# Why is this force the mgCos while the other is mgSin?

1. Feb 8, 2017

### EthanVandals

1. The problem statement, all variables and given/known data
A block is placed on a ramp that is inclined to 60 degrees to the horizon. If the block weighs 10kg and the coefficient of kinetic friction is 0.3, how fast does the block accerlate down the ramp?

2. Relevant equations
Theta=60 degrees
m=10kg
g=10m/s^2
mu=0.3

3. The attempt at a solution
So far, all I have is the diagram that he gave us in class. I can somewhat tell why we're using the trig functions, but what I'm curious about is why mgsin(theta) is the block sliding down the ramp and mgcos(theta) is the reactionary force of gravity. Why not the other way around?

2. Feb 8, 2017

### QuantumQuest

Draw the vertical lines from the down edge of mg to the opposite sides of plane and see which angle is $\theta$ in the right triangle that is formed. In this triangle, find each vertical side using trigonometry.

3. Feb 8, 2017

### Comeback City

This is where theta is to be placed. If you understand why, try to figure it out from there.

4. Feb 9, 2017

### CWatters

You can also ask yourself what happens to the normal force as theta approaches zero? Would Sin or Cos do that?