1. ## Resultant Forces

In the following question, Why can't we resolve horizontally like this:
$\displaystyle 6 \cos 55 = 3.44N$?

Can someone explain, how this resolving is done. Thanks in advance.

2. Every diagonal vector can be broken down into horizontal and vertical components. For vector a, the horizontal component is found by taking the magnitude times cos 55. Thus 3 cos 55 gives you its horizontal force.

Since vector b is horizontal only, it gets added to 3 cos 55. Is that what you're asking?

3. Originally Posted by mathdeity
Every diagonal vector can be broken down into horizontal and vertical components. For vector a, the horizontal component is found by taking the magnitude times cos 55. Thus 3 cos 55 gives you its horizontal force.

Since vector b is horizontal only, it gets added to 3 cos 55. Is that what you're asking?
Yes. I was confused earlier as I was told that:

Horizontal: $\displaystyle R \cos \theta$
Vertical: $\displaystyle R \sin \theta$

So I though for horizontal it was $\displaystyle 6 \cos 55$. I now realise the Resultant force is 3N.

4. Originally Posted by Air
Yes. I was confused earlier as I was told that:

Horizontal: $\displaystyle R \cos \theta$
Vertical: $\displaystyle R \sin \theta$

So I though for horizontal it was $\displaystyle 6 \cos 55$. I now realise the Resultant force is 3N.
I think maybe your terminology might be going against you. The "resultant" is the result of your vector addition. This is the 8.1 N at 17.656 degrees.

To find the components of vector a,
$\displaystyle a_x = | \vec{a} | ~ cos(\theta)$
and
$\displaystyle a_y = | \vec{a} | ~ sin(\theta)$
where $\displaystyle | \vec{a} |$ is the magnitude (size) of vector a and $\displaystyle \theta$ is the angle between the vector and the +x axis.

Does this help?

-Dan