag force against its velocity. At low speeds, or for very small bodies, drag is nearly proportional to the speed, \(F_D = cv\); for cars, balls, boats and people at everyday speeds it is closer to the square of the speed, \(F_D = kv^2\).

Example 5.3 — A boat coasts

A \(500\ \text{kg}\) boat moving at \(10\ \text{m/s}\) cuts its engine. The water's drag is \(F_D = 20v^2\ \text{N}\), with \(v\) in m/s. How long does it take to slow to \(5\ \text{m/s}\), and how far does it go?

Show solution

For the time, \(a = dv/dt\):

\[ 500\frac{dv}{dt} = -20v^2 \quad\Rightarrow\quad \int_0^t dt = -25\int_{10}^{5}\frac{dv}{v^2} \quad\Rightarrow\quad t = 25\left(\frac15 - \frac1{10}\right) = 2.5\ \text{s} \]

For the distance, \(a = v\,dv/ds\):

\[ 500\,v\frac{dv}{ds} = -20v^2 \quad\Rightarrow\quad \int_0^s ds = -25\int_{10}^{5}\frac{dv}{v} \quad\Rightarrow\quad s = 25\ln 2 = 17.3\ \text{m} \]

Quadratic drag never quite stops the boat: \(s\) grows without limit as \(v \to 0\), though ever more slowly.

Falling against drag: terminal velocity

A skydiver of mass \(m\) falls with drag \(kv^2\). Take down as positive:

\[ m\frac{dv}{dt} = \colW{mg} - \colP{kv^2} \]

At first the drag is small and the skydiver accelerates at nearly \(g\). As the speed grows, so does the drag, and the acceleration shrinks. It reaches zero when the drag equals the weight:

Terminal velocity, and the approach to it from rest

\[ v_t = \sqrt{\frac{mg}{k}} \qquad\qquad v = v_t\tanh\frac{gt}{v_t} \]

Separate \(dt = dv/[g(1 - v^2/v_t^2)]\) and integrate to get the second result. Starting from rest, \(v\) creeps up on \(v_t\) but never reaches it. Starting faster than \(v_t\), the drag exceeds the weight, the acceleration points up, and the body slows down to \(v_t\).

Figure 5.2 A skydiver falling against air drag \(\colP{F_D = kv^2}\). The view falls with the skydiver; the scale on the left shows the distance fallen. The weight \(\colW{W}\) ends at the center of gravity, and the drag pushes up on the side facing the oncoming air. The bars add the two forces to give \(\colMA{ma}\), which shrinks as the drag grows, and the chart shows the speed approaching \(v_t\), dashed. Start the skydiver faster than \(v_t\) and the drag wins: the body slows down.

A spread-eagle skydiver of \(80\ \text{kg}\) has \(k \approx 0.25\ \text{kg/m}\), so \(v_t = \sqrt{80(9.81)/0.25} = 56\ \text{m/s}\), about \(200\ \text{km/h}\). Five seconds after jumping, \(v = 56\tanh(9.81 \times 5/56) = 39.5\ \text{m/s}\); it takes about \(10\ \text{s}\) and \(370\ \text{m}\) of fall to reach \(95\%\) of \(v_t\). Diving head first cuts \(k\) to about \(0.08\ \text{kg/m}\) and raises \(v_t\) to nearly \(100\ \text{m/s}\).

Check your understanding

Key takeaways

When the force varies, so does the acceleration: integrate, and never use the constant-acceleration formulas.
\(F(t)\): use \(a = dv/dt\). \(F(s)\): use \(a = v\,dv/ds\). \(F(v)\): use either, for the time or for the distance.
Static friction rises with the push until it reaches \(\mu_s N\); then it drops to \(\mu_k N\), and the acceleration jumps.
A body falling against drag approaches its terminal velocity, where the drag equals the weight: \(v_t = \sqrt{mg/k}\) for \(F_D = kv^2\).
Next, Lesson 6 writes the equation of motion along and across a curved path.