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

[Thermofox]

Homework Statement

A block of mass $m$ is placed on an inclined plane of height, $h$. In particular the block is placed at a height of $h/2$. The surface of the inclined plane is rough, characterized by a coefficient of static friction, $\mu_s$. Initially both the inclined plane and the block are at rest.
1) Knowing that $\theta = 30°$, determine the minimum value of $\mu_s$ such that the system doesn't move.
2) Now suppose to drag the system (block + inclined plane) to the left so that it has a constant acceleration $a_c$. Knowing that $\mu_s$ is the same as found in point 1 and that there is no friction between the inclined plane and the horizontal plane, Determine:
the maximum value of $a_c$ that allows the block to remain in contact with the inclined plane.

Relevant Equations

$\Sigma F= ma$

For the first point I need to draw a free body diagram of the block and balance the forces:
I chose to use as axis the ones that have the same direction of the components of the weight force; y-positive upwards, x-positive leftwards
$\begin{cases} \Sigma F_y=0 \\ \Sigma F_x=0 \end{cases}$ $\Rightarrow \begin{cases} N-mg\cos(\theta)=0 \\ mg\sin(\theta)-f=0 \\ f= N\mu_s \end{cases}$ $\Rightarrow mg\sin(\theta) = mg\cos(\theta) \mu_s$

The block won't move only if $mg\sin(\theta) \leq mg\cos(\theta) \mu_s$ Which means that $\mu_s \geq \tan(\theta)$. Thus $\mu_{s,min}= \tan(30°)= \frac {\sqrt {3}} 3$.

As for point 2, I don't understand how I should proceed forward. I don't know how to analyze the system if the inclined plane is moving.

 

[kuruman]

Proceed the same way as before except that you have an acceleration on the other side instead of zero.

 

[Thermofox]

So $\begin{cases} N-mg\cos(\theta)=ma_y \\ mg\sin(\theta)-f=ma_x\\ f= N\mu_s \end{cases}$ ?

 

[haruspex]

$f= mgN\mu_s$

That makes no sense dimensionally. What did you mean?

 

[Thermofox]

That makes no sense dimensionally. What did you mean?

I meant $N\mu_s$

 

[haruspex]

So $\begin{cases} N-mg\cos(\theta)=ma_y \\ mg\sin(\theta)-f=ma_x\\ f= N\mu_s \end{cases}$ ?

Yes. And what relates the two accelerations?

 

[Thermofox]

Yes. And what relates the two accelerations?

That they are the components of $a_c$.
$\Rightarrow \begin{cases} a_x= a_c \sin\theta \\ a_y= a_c \cos\theta \end{cases}$

 

[Steve4Physics]

That they are the components of $a_c$.
$\Rightarrow \begin{cases} a_x= a_c \sin\theta \\ a_y= a_c \cos\theta \end{cases}$

Are you sure? Consider a limiting case such as $\theta = 0$ (so that $\sin \theta = 0$) and check this gives the expected behaviour.

 

[Thermofox]

Are you sure? Consider a limiting case such as $\theta = 0$ (so that $\sin \theta = 0$) and check this gives the expected behaviour.

$\begin{cases} a_y= a_c \sin\theta \\ a_x= a_c \cos\theta \end{cases}$ Now the block doesn't fly up in the sky if there is no inclined plane.

But even then, $a_c$ is the acceleration of the whole system. Wouldn't the block have a different acceleration when compared to the inclined plane?

 

[kuruman]

But even then, $a_c$ is the acceleration of the whole system. Wouldn't the block have a different acceleration when compared to the inclined plane?

The problem is asking you to find the maximum value of $a_c$ such that the block and the incline move as one.

I think the algebra is simplified if you rotate the axes with respect to which you write the vectors. Consider a vertical-horizontal set of axes in which the acceleration has zero vertical component. Then the normal force and friction will have vertical and horizontal components but the vertical component of the net force will be zero.

 

[Thermofox]

The problem is asking you to find the maximum value of ac such that the block and the incline move as one.

I think the algebra is simplified if you rotate the axes with respect to which you write the vectors. Consider a vertical-horizontal set of axes in which the acceleration has zero vertical component. Then the normal force and friction will have vertical and horizontal components but the vertical component of the net force will be zero.

Ok. So this should imply that:
$\begin{cases} N_y + f_y= mg \\ N_x - f_x = ma_c \end{cases}$
I'm wondering if, in the second equation, it is fair to use $m$. If the acceleration $a_c$ refers to the plane and the block, shouldn't it be $(M_{plane}+m)ac$?

