Trig Help: Deriving a Friction Equation from Circular Motion Example

[CINA]

Homework Statement

Well I'm following an example problem on this site:
http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Circular Motion/banked_with_friction.htm
I just need some help on the derivation math. I'll just host the image below.

Homework Equations

This is the problem equation:
http://img105.imageshack.us/img105/3449/frictioneqn4.th.gif

On the last line when it goes from sines and cosines to tangants I can't follow what the person did.

The Attempt at a Solution

I'm assuming he used some type of trig Identity but I can't find one that is apt. I thinking it's a sum/difference trig identity but it just doesn't help. What could he be doing?

Any help at all is GREATLY apperciated!

Thanks

 

[LowlyPion]

Multiply top and bottom by 1/cosθ

 

[nlightner]

I understand the importance of being able to derive equations from given examples. In this case, the friction equation is derived from circular motion. It appears that the person in the example is using the trigonometric identity of tangent (tan) being equal to sine (sin) over cosine (cos). This is a common identity used in solving problems involving circular motion and forces.

To understand this, let's look at the diagram provided in the example. The object is moving in a circular path, and the forces acting on it are the normal force (N), weight (mg), and friction force (F). The normal force and weight are acting perpendicular to each other, while the friction force is acting at an angle to the normal force.

To solve for the friction force, we need to use the trigonometric identity of tan = sin/cos. In this case, the angle we are interested in is the angle between the normal force and the friction force, which is the same as the angle between the velocity vector and the horizontal axis.

Using this identity, we can write the equation for the friction force as F = μN = μmgcosθ. However, since we are solving for the coefficient of friction (μ), we need to rearrange the equation to isolate μ. This is where the substitution of tanθ = sinθ/cosθ comes in. By substituting this into the equation, we can rearrange it to get μ = tanθ/cosθ.

I hope this explanation helps you understand the derivation process better. Keep in mind that trigonometric identities are commonly used in physics and it's important to be familiar with them to solve problems involving circular motion. Keep practicing and you will become more comfortable with them. Good luck!