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I recently was assigned this problem as part of my homework:

"A block sits on a plane that is inclined at an angle $\theta$. Assume that the friction force is large enough to keep the block at rest. What are the horizontal components of the friction and normal forces acting on the block? For what $\theta$ are these horizontal components maximum?"

To solve the first sub-question within this problem, I drew the following Free-Body Diagram:

enter image description here

Since the block is at rest, I concluded that $F_n = mg\cos(\theta)$ and that $F_{fr} = mg\sin(\theta)$. Then, to find the horizontal component of each of these forces, I realized that

$(F_n)_x = (F_n) * \sin(\theta) = mg\cos(\theta)\sin(\theta) = \frac{1}{2} mg \sin(2\theta)$, and

$(F_{fr})_x = (F_{fr}) * \cos(\theta) = mg\sin(\theta)\cos(\theta) = \frac{1}{2} mg \sin(2\theta)$.

But now I'm very confused, because I don't see how these two quantities can be the same! When solving the FBD, I assumed that the frictional force would cancel out with the $mg\sin(\theta)$ and the normal force would cancel out with the $mg\cos(\theta)$, meaning that those x-components would be the same.

For instance, if we assume that $\theta$ is some very small angle, then the x-component of $F_n$ will be very close to $\theta$ while the x-component of $F_{fr}$ will reach its max, but my derivation shows that they would still be the same in that case, which makes $\theta$ sense to me. I'm pretty confused by all of this, so can someone tell me what mistake I've been making please?

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If the body is in static equilibrium the net force is zero. This means that the net force has zero component on any direction, including the horizontal direction. In this setup you have 3 forces. One of them, the weight, has no horizontal component. So it is expected that the other two, which have horizontal components, will cancel out along this direction. If they didn't the body will accelerate along the horizontal direction.

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