The Prodigal Physicist

Finally, some nuts and bolts physics! We’ll fly through this grade school stuff post-haste and I’ll try to stick to the heart-pounding, edge-of-your-seat, action-packed examples so you don’t fall asleep.

There are three laws at the heart of Newtonian mechanics:

An object keeps its state of motion unless acted upon by an external force
$\sum{\textbf{F}}=m\textbf{a}$
For every force, there is an equal and opposite force

The second of these laws is the omnipotent and ubiquitous “eff-equals-em-ey” which just means that the sum of all vector forces equals the scalar mass times the sum of all vector accelerations.

For absolutely any classical problem, set up a free body diagram (a simple illustration of the magnitude and direction of all the vectors), add up the forces in each dimension and evaluate. Check it out:

example: ice block

Viewed from above, a 10kg block of ice is resting on a sheet of ice. (Forgive the contrived example. The point is, it’s frictionless!) What is the acceleration in the x and y directions?

First, the x component:

$\sum F_x=ma_x$

Then the y component:

$\sum F_y=ma_y$

Then the magnitude of the vector a is:

$|\textbf{a}|=\sqrt{{a_x}^2+{a_y}^2}=\sqrt{(4.4)^2+(-1.22)^2}=\sqrt{19.36+1.49}=4.57m/s^2$

And the angle from the x axis is:

$tan\theta_a=\frac{a_y}{a_x}=\frac{-1.22}{4.4}=-0.277$

$\theta_a=tan^{-1}(-.277)\approx-51^\circ$

example: frictionless pulley

What is the acceleration of both weights?

For this problem, our (frictionless) pulley only allows forces in the ±y direction. Let’s look at the problem as a free body diagram:

T here is the tension on the rope and g is always the acceleration due to gravity. On the left we have:

$\sum\textbf{F}=m\textbf{a}=10\textbf{g}-\textbf{T}=10\textbf{a}$

And on the right we have:

$\sum\textbf{F}=m\textbf{a}=\textbf{T}-5\textbf{g}=5\textbf{a}$

Solving the above for T, then plugging it into the first equation, we get:

$10\textbf{g}-(5\textbf{a}+5\textbf{g})=10\textbf{a}\\$

$98-49=10\textbf{a}+5\textbf{a}$

$\textbf{a}=\frac{49}{15}=3.27m/s^2$

So, the 10kg weight accelerates down (+y) on the left and the 5kg weight accelerates up (+y) on the right, both with the acceleration a.

You may say, “Of course they accelerate at the same rate, jerk.” Or “The heavier one has to pull the lighter one down.” But these are easy things to overlook. Anytime you’re working out a problem, take a step back and try to think intuitively about the (no pun intended) forces at work. “What’s happening in this situation?” is a great question to keep asking yourself.

The thicket of mathematics has many brambles. Don’t get stuck in the equations!

friction

For some more realistic calculations, let’s work friction into the mix. The two types of friction are static and kinetic and the equations of force are as follows:

$\textbf{F}_s\le\mu_s\textbf{N}\hspace{20 mm}\textbf{F}_k=\mu_k\textbf{N}\hspace{20 mm}\mu_k<\mu_s$

Here μ_s is the coefficient of static friction, μ_k is the coefficient of kinetic friction and N is the normal (the “equal and opposite force” to the block’s weight).

μ_s is the force a body must overcome to move from a (static) state of rest. μ_k is the force that decelerates a moving (kinetic) body. The inequality of the coefficients represents the “bump” that you feel when you decelerate and stop completely.

example: inclined plane

What is the acceleration of the blocks?

Like always, let’s take each block separately, create a free body diagram to evaluate the x and y contributions and then add it all up.

For the 10kg block, which only moves in the y direction, we have:

free body diagram for the 10kg block

$x:\emptyset\hspace{10 mm}y:10\textbf{g}-\textbf{T}_2=10\textbf{a}$

For the 3kg block, we have:

free body diagram for the 3kg block

$x:\textbf{T}_1-.2\textbf{N}_1=3\textbf{a}$

$y:\textbf{N}_1-3\textbf{g}=0\hspace{10 mm}so,\hspace{10 mm}\textbf{N}_1=3\textbf{g}=(3)(9.8)=29.4N$

And for the 4kg block, we see the true beauty of the free body diagram. We can set our axes in any direction we choose so the equations don’t get mucked up with superfluous sines and cosines. In this case, we’ll rotate the xy-axes by -30º and call the axes x´and y´:

free body diagram for the 4kg block

$x\prime: \textbf{T}_2-\textbf{T}_1-.1\textbf{N}_2-4\textbf{g}sin(30)=4\textbf{a}$

$y\prime: \textbf{N}_2-4\textbf{g}cos(30)=0\hspace{10 mm}so,\hspace{10 mm}\textbf{N}_2=4\textbf{g}cos(30)=33.94N$

We get values for N₁ and N₂ right off the bat where N is the unit of force called the Newton (not to be confused with N, the normal vector).

