Chapter 9: Force

9.2 Newton’s laws

In the late 1600s and early 1700s, Isaac Newton published the Principia Mathematica in which he built upon discoveries of scientists (at the time called “natural philosophers”) before him, laying the foundation of classical mechanics. In the Principia, he described three laws of motion:

  1. An object’s momentum will only change if the object is acted on by some force. This is known as the principle of inertia.
  2. The total force acting on an object is equal to the rate of change of the object’s momentum: $$\vec{F}_\textit{net} = \frac{\Delta\vec{p}}{\Delta t} \tag{9.1}$$
  3. For every force acting on an object, there is a reaction force equal in magnitude and opposite in direction acting on another object.

We have already discussed the first law in the context of energy; if you understand inertia, you can explain many phenomena you observe. The second law is a mathematical statement which we can use to quantitatively solve problems. The third law, as we’ll see later, is a result of conservation of momentum.

Let’s consider situations where an object’s mass remains constant (which is often the case), and apply Newton’s second law:

\[
\begin{align}
\vec{F}_\textit{net} &= \frac{\Delta\vec{p}}{\Delta t} \\
&= \frac{\vec{p}_f – \vec{p}_i}{\Delta t} \\
&= \frac{m\vec{v}_f – m\vec{v}_i}{\Delta t}
\end{align}
\]

Since the mass is not changing, we can factor it out:

\[
\begin{align*}
\vec{F}_\textit{net} &= \frac{m\left(\vec{v}_f – \vec{v}_i\right)}{\Delta t} \\
&= m\frac{\Delta\vec{v}}{\Delta t}
\end{align*}
\]

The rate of change of velocity is acceleration:

\[
\vec{F}_\textit{net} = m\vec{a} \tag{9.2}
\]

This is a specific case of Newton’s second law; it only applies when the mass of a system is not changing.

Remember that force, momentum and acceleration are all vector quantities. Breaking them up into components gives us:

\[
\begin{align}
F_\textit{net,x} &= \frac{\Delta p_x}{\Delta t} & F_\textit{net,y} &= \frac{\Delta p_y}{\Delta t} \tag{9.3} \\
F_\textit{net,x} &= ma_x & F_\textit{net,y} &= ma_y \tag{9.4}
\end{align}
\]

The unit of force is named the newton (N) in reference to Isaac Newton. Using Newton’s second law, we can break the newton down into base SI units: N = kg·m/s2. One newton is approximately the force required to push down one key on a computer keyboard.

Example

A 1380 kg car is moving due East with an initial speed of 27 m/s. After 8 s, the car has slowed down to 17 m/s. Find the net force acting on the car.

Let’s call due East the \(+x\) direction. There are no \(y\) components of the car’s motion. Since this is a one-dimensional problem, we simply use + and – signs to indicate direction instead of using vector notation.

\[
\begin{align*}
F_\textit{net,x} &= \frac{\Delta p_x}{\Delta t} \\
&= m\frac{\Delta v}{\Delta t} \\
&= m\frac{v_f – v_i}{\Delta t} \\
&= (1380\ \textrm{kg})\left(\frac{17 – 27\ \textrm{m/s}}{8\ \textrm{s}}\right) \\
&= -1725\ \textrm{N}
\end{align*}
\]

The negative value for the net force indicates that it is applied in the opposite direction as the car’s motion.

Practice

Practice

Example

Newton’s third law, examined

Two carts are on a nearly-frictionless track. Cart \(A\) has a mass of 0.5 kg and is traveling towards cart \(B\) with a speed of 2.0 m/s. Cart \(B\) is initially at rest, and has a mass of 0.25 kg. The collision is perfectly inelastic. Determine the final velocity of the two carts, and the force that each cart exerts on the other.

We’ll start by analyzing conservation of momentum (revisit chapter 7 for a refresher) to determine the final velocity of the two carts. Let’s say cart \(A\) is traveling in the \(+x\) direction with speed \(v_i\).

\[
\begin{align*}
p_{ix} &= p_{fx} \\
p_a &= p_f \\
m_av_i &= (m_a + m_b)v_f \\
\hookrightarrow v_f &= \frac{m_av_i}{m_a + m_b} \\
&= \frac{(0.5\ \textrm{kg})(2.0\ \textrm{m/s})}{(0.5 + 0.25)\ \textrm{kg}} \\
&= 1.3\ \textrm{m/s}
\end{align*}
\]

Now we’ll apply Newton’s second law to each cart individually. Let’s say the collision lasts for a time of 0.2 s. This is the period of time during which the carts’ velocities are changing, and the period of time that separates “initial” from “final” in our momentum analysis above. Note that \(v_{i,b} = 0\) and \(v_f = 1.3\ \textrm{mps}\) for both carts.

\[
\begin{align*}
\text{Cart $A$} & & \text{Cart $B$} & \\
F_\textit{on a} &= m_a\frac{\Delta p_a}{\Delta t} & F_\textit{on b} &= m_b\frac{\Delta p_b}{\Delta t} \\
&= m_a\frac{v_f – v_i}{\Delta t} & &= m_b\frac{v_f – 0}{\Delta t} \\
&= \left(0.5\ \textrm{kg}\right)\left(\frac{(1.3 – 2.0)\ \textrm{m/s}}{0.2\ \textrm{s}}\right) & &= \left(0.25\ \textrm{kg}\right)\left(\frac{1.3\ \textrm{m/s}}{0.2\ \textrm{s}}\right) \\
&= -1.7\ \textrm{N} & &= 1.7\ \textrm{N}
\end{align*}
\]

So, the force that cart \(B\) exerted on cart \(A\) has the exact same magnitude as the force that \(A\) exerted on \(\), but in the opposite direction—this is Newton’s third law!

As seen in the previous example, Newton’s third law is a direct consequence of conservation of momentum. During the collision, cart \(A\) transferred some of its momentum to cart \(B\); since momentum is conserved, cart \(B\) gained the exact same amount of momentum that cart \(A\) lost.

Speaking in terms of force, we say that cart \(A\) exerted a force on cart \(B\), which caused cart \(B\)’s momentum to change. Through this interaction, cart \(A\)’s momentum also changed. In order for this to happen, cart \(B\) needed to exert a force on cart \(A\). Since momentum is conserved, cart \(B\) must have exerted the exact same amount of force on cart \(A\) that cart \(A\) exerted on cart \(B\).

Remember: all forces describe interactions between two objects; all forces come in pairs.

A line drawing of two pears sitting side-by-side, touching each other. Written on each pear is the expression delta p divided by delta t.
All forces come in pears