 

[kuruman]

Ok. So this should imply that:
$\begin{cases} N_y + f_y= mg \\ N_x - f_x = ma_c \end{cases}$
I'm wondering if, in the second equation, it is fair to use $m$. If the acceleration $a_c$ refers to the plane and the block, shouldn't it be $(M_{plane}+m)ac$?

No, because the system is the only the block and you are considering the forces acting on it, not the incline.

Check your equations for correct signs. You are seeking the maximum acceleration $a_c$. How will the block move if the acceleration of the incline exceeds this maximum value?

 

[Thermofox]

No, because the system is the only the block and you are considering the forces acting on it, not the incline.

Ok, now I understand why we use $ma_c$.

Check your equations for correct signs. You are seeking the maximum acceleration $a_c$. How will the block move if the acceleration of the incline exceeds this maximum value?

I checked again, but I cannot find the sign mistakes. I suppose that if the acceleration exceeds the maximum limit, the friction force won't be able to maintain the block still. Thus the block will move to the left.

 

[kuruman]

Please post your free body diagram so that we have something to talk about.

 

[Thermofox]

Please post your free body diagram so that we have something to talk about.

Approximately it should be like this.

 

[erobz]

The frictional force is providing the acceleration of the small block as seen in the inertial frame. I think it will
Be pointing the other way, but you’ll get a negative result.

I’d resolve forces (and acceleration) parallel and normal to the slope, but as long as you are careful that is personal preference.

 

[Thermofox]

I’d resolve forces (and acceleration) parallel and normal to the slope, but as long as you are careful that is personal preference.

$\begin{cases} N-mg\cos(\theta)=ma_c \sin\theta \\ mg\sin(\theta)-f=m a_c \cos\theta\\ f= N\mu_s \end{cases}$ $\Rightarrow N= ma_c\sin\theta + mg\cos\theta$
Therefore, $mg\sin(\theta)-\mu_s ma_c\sin\theta -\mu_s mg\cos\theta=m a_c \cos\theta \Rightarrow mg\sin(\theta) - \mu_s mg\cos\theta = a_c (\mu_s m\sin\theta + m \cos\theta)$ $$\text{Finally}\space a_c= \frac {g\sin(\theta) - \mu_s g\cos\theta} {\mu_s \sin\theta + \cos\theta}$$

 

[haruspex]

I checked again, but I cannot find the sign mistakes. I suppose that if the acceleration exceeds the maximum limit, the friction force won't be able to maintain the block still. Thus the block will move to the left.

If the acceleration is too great, which way will the block move relative to the slope?

 

[Thermofox]

If the acceleration is too great, which way will the block move relative to the slope?

The block will slide down the slope

 

[haruspex]

The block will slide down the slope

Sure?
Consider zero slope to start with. The (flat) "wedge" is accelerating to the left. If the block slips, which way will it move relative to the wedge?

 

[Thermofox]

Sure?
Consider zero slope to start with. The (flat) "wedge" is accelerating to the left. If the block slips, which way will it move relative to the wedge?

Ohhh now I see it. If the slope is zero the block will move to the right, if the slope is $\theta$ the block will move up the slope.

 

[kuruman]

Ohhh now I see it. If the slope is zero the block will move to the right, if the slope is $\theta$ the block will move up the slope.

To complete the picture.
The free body diagram in post #15, as drawn, results in two equation that you have correctly written down. For part 2, we know that the acceleration increases. We also know that the normal force on the block increases as the acceleration increases. The question is what happens to the force of static friction as the acceleration increases? We know that it adjusts itself to provide the observed acceleration but how?

To answer that question, we solve the two equations, for the force of static friction by eliminating the normal force. With $\mu_s=\tan\theta$, we get $$f_s=m(\mu_sg-a_c)\cos\theta.$$The expression above says that

• When $a_c=0$, $f_c=\mu_smg\cos\theta.$ We already knew that from part (1);
• As $a_c$ increases, the magnitude of $f_s$ decreases until it reaches zero when $a_c=\mu_sg.$ This is the value of the incline plane's acceleration at which the block will stay on if the incline were frictionless.
• When $a_c>\mu_sg$, the force of static reverses direction and points down the incline. As the acceleration, increases, $f_s$ also increases but up to a point. After that point, the block will start sliding uphill.

Now you see that the correct FBD for when the block is on the verge of sliding uphill must have the force of static friction pointing down the incline. This is what I meant in post #12 when I asked you to check your signs.

