Given an object resting on an inclined plane, we have the following equations to determine the forces acting on said object:
$$\sum f_x = mg\sin(\theta) - f_s = 0 $$
Which says that the sum of the forces acting on the object parallel with the incline is equal to zero, and where $f_s$ represents the maximum force due to static friction. We also have:
$$\sum f_y = mg\cos(\theta) - n = 0 $$
Which says that the sum of the forces acting on the object perpendicular to the surface of the incline is equal to zero, and where $n$ represents the normal force acting perpendicular to the surface of the incline.
Note that $m,g$ and $\theta$ represent the mass of our object, acceleration due to gravity, and the angle that the incline makes with the horizontal, respectively.
Now then, given a particular coefficient of static friction $\mu_s$, masses $m_1$, and having $g=9.8 \frac{m}{s^2}$, it should be possible to calculate the angle of the incline, along the following lines:
First, recall that the force of an object due to static friction is proportional to the normal force acting on the object:
$$f_s = \mu_sn $$
where $\mu_s$ represents the coefficient of static friction.
Thus, solving $\sum f_y$ for $mg$, we have:
$$mg = \frac{n}{\cos(\theta)}$$
Substitute the result for $mg$: $$\sum f_x = \frac{n\sin(\theta)}{\cos(\theta)} - f_s = 0$$
Recall that $f_s = n\mu_s$, and that $n=mg\cos(\theta)$, and substitute appropriately:
$$ \frac{mg\cos(\theta)\sin(\theta)}{\cos(\theta)} - mg\cos(\theta)\mu_s = 0 $$ $$ mg\cos(\theta)(\tan(\theta) - \mu_s) = 0 $$
If we follow the logic to the bitter end, then solving for $\theta$, we have:
$$ \cos(\theta)\tan(\theta) - \cos(\theta)\mu_s = \frac{0}{mg} $$ $$ \sin(\theta) - \cos(\theta)\mu_s = 0 $$ $$ \sqrt{1-\cos^{2}(\theta)} - \cos(\theta)\mu_s = 0 $$ $$ \sqrt{1-\cos^{2}(\theta)} = \cos(\theta)\mu_s $$ $$ 1-\cos^{2}(\theta) = \cos^{2}(\theta)\mu^{2}_s $$ $$ 1 = 2\cos^{2}(\theta)\mu^{2}_s $$ $$ \frac{1}{2\mu^{2}_s} = \cos^{2}(\theta) $$ $$ \frac{1}{\mu_s\sqrt{2}} = \cos(\theta) $$
Such that, finally, the angle of the incline is equal to:
$$ \theta = \arccos(\frac{1}{\mu_s\sqrt{2}}) $$
It seems reasonable enough, yes? BUT IT DOESN'T WORK!
Recall the following:The force of an object due to static friction is proportional to the normal force acting on the object:
$$f_s = \mu_sn $$
And as mentioned above, the normal force of an object is equal and opposite to the force acting upon that object, which in this case would be the force due to gravity acting perpendicular to the surface of the incline. This would imply that *an object of greater mass would experience a greater normal force.
In turn, that would imply that an object of greater mass would have a higher maximum resistant force due to static friction: $f_s$. A higher $f_s$ would necessarily require a greater force parallel with the surface of the incline to reach and overcome the force due to static friction.
Recall that the amount of force acting parallel with the surface of the incline is determined as $mg\sin(\theta)$, where $\theta$ is the angle of inclination of the surface.
Thus, for a greater amount of mass, we should expect a greater amount of force needed to act parallel with the surface of the incline, but the only way to do this is to increase the angle of inclination.
Look again at the final equation:
$$ \theta = \arccos(\frac{1}{\mu_s\sqrt{2}}) $$
It is completely determined by the coefficient of friction, and isn't dependent on the mass of the object.
But if the final angle is not dependent on the mass of the object, then that would imply that the force needed to move an object is independent of that object's mass.
Which is completely ludicrous.
This is really bugging me, and I don't know exactly where to go from here. Also, I should probably point out that this is not a homework problem.
Thank you very much for taking the time to read this. I appreciate it. :D
:edit: p.s. I'm probably doing something horribly wrong. In which case, I apologize in advance. :(