If an object is at rest on an incline, being pushed up in place by a horizontal force.
Does the component of the vector holding it up (relative to the incline), add to the normal force of the object, therefore increasing its static force of friction?
Does this mean that the static friction of an object is not a constant?
This is a sample question related to this idea that I am trying to understand if anyone is confused as to what I am asking:
A 5kg block rests on a 30 degree incline. The coefficient of static friction between the block and the incline is 0.2. How large of a horizontal force must push on the block if the block is to be on the verge of sliding up?
This is my free body diagram to try and understand this:
$$ F_{\text{net}}=0\,\mathrm N\quad m=5\,\mathrm {kg}\\ \begin{align} F_F=0.2(42.4\,\mathrm N)&=8.48\,\mathrm N\\ 24.5+8.5&=33\,\mathrm N \end{align}\\[16pt] F_N=(5\times9.8)\cos(30)\\ \text{Static friction}=0.2(42.4)=8.5\,\mathrm N\\ 24.5+8.5=33\,\mathrm N\\[12pt] \frac x{\sin90}=\frac{33}{\sin60}\quad x=38\,\mathrm N\\[16pt] \text{Answer is}\,43\,\mathrm N $$