Lesson 8 · 40 min
Systems of Particles and Choosing a Method
A train, a truck and its trailer, a stack of crates: many problems involve several bodies that move together. Draw a free-body diagram around all of them and the forces between them disappear, leaving one equation for the whole system. Draw one around a part and those forces come back, as external forces on that part. Choosing where to cut is the skill; this lesson also looks ahead to the other methods that solve kinetics problems.
Learning objectives
- Explain why internal forces cancel, and write \(\sum\Fvec_{\text{ext}} = m\avec_G\) for a system of particles.
- Find the acceleration of bodies that move together from a free-body diagram of the whole system.
- Find contact and coupling forces by isolating part of the system.
- Choose between \(\sum\Fvec = m\avec\), work and energy, and impulse and momentum for a given question.
The equation of motion for a system
Apply \(\sum\Fvec = m\avec\) to each particle \(i\) of a system. The forces on it are external, \(\Fvec_i\), exerted by things outside the system, and internal, \(\mathbf f_i\), exerted by the other particles:
\[ \Fvec_i + \mathbf f_i = m_i\avec_i \]Add these equations over the whole system. By Newton's third law, internal forces come in equal, opposite, collinear pairs, so \(\sum\mathbf f_i = \mathbf 0\). What is left involves the mass center \(G\), whose position is \(m\rvec_G = \sum m_i\rvec_i\), with \(m = \sum m_i\):
A system of particles
\[ \textstyle\sum\Fvec_{\text{ext}} = \sum m_i\avec_i = m\colMA{\avec_G} \]The mass center moves as if it were a single particle carrying the whole mass, acted on by all the external forces. Internal forces cannot change the motion of \(G\), however large they are.
- When the bodies move together with the same acceleration, \(\sum F_{\text{ext}} = (\sum m_i)\,a\): one equation gives \(a\).
- A car's engine is internal to the car. What accelerates the car is the external friction of the road on the driven tires.
- To find a force between two parts, cut the system there: the force becomes external to each part.
Where to cut
Three crates stand in a row on a rough floor. A worker pushes the first with a horizontal force \(\colP{P}\), and all three slide together. Figure 8.1 lets you draw the free-body diagram around any group of them.
Example 8.1 — Three crates in a row
In Figure 8.1, \(P = 60\ \text{N}\) and the floor is smooth. Find the acceleration and the force each crate exerts on the next.
Show solution
The whole system: the only external horizontal force is \(P\).
\[ 60 = (2 + 3 + 5)\,a \quad\Rightarrow\quad a = 6\ \text{m/s}^2 \]Cut between \(A\) and \(B\), and take \(B\) and \(C\) together: the only horizontal force on them is the push of \(A\).
\[ F_{AB} = (3 + 5)(6) = 48\ \text{N} \qquad\text{and, with } C \text{ alone,}\qquad F_{BC} = 5(6) = 30\ \text{N} \]Check with \(A\) alone: \(60 - 48 = 2(6)\). Each contact force pushes only the crates ahead of it, so it is \(P\) times their share of the mass: \(F_{AB} = \tfrac{8}{10}P\). With the same \(\mu_k\) under every crate, friction scales with mass too, and these shares do not change.
Example 8.2 — A rig whose trailer brakes have failed
An \(8000\ \text{kg}\) tractor tows a \(4000\ \text{kg}\) trailer whose brakes have failed. The tractor's brakes slow the whole rig at \(3\ \text{m/s}^2\). Find the braking force of the road on the tractor, and the force in the coupling.
Show solution
The whole rig: the road's braking friction on the tractor's tires is the only external horizontal force (ignoring drag).
\[ F_b = (8000 + 4000)(3) = 36\,000\ \text{N} \]The trailer alone: only the coupling can slow it, so the coupling pushes back on the trailer with
\[ F_c = 4000(3) = 12\,000\ \text{N} \]and, by the third law, the trailer pushes forward on the tractor with \(12\ \text{kN}\). The coupling is in compression; if the rig is also turning, that push from behind is what can jackknife it.
Choosing a method
\(\sum\Fvec = m\avec\) relates forces to accelerations at an instant. To get speeds, distances and times, you combine it with kinematics, as in every lesson of this module. Integrating it once, as in Lesson 5, gives two other methods, each with its own module, that skip the acceleration altogether.
| Method | Relates | Best when |
|---|---|---|
| Force and acceleration, \(\sum\Fvec = m\avec\) | Forces and acceleration, at an instant | You need a force, a tension or a normal force, or the acceleration itself; curved paths (\(\sum F_n = mv^2/\rho\)); checking whether something slips or leaves a surface |
| Work and energy | Forces, distances and speeds | Speeds at known positions, with forces that depend on position (springs, gravity); no time involved |
| Impulse and momentum | Forces, times and velocities | Speeds after a known time; forces that depend on time; impacts and collisions |
The methods never disagree: they are the same law, integrated in different ways. Many problems need two. A pendulum's speed at the bottom comes from energy; the cord's tension there still needs \(\sum F_n = mv^2/\rho\).
Example 8.3 — One problem, three ways
A \(10\ \text{kg}\) crate starts from rest and is pushed \(4\ \text{m}\) across a floor by a horizontal force of \(50\ \text{N}\), with \(\mu_k = 0.2\). Find its speed at the end, and the time it takes.
Show solution
Force and acceleration. \(N = 98.1\ \text{N}\) and \(F_f = 19.62\ \text{N}\), so
\[ a = \frac{50 - 19.62}{10} = 3.038\ \text{m/s}^2 \qquad v = \sqrt{2as} = \sqrt{2(3.038)(4)} = 4.93\ \text{m/s} \qquad t = \frac{v}{a} = 1.62\ \text{s} \]Work and energy (next module): the net force does work \((50 - 19.62)(4) = 121.5\ \text{J}\), which becomes kinetic energy, \(\tfrac12(10)v^2\), so \(v = 4.93\ \text{m/s}\), with no acceleration in sight; it cannot give the time.
Impulse and momentum (the module after): the net force's impulse \((50 - 19.62)\,t\) becomes momentum, \(10(4.93)\), so \(t = 1.62\ \text{s}\), once the speed is known.
Check your understanding
Key takeaways
- Internal forces cancel in pairs: \(\sum\Fvec_{\text{ext}} = m\avec_G\) for any system of particles.
- For bodies that move together, the free-body diagram of the whole system gives the acceleration.
- Cut the system where you want a force between parts; it becomes an external force on each side.
- \(\sum\Fvec = m\avec\) for forces and instants; work and energy for speeds at positions; impulse and momentum for speeds after times, and impacts.
- The Practice Lab and the Force Lab put the whole module to work.