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How do we formulate Classical Mechanics?

In the past, you have learned about Newton’s Laws of Motion. These laws are the foundation of classical mechanics. They are used to describe the motion of objects in the universe. In this course, we will use these laws to describe the motion of particles and rigid bodies. As we progress, we will learn other formulations of classical mechanics, such as the Lagrangian and Hamiltonian reformulations.

Newton’s Laws of Motion are a vector formulation of classical mechanics. The laws are as follows:

  1. First Law: An object at rest will remain at rest, and an object in motion will remain in motion at a constant velocity unless acted upon by an external force.

  2. Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The direction of the acceleration is in the direction of the net force.

  3. Third Law: For every action, there is an equal and opposite reaction.

These are the classic laws of motion that you have learned in other courses. How we formulate these laws mathematically is the subject of this course.

Newton’s Second Law

Newton’s Second Law provides the mathematical foundation for classical mechanics. It provides a vector relationship between the net force acting on an object and its acceleration. The law is given by the equation:

F⃗net=ma⃗\vec{F}_{net} = m\vec{a}

where F⃗net\vec{F}_{net} is the net force acting on the object, mm is the mass of the object, and a⃗\vec{a} is the acceleration of the object. This equation is a vector equation, meaning that it must be satisfied in each direction.

Fnet,x=maxFnet,y=mayFnet,z=maz\begin{aligned} F_{net,x} &= m a_x \\ F_{net,y} &= m a_y \\ F_{net,z} &= m a_z \end{aligned}

Each push in a Cartesian direction results in a proportional response -- an acceleration in the same direction of the net push. Let’s go through a common example to illustrate this relationship.

Example: A Block on an Inclined Ramp

Consider the box (mass, mm) above on an inclined ramp (angle, θ\theta). The box is at rest subject to static friction, μs\mu_s. What angle of inclination will cause the box to start sliding down the ramp?

We start by drawing the Free Body Diagram (FBD) of the box.

Free Body Diagram of Box on a Inclined Ramp; the arrows label the direction of forces acting on the box

The FBD is a diagram that shows all the forces acting on the object. In this case, the forces acting on the box are:

We tilt our coordinate system to align with the ramp. This makes the normal force, NN, act in the yy-direction and the frictional force, ff, act in the xx-direction. The force of gravity is split into two components: mgsin⁡(θ)mg\sin(\theta) and mgcos⁡(θ)mg\cos(\theta).

The net force acting on the box is the sum of the forces acting on it, and is zero up to the point where the box starts sliding. At this point, the frictional force is at its maximum value, μsN\mu_s N. Taking the sum of the forces acting on the box in each direction we have:

F⃗net=f⃗+N⃗+F⃗gravity=0\vec{F}_{net} = \vec{f} + \vec{N} + \vec{F}_{gravity} = 0
∑Fx,i=f−mgsin⁡(θ)=0\sum F_{x,i} = f - mg\sin(\theta) = 0
∑Fy,i=N−mgcos⁡(θ)=0\sum F_{y,i} = N - mg\cos(\theta) = 0

Thus,

f=mgsin⁡(θ)f = mg\sin(\theta)
N=mgcos⁡(θ)N = mg\cos(\theta)

At the point where the box starts sliding, the frictional force is at its maximum value, μsN\mu_s N. Thus, the box starts sliding when:

f=μsNf = \mu_s N

Substituting the expressions for ff and NN into the equation above, we have:

mgsin⁡(θ)=μsmgcos⁡(θ)mg\sin(\theta) = \mu_s mg\cos(\theta)

Solving for θ\theta:

tan⁡(θ)=μs\tan(\theta) = \mu_s
θ=tan⁡−1(μs)\theta = \tan^{-1}(\mu_s)

Thus, the box starts sliding when the angle of inclination is equal to the arctangent of the coefficient of static friction.

The relationship θ=tan⁡−1(μs)\theta = \tan^{-1}(\mu_s) holds for any coefficient of static friction. The plot below shows how the critical sliding angle grows with μs\mu_s, with the steel and rubber examples marked.

<Figure size 600x400 with 1 Axes>

Let’s work an example that is dynamic, where the net force is not zero.