Now just add up all the results (”eff-equals-em-ey” is the sum of all forces) and solve for a:

$\textbf{T}_1-(.2)(29.4)=3\textbf{a} \\(10)(9.8)-\textbf{T}_2=10\textbf{a} \\\underline{\textbf{T}_2-\textbf{T}_1-(.1)(33.94)-(4)(9.8)(.5)=4\textbf{a}} \\98-5.88-3.39-19.6=17\textbf{a}$

$\textbf{a}=\frac{69.13}{17}=4.07m/s^2$

newtonian gravity

The equation of force for Newtonian Gravity is:

$\textbf{F}_g=G\frac{m_1m_2}{\textbf{r}^2}$

Where,

F_g is the force between a body of mass m₁ and a body of mass m₂,

G is the gravitational constant (6.67 × 10^-¹¹ N·m²·kg^-²),

and r is the distance between the two bodies.

If an orbiting body with mass M has a period τ, the relationship between the period and the radius of two bodies is:

$\textbf{v}=\frac{2\pi\textbf{r}}{\tau}=\sqrt{\frac{GM}{\textbf{r}}}\hspace{5 mm}so,\hspace{5 mm}GM=\frac{4\pi^2\textbf{r}^3}{\tau^2}\hspace{5 mm}thus,\hspace{5 mm}\textbf{r}^3=(\frac{GM}{4\pi^2})\tau^2$

Which brings us to:

Kepler’s Laws:

Orbits are elliptical with two foci; one being the mass M
An orbiting body covers equal area in equal time
$\frac{\textbf{r}_1^3}{\textbf{r}_2^3}=\frac{\tau_1^2}{\tau_2^2}$

ex: Bang, zoom, straight to the moon!

At what distance between the Earth and the Moon does a body have no net acceleration due to gravity?

So the free body diagram at the point looks very simple:

Earth/Moon free body diagram

Since there is no net acceleration, the forces to the left and right should be equal, giving:

$\textbf{F}_E=\textbf{F}_M=G\frac{M_E\hspace{1 mm}m}{\textbf{x}^2}=G\frac{M_M\hspace{1 mm}m}{(\textbf{R}-\textbf{x})^2}\hspace{5 mm}so,\hspace{5 mm}\frac{M_E}{\textbf{x}^2}=\frac{M_M}{(\textbf{R}-\textbf{x})^2}$

Here, m is the mass of the thing at the point (it could be an asteroid, a taco, whatever) but you don’t have to worry about it since it cancels. Plugging in the values for the mass of the Earth, the mass of the Moon and the distance between their centers, we get:

$(1-\frac{M_M}{M_E})\textbf{x}^2-2\textbf{Rx}+\textbf{R}^2=0$

$(1-\frac{7.35\times10^{22}kg}{5.97\times10^{24}kg})\textbf{x}^2-2(3.84\times10^8m)\textbf{x}+(3.84\times10^8m)^2=0$

$(.988)\textbf{x}^2-(7.68\times10^8)\textbf{x}+(1.47\times10^{17})=0$

Using the quadratic formula, we get the positive and negative roots:

$x_{\pm}=(7.68\times10^8)\pm\frac{\sqrt{(5.9\times10^{17})-4(.988)(1.47\times10^{17})}}{2(.988)}$

$x_+=4.36\times10^8m\hspace{10 mm}and\hspace{10 mm}x_-=3.41\times10^8m$

The answer must be the negative root since the positive root is larger than R. That means that the position at which an asteroid, taco, whatever would feel no net acceleration from the Earth or the Moon is:

$x=3.41\times10^8(\frac{R}{3.84\times10^8})=.89R$

That’s nine-tenths the distance from the Earth to the Moon. POW, right in the kisser!

Classical Mechanics

Newton’s Laws

example: ice block

example: frictionless pulley

friction

example: inclined plane

newtonian gravity

ex: Bang, zoom, straight to the moon!

Problem Set 3 » Newton’s Laws