Block on top of an inclined plane, with a rough surface, that moves with constant acceleration

[Thermofox]

Why can it not increase past that point?

Because friction can no longer hold the block still.

 

[kuruman]

Because friction can no longer hold the block still.

Right. So friction has reached its upper limit. Remeber post #25

If I can say that $f=N\mu_s$ then yes. I have 2 unknowns and 2 equations.

Can you say that $f=N\mu_s$?

 

[Thermofox]

Right. So friction has reached its upper limit. Remeber post #25

Can you say that $f=N\mu_s$?

Yes, because $f_{s,max}=N\mu_s$

 

[kuruman]

So go ahead and solve the system. The algebra is much simpler if you replace the values
$\sin\theta =\sin(30^{\circ})=\frac{1}{2}~;~~\cos\theta =\frac{\sqrt{3}}{2}~;~~\mu_s=\tan\theta =\frac{1}{\sqrt{3}}~$ right from the start.

 

[Thermofox]

$\begin{cases} N_y=mg+f_y \\ N_x + f_x = ma_c \\ f=\mu_s N \end{cases}$ ; $\begin{cases} N \frac {\sqrt{3}} 2=mg+ \frac f 2 \\ \frac N 2+ f\frac {\sqrt{3}} 2 = ma_{c,max} \\ f=\mu_s N \end{cases}$ $\Rightarrow N \frac {\sqrt{3}} 2 - N \frac {\mu_s } 2 =mg$ $\Rightarrow N= \frac {2mg} {\sqrt{3}-\mu_s}$

$\begin{cases} N= \frac {2mg} {\sqrt{3}-\mu_s} \\ f= \mu_s \frac {2mg} {\sqrt{3}-\mu_s} \\ \frac {2mg} {\sqrt{3}-\mu_s} \frac 1 2 + \mu_s \frac {2mg} {\sqrt{3}-\mu_s} \frac {\sqrt{3}} 2 = ma_{c,max} \end{cases}$ $\Rightarrow a_{c,max}= \frac {\frac {mg} {\sqrt{3}-\mu_s} + \mu_s \frac {\sqrt{3}\space mg} {\sqrt{3}-\mu_s}} m = \frac g {\sqrt{3}-\mu_s} (1 + \mu_s \sqrt{3}) \approx 17.0\space m/s^2$.

If I solve the system using the frame of reference where the x-axis is parallel to the slope I should obtain the same $a_{c,max}$ right?

 

[kuruman]

I agree with your solution. More presicely $a_{c,max}=\sqrt{3}g.$

Yes, the magnitude of the acceleration is a scalar and does not depend on how you choose your axes. Usually, the algebra is simpler if you choose them so that the acceleration is along one of the principal axes.

 

[Thermofox]

I agree with your solution. More presicely $a_{c,max}=\sqrt{3}g.$

Yes, the magnitude of the acceleration is a scalar and does not depend on how you choose your axes. Usually, the algebra is simpler if you choose them so that the acceleration is along one of the principal axes.

That's great to hear. Thanks for your immense patience!!