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

[kuruman]

$f= \text {it's maximum value} = N\mu_s$.

I am not asking about the the maximum value of the force of friction. I am asking about the maximum value of $a_c$ which is the acceleration of the incline plus the block on it. Why should there be a maximum value of this acceleration? What happens when this value is reached?

If $f_s\leq \mu_sN$ isn't that the reason why it can't go over $\mu_s N$?

Writing $f_s\leq \mu_sN$ says with a mathematical equation that "the force of static friction can't go over the product of the coefficient of static friction and the normal force."

 

[Thermofox]

I am not asking about the the maximum value of the force of friction. I am asking about the maximum value of $a_c$ which is the acceleration of the incline plus the block on it. Why should there be a maximum value of this acceleration? What happens when this value is reached?

Writing $f_s\leq \mu_sN$ says with a mathematical equation that "the force of static friction can't go over the product of the coefficient of static friction and the normal force."

Since there is an acceleration on the block, there is a force, $F$, that pushes the block up.
$\Rightarrow \text{if}\space f - F = 0$ then the acceleration is maximum, because if $F\gt f \Rightarrow$ the block has an acceleration that moves it up the slope.

$\Rightarrow \begin{cases} N_y + F_y =mg+f_y \\ N_x + f_x - F_x = ma_{c,max}\\ f=F \end{cases}$ ; $\begin{cases} N_y=mg \\ N_x = ma_{c,max} \end{cases}$

 

[kuruman]

Since there is an acceleration on the block, there is a force, $F$, that pushes the block up.

Where does this force $F$ come from? There are only two entities exerting forces on the block, the Earth and the incline. Therefore, there are only two separate forces to consider.
The Earth force $~\mathbf F_g ~$ is directed straight down and has fixed magnitude $mg$.
The incline force $~\mathbf F_c ~$ is a contact force and adjusts itself to provide the observed acceleration. This means that both its direction and magnitude depend on the observed acceleration. Note that for easy identification, the component of $~\mathbf F_c ~$ that is perpendicular to the surface is given the name "normal force" and the symbol $N$. The component of $~\mathbf F_c ~$ that is parallel to the surface is given the name "friction" and the symbol $f$. Both components can increase up to a certain limit.

So I will ask you the question again. Why should there be a maximum value of the acceleration in this problem? What happens when this value is reached?

 

[Thermofox]

What happens when this value is reached?

The block starts moving.

 

[kuruman]

Right. This means that the block + incline system have reached a point where they no longer can have the same acceleration and move as one. In other words, the common acceleration has reached a point above which it can longer increase. That's what defines as "maximum". Why can it not increase past that point?

 

[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!!