I'm currently studying forces in physics and in this example, it states that Fn +Fay = mg. Why is this so? Is it because they are both in the same direction?
[Physics] Calculation of a normal force with an angle
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At the beginning I found this question a bit naive. But now I think it is worth to think a little bit about it.
They definitely share some properties, at least at the microscopic level. Both have the same microscopic origin: the electromagnetic interaction. The upthrust requires gravity to create a pressure gradient on the fluid, resulting in an upwards net force. The normal force that a surface does on a body resting on it also has gravity as a "trigger", though this is not always required. If someone presses the body against the surface then the correspondent parcel of the normal force has nothing to do with gravity.
From the macroscopic point of view they are quite different. The upthrust has always the direction of the pressure gradient. The direction of the normal force can be any, defined by the normal to the surface. The upthrust on a (completely immersed) body of volume $V$ depends only on its volume and the density of the agent (fluid). The normal force acting on the same body when resting over a surface depend on its mass instead and does not depend on physical properties of the agent (surfaces).
I think this is enough to see differences and similarities.
Here's how I interpret this question:
We have two equations: The horizontal projection of the 2 kg force equals $\mu$ times the sum of gravity and the vertical projection, i.e. $$2 \cos(\theta) = \mu(4-2\sin(\theta));$$ and similarly for the 4 kg force, except it's projection on the vertical direction is $+\sin(\theta)$, so $$4\cos(\theta) = \mu(4+4\sin(\theta)).$$
Taking a ratio of these two equations (latter over former) gives $2 = \frac{2+2\sin(\theta)}{2-sin(\theta)}$, which can be solved to get $\sin(\theta) = \frac{1}{2}$, i.e. $\theta=30^{\circ}$. Thus $\cos(\theta) = \frac{\sqrt{3}}{2}$, so the first of the original two equations becomes $$\sqrt(3) = 3 \mu \implies \mu = \frac{\sqrt{3}}{3}$$.
Best Answer
That equation expresses the fact that the forces on the block in the vertical direction add up to zero. $F_n$ and $F_{ay}$ are both forces pulling the block upwards, while the force of gravity on the block is $mg$ downwards. We don't want the block to accelerate up or down, and so the net forces in the vertical direction must cancel out.