 

[Thermofox]

To complete the picture.
The free body diagram in post #15, as drawn, results in two equation that you have correctly written down. For part 2, we know that the acceleration increases. We also know that the normal force on the block increases as the acceleration increases. The question is what happens to the force of static friction as the acceleration increases? We know that it adjusts itself to provide the observed acceleration but how?

To answer that question, we solve the two equations, for the force of static friction by eliminating the normal force. With $\mu_s=\tan\theta$, we get $$f_s=m(\mu_sg-a_c)\cos\theta.$$The expression above says that

• When $a_c=0$, $f_c=\mu_smg\cos\theta.$ We already knew that from part (1);
• As $a_c$ increases, the magnitude of $f_s$ decreases until it reaches zero when $a_c=\mu_sg.$ This is the value of the incline plane's acceleration at which the block will stay on if the incline were frictionless.
• When $a_c>\mu_sg$, the force of static reverses direction and points down the incline. As the acceleration, increases, $f_s$ also increases but up to a point. After that point, the block will start sliding uphill.

Now you see that the correct FBD for when the block is on the verge of sliding uphill must have the force of static friction pointing down the incline. This is what I meant in post #12 when I asked you to check your signs.

Okay, I get it now. If I want to determine the maximum acceleration $a_c$ then I have to study the extreme case, which is the block on the verge of moving. In this instance, the friction will be directed towards the left.

$\Rightarrow \begin{cases} N_y=mg+f_y \\ N_x + f_x = ma_c \\ f=\mu_s N ?? \end{cases}$
Is the system where $a_c = a_{c,max}$ so that the 2 bodies have the same acceleration.

If I chose to use the axis parallel and orthogonal to the slope would then the system be:$$
\begin{cases}



N-mg\cos(\theta)=ma_c \sin\theta \\



mg\sin(\theta)+f=m a_c \cos\theta\\



f= N\mu_s



\end{cases}
~~~?$$

 

[kuruman]

OK, can you solve for $a_c$?

 

[Thermofox]

OK, can you solve for $a_c$?

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

 

[kuruman]

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

Understanding what you are trying to calculate should guide your thinking whether you can or cannot say that $f=\mu_sN.$ What are you trying to calculate?

 

[Thermofox]

Understanding what you are trying to calculate should guide your thinking whether you can or cannot say that $f=\mu_sN.$ What are you trying to calculate?

I want to calculate $a_{c,max}$. I think that in this case friction should count in the fact that there is an acceleration, after all the friction force changes direction because there is this acceleration. Therefore I feel like that $f= m\cos(\theta)(g\mu_s -a_c)$. The only thing is that I used the system I already have to find this $f$, so if I try and solve the system with this $f$ I won't find anything.

 

[Thermofox]

Understanding what you are trying to calculate should guide your thinking whether you can or cannot say that $f=\mu_sN.$ What are you trying to calculate?

On the other hand, static friction is $\mu_s$ times the force that pushes the stationary mass against the surface where friction is acting, so $N$. I know that $N$ already counts in acceleration so I don't see why in this case $f$ can't equal $N\mu_s$

 

[kuruman]

I want to calculate $a_{c,max}$.

How would you know that $a_c$ has reached its maximum value? Physically, what happens then? What is the criterion?

Therefore I feel like that $f= m\cos(\theta)(g\mu_s -a_c)$.

I derived this expression in post #22 to show how the force of friction changes with acceleration. To do that, I drew the friction as you did, up the incline. I also argued that, as the acceleration increases the force of friction changes direction and starts pointing down the incline. You need to (a) write down a new set of equations based on a redrawn friction down the incline; (b) apply the criterion that says that the acceleration has reached its maximum value and replace $a_c$ with $a_{c,max}~$; (c) solve the system of equations for $a_{c,max}.$

On the other hand, static friction is $\mu_s$ times the force that pushes the stationary mass against the surface where friction is acting, so $N$.

That is incorrect. I will say it one more time, "the force of static friction adjusts itself to provide the observed acceleration, but only up to a certain point." If static friction is not needed it will be zero. If it is needed, it will be what it needs to be. However, it cannot be larger than $\mu_sN.$ That is why one writes $f_s\leq \mu_sN.$ So if you want to substitute $f_s=\mu_sN$, you need to supply a reason why the force of static friction cannot increase any more.

 

[Thermofox]

How would you know that $a_c$ has reached its maximum value? Physically, what happens then? What is the criterion?

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

So if you want to substitute $f_s=\mu_sN$, you need to supply a reason why the force of static friction cannot increase any more.

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

 

[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?