Example: Falling Object in One Dimension

Consider an object of mass mm falling, but it is subject to air resistance. The free body diagram of the object shows that the forces acting on the object are:

Free Body Diagram of Falling Object; the arrows label the direction of forces acting on the object

Here we have chosen positive yy to be the downward direction. We want to predict the motion (aa, vv, yy) of the object as a function of time. This is a very common problem for classical mechanics.

Air Drag?

First, we notice that we do not know the force due to air resistance. We do know that the force is related to the velocity of the object. So let’s start by writing the air resistance force as a function of velocity:

Fair=F(v)F_{air} = F(v)

where F(v)F(v) is some function of velocity.

Taylor Series Expansion

Because we are modeling low-speed motion, we can approximate the drag law near v=0v=0 using a Taylor Series. Classical mechanics itself does not require objects to move slowly. In general, the Taylor Series of a function f(x)f(x) about a point aa is given by:

f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

where f(n)(a)f^{(n)}(a) is the nn-th derivative of f(x)f(x) evaluated at x=ax=a. We can write the first few terms out,

f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+…f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots

Because we know that the object is moving slowly, we can expand the function F(v)F(v) about v=0v=0:

F(v)=F(0)+F′(0)v+F′′(0)2!v2+…F(v) = F(0) + F'(0)v + \frac{F''(0)}{2!}v^2 + \ldots

We can assume that the first term is zero, F(0)=0F(0)=0, because there is no air resistance when the object is at rest. Thus, the force due to air resistance is approximately:

F(v)≈F′(0)v+F′′(0)2!v2F(v) \approx F'(0)v + \frac{F''(0)}{2!}v^2

We call the first term the linear drag term and the second term the quadratic drag term. The linear drag term is proportional to the velocity of the object, and the quadratic drag term is proportional to the square of the velocity of the object. We also typically replace the evaluated derivatives with constants, bb and cc -- because they are constants that depend on the object and the fluid it is moving through. And thus,

For a signed one-dimensional velocity, the resistive force must oppose the motion. A common model is Fair=bv+c∣v∣vF_{air}=b v+c|v|v when positive yy is downward; if FairF_{air} denotes only the upward resistance magnitude, the simpler bv+cv2bv+cv^2 applies for v≥0v\ge 0.

The plot below shows a representative nonlinear drag force alongside its Taylor approximations. Keeping only the linear term matches well near v=0v=0, but adding the quadratic term extends the range over which the approximation stays accurate.

<Figure size 600x400 with 1 Axes>

Back to Newton’s Second Law

In the yy-direction, the net force acting on the object is:

Fy=W−Fair=mg−bv−cv2F_y = W - F_{air} = mg - bv - cv^2

And thus, the acceleration of the object is:

a=g−bmv−cmv2a = g - \frac{b}{m}v - \frac{c}{m}v^2

This differential equation can be written in a variety of ways. One common way is to write the equation as a second-order differential equation.

d2ydt2=g−bmdydt−cm(dydt)2\frac{d^2y}{dt^2} = g - \frac{b}{m}\frac{dy}{dt} - \frac{c}{m}\left(\frac{dy}{dt}\right)^2

Note that this is a nonlinear differential equation (i.e., there’s a (dny/dtn)m(d^ny/dt^n)^m term where m>1m > 1), which are notoriously difficult to solve in general. We can write it using the dot notation for derivatives (i.e., y˙=dy/dt\dot{y} = dy/dt, y¨=d2y/dt2\ddot{y} = d^2y/dt^2):

y¨=g−bmy˙−cmy˙2\ddot{y} = g - \frac{b}{m}\dot{y} - \frac{c}{m}\dot{y}^2

We can also use the velocity as the independent variable, v=y˙v = \dot{y}. Both equations below are equivalent:

dvdt=g−bmv−cmv2\frac{dv}{dt} = g - \frac{b}{m}v - \frac{c}{m}v^2
v˙=g−bmv−cmv2\dot{v} = g - \frac{b}{m}v - \frac{c}{m}v^2

How do we solve this equation to find the motion of the object as a function of time? We will come back to this, but solving differential equations is the primary tool of classical mechanics. We will learn how to solve these equations analytically and numerically in this course.