A Physical Trig Identity

Can a basic physics problem give you insight into math?

For example, mathematically

\cos(x) - \sin(x) = \sqrt{2}\cos(x + \pi/4)

which is easy to verify using the angle addition formula.

I came across this formula while solving a simple problem in statics.

Imagine the classic “block on an inclined plane”. Gravity (F_g) pulls the block down, and you push (F_p) on it sideways, like this:

Gravity pulls down on the block and you push on it to the right.

What is the minimum coefficient of static friction to keep the block stationary? In order to calculate this, we need to know the component of force parallel to the plane.

First look at gravity. We want to find the green component F_{g1}.

The force of gravity can be decomposed into two components. One points along the plane and the other is normal to it.

Let’s say the positive direction is to the right. Then gravity is pulling backwards some, so F_{g1} is negative. I know it’s either a sine or cosine of \theta, and in the limit as \theta \to 0, I see that F_{g1} = 0, so

F_{g1} = -F_g \sin\theta .

Then we look at the pushing force.

The force from pushing is likewise decomposed.

A similar procedure gives

F_{p1} = F_p \cos\theta .

So the total force in the direction of the ramp is

-F_g \sin\theta + F_p \cos\theta .

In the special case where F_g = F_p = 1 the force is

\cos\theta - \sin\theta .

Now we will find this component of the force another way. We start by tip-to-tail adding the force of gravity and the force of the push. They’re at right angles, and assuming they’re equal in magnitude we get a resultant force with length \sqrt{2} that bisects the angle between the gravity and pushing forces.

We can also find the component of force along the plane by first adding the two vectors...

The angle between this resultant force and the plane is \pi/4 + \theta.

...and then finding a component of the sum.

The component of this force along the plane is then the cosine of \pi/4 + \theta, so the force along the direction of the plane is

\sqrt{2} \cos(\pi/4 + \theta)

and since it’s the same quantity we calculated before, we have

\cos\theta - \sin\theta = \sqrt{2}\cos(\pi/4 + \theta) .

There is no physics in this calculation, but if you had simply asked me to write \cos\theta-\sin\theta as a single trig function, I wouldn’t have thought to approach it like this.

Tags: ,

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: