Lesson 4 · 35 min

Forces That Vary with Time

A motor ramps up, a push fades, a thrust pulses. When a force is given as a function of time, impulse and momentum is the natural method: integrate the force over the interval and the velocity follows, with no need to solve a differential equation for the acceleration. The care goes into choosing the interval, especially when friction holds the body still at first.

Learning objectives

Impulse of a force that varies

When a force changes during the interval, its impulse is the integral, or the area under its graph, and the principle becomes

\[ m v_1 + \int_{t_1}^{t_2} F(t)\,dt + \text{(the impulses of the other forces)} = m v_2 \]

Integrals of polynomials and sines are all you need for most problems:

Impulses of some common forces, from \(t = 0\) to \(t\)
ForceImpulse \(\int_0^t F\,dt\)Graph
\(F_0\) (constant)\(F_0 t\)rectangle
\(k t\) (growing steadily)\(\tfrac12 k t^2\)triangle
\(c\,t^2\)\(\tfrac13 c\,t^3\)area under a parabola
\(F_0\sin(\pi t/T)\), \(0 \le t \le T\)\(\dfrac{F_0 T}{\pi}\left[1 - \cos(\pi t/T)\right]\), which is \(2F_0T/\pi\) at \(t = T\)half-sine pulse

A force made of pieces, like "grows for \(4\ \text{s}\), then stays constant", is integrated piece by piece, and each piece's area adds.

Example 4.1 — A polynomial force

A \(3\ \text{kg}\) block slides at \(2\ \text{m/s}\) along a smooth horizontal surface when a force \(F = 6t^2\ \text{N}\), with \(t\) in seconds, starts to push it forward. Find its speed at \(t = 3\ \text{s}\).

Show solution
\[ \colP{\int_0^3 6t^2\,dt} = \colP{2t^3\Big|_0^3 = 54\ \text{N·s}} \qquad 3(2) + 54 = 3\,v_2 \quad\Rightarrow\quad v_2 = 20\ \text{m/s} \]

The weight and the normal force cancel vertically, as before. Newton's second law would need \(a = 2t^2\) integrated for the velocity: the same integral, in different clothes.

Example 4.2 — A half-sine pulse

A \(2\ \text{kg}\) puck at rest on smooth ice is struck by a force \(F = 20\sin(\pi t/2)\ \text{N}\) for \(0 \le t \le 2\ \text{s}\). Find its speed afterward.

Show solution
\[ \colP{\int_0^2 20\sin\frac{\pi t}{2}\,dt} = \colP{\frac{40}{\pi}\left[1 - \cos\frac{\pi t}{2}\right]_0^2 = \frac{80}{\pi} = 25.46\ \text{N·s}} \] \[ 0 + 25.46 = 2\,v_2 \quad\Rightarrow\quad v_2 = 12.7\ \text{m/s} \]

The average force is \(25.46/2 = 12.7\ \text{N}\), \(2/\pi\) of the peak, as for the half-sine in Figure 2.1.

When does it start to move?

A crate at rest on a rough floor does not move the instant a push begins. While the push is less than the largest static friction, \(\mu_s N\), the floor's friction matches it exactly, and their impulses cancel. Only from the instant the push exceeds \(\mu_s N\) does the crate slide, and from then on kinetic friction, \(\mu_k N\), opposes it.

Start the clock when the motion starts

Find \(t_1\), the instant the push reaches \(\mu_s N\). Apply the principle from \(t_1\), with \(v_1 = 0\), and integrate the push from \(t_1\), not from zero.

Example 4.3 — A crate pushed by a growing force

A \(20\ \text{kg}\) crate rests on a horizontal floor, with \(\mu_s = 0.4\) and \(\mu_k = 0.3\). A horizontal force \(P = 30t\ \text{N}\), with \(t\) in seconds, is applied until \(t = 4\ \text{s}\), after which \(P\) stays at \(120\ \text{N}\). Find the speed of the crate at \(t = 6\ \text{s}\).

Show solution

Vertical. Nothing moves vertically, so \(N = 20(9.81) = 196.2\ \text{N}\).

When does it slide? When \(P\) reaches \(\mu_s N = 0.4(196.2) = 78.48\ \text{N}\):

\[ 30\,t_1 = 78.48 \quad\Rightarrow\quad t_1 = 2.616\ \text{s} \]

Until then the crate stays at rest. After that, kinetic friction \(\colF{F_f = \mu_k N = 58.86\ \text{N}}\) opposes the motion.

Impulse of \(P\) from \(t_1\) to \(6\ \text{s}\), in two pieces:

\[ \colP{\int_{2.616}^{4} 30t\,dt + 120(6 - 4)} = \colP{15\left(4^2 - 2.616^2\right) + 240 = 377.3\ \text{N·s}} \]

Horizontal, positive to the right, from \(t_1\) to \(6\ \text{s}\):

\[ 0 + \colP{377.3} \colF{{} - 58.86(6 - 2.616)} = 20\,v_2 \quad\Rightarrow\quad 377.3 - 199.2 = 20\,v_2 \quad\Rightarrow\quad v_2 = 8.91\ \text{m/s} \]

The most common error is to start the impulse at \(t = 0\). That would credit the push with \(\int_0^{2.616} 30t\,dt = 102.7\ \text{N·s}\) that static friction cancels exactly.

Could work and energy solve this? Not directly: \(P\) is known as a function of time, and the distance the crate slides is unknown.

Figure 4.1 Example 4.3 in motion. The chart under the crate plots the push \(\colP{P}\) and friction \(\colF{F_f}\) against time. Until the crate slips, friction is the mirror image of the push and the two areas cancel; after it slips, friction drops to \(\mu_k N\) and stays there. The impulse ledger adds the two areas up to the crate's momentum. Change the rate at which the push grows, or the friction.

In Figure 4.1, look at the kink in the friction curve at \(t_1\). Static friction can be anything from zero to \(\mu_s N\); it takes whatever value holds the crate still, until it cannot. Then the crate slips and friction drops to the smaller kinetic value, which is why a crate that has just broken loose lurches forward.

A push that fades: the largest speed

When a push decreases while friction stays constant, the body keeps speeding up as long as the push is larger than friction, and slows down after. Its largest speed is at the instant the net force is zero: there, \(dv/dt = 0\).

Example 4.4 — The largest speed, and when it stops

A \(10\ \text{kg}\) crate at rest on a floor with \(\mu_s = 0.25\) and \(\mu_k = 0.2\) is pushed by a horizontal force \(P = 100 - 25t\ \text{N}\) until \(P\) reaches zero at \(t = 4\ \text{s}\). Find (a) the largest speed of the crate and when it occurs, and (b) when the crate stops.

Show solution

\(N = 98.1\ \text{N}\). At \(t = 0\), \(P = 100\ \text{N}\) is far more than \(\mu_s N = 24.5\ \text{N}\), so the crate slides at once, against \(\colF{F_f = 19.62\ \text{N}}\).

(a) The speed is largest when \(P = F_f\): \(100 - 25t_m = 19.62\), so \(t_m = 3.215\ \text{s}\). From \(0\) to \(t_m\):

\[ 0 + \colP{\int_0^{t_m}(100 - 25t)\,dt} \colF{{} - 19.62\,t_m} = 10\,v_{\max} \] \[ \colP{100(3.215) - 12.5(3.215)^2} \colF{{} - 19.62(3.215)} = 129.2 = 10\,v_{\max} \quad\Rightarrow\quad v_{\max} = 12.9\ \text{m/s} \]

(b) At \(t = 4\ \text{s}\), the push has given its whole impulse, \(\colP{\tfrac12(4)(100) = 200\ \text{N·s}}\), and friction has taken \(\colF{19.62(4) = 78.5\ \text{N·s}}\), so \(10v = 121.5\) and \(v = 12.2\ \text{m/s}\). After that only friction acts, and the crate stops when

\[ 121.5 \colF{{} - 19.62\,(t_2 - 4)} = 0 \quad\Rightarrow\quad t_2 = 4 + 6.19 = 10.2\ \text{s} \]

Or in one equation from \(0\) to \(t_2\): \(\colP{200} \colF{{} - 19.62\,t_2} = 0\), which is the same answer, since friction never changed direction.

Figure 4.2 Example 4.4 in motion, with the velocity against time under the crate. The speed peaks where the falling push crosses the level of friction, and the velocity curve is flat there. After the push ends, friction alone brings the crate to rest along a straight line: a constant force gives a constant rate of change of momentum.

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Key takeaways