Block at rest on an inclined surface

[Akash47]

Homework Statement

Homework Equations

Component of force,equilibrium of force.

The Attempt at a Solution

In the problem (ii),the friction force is $Mgcosθμ$,the component of weight in the inclined surface is $Mgsinθ$ and the component of the applied force $F$ along the inclined surface is $Mgcosθ$.Here the component of $F$ is acting opposite to the direction of the component of weight.Then what will be the direction of friction force.Please clear me this without any question.If it would be a plane surface,then there would work only $F$ and the friction force and to remain the block in rest for that case,the friction force had to be greater or equal to $F$.But being the surface inclined,weight is acting here.So how can I relate the terms(i.e $Mgsinθ$ etc.) in this case to get the range of angles $θ$?

 

[BvU]

In the problem (ii),the friction force is $Mg\cos θμ$

How do you calculate that ?

 

[Akash47]

How do you calculate that ?

The friction force $F_f=μN$.

 

[Hen Sharir]

the component of the applied force $F$ along the inclined surface is $Mgcosθ$.

correct me if I'm wrong but I think the component of the applied force along the inclined surface should be $Mgsinθ$ as well (just to the opposite direction).

 

[BvU]

The friction force $F_f=μN$.

What is N ?

 

[Hen Sharir]

What is N ?

N is the normal force. (the force that Earth applies)

 

[RPinPA]

Then what will be the direction of friction force.

Is that your question? If it moves, it will slide down the slope. Therefore the friction acts to oppose that motion. It is a vector pointing up the slