Vertical and horizontal components of forces and vectors

I'm getting a bit confused when finding components of vectors and forces.

In problems for vectors, I've always known that if you want to get the components of a vector, you would use the following:

Recently we've started working with Forces in homework problems.

This diagram is from one of my homework problems, which they want me to draw the free body diagram:

Here is the free body diagram given by the homework solution:

What I don't understand is why would I be using $m_{2}gsin(\theta)$ instead of $m_{2}gcos(\theta)$ when finding the X-component of the gravitational force of $m_{2}$.

Force is a vector, so why wouldn't you use $m_{2}gcos(\theta)$? Are the rules different when dealing with forces?

It depends how you define the angle. In this diagram you define the angle with respect to the horizontal and take the x-axis along the slope. So the x-component of of gravitational force comes out to be $m_2gsin\theta$. If you define the angle $\theta$ with respect to the vertical, then you would see $m_2gcos\theta$ as the x-component of the gravitational force. So it all depends on how you define the angle of slope.

• Oh I see, so if I defined $\theta$ as being between $m_{2}g$ and $m_{2}gsin(\theta)$, then I would treat $m_{2}gcos(\theta)$ as the horizontal component of the force vector. Correct? Nov 1 '13 at 3:18
• yes, that is correct
– nole
Nov 1 '13 at 3:20

The formula for horizontal or vertical will depend on the angle given. In the diagram:

-The horizontal component can be found using the two angles: Using angle A:

                CosA = adjacent/hypotoneuse= b/c
Therefore, b=c*cosA


Using angle B:

                 SinB= opposite/hypotoneuse= b/c
Therefore, b=c*sinA


As you can see, the horizontal component can be found using any of the two formulas. Horizontal component does NOT always use the cos function

That's an easy mistake to make! Think about the definition of $cos$ that you learned in freshman/sophomore geometry class - Adjacent over Hypotenuse. Notice in the freebody diagram that $m_2g \cos\theta$ is indeed adjacent to $\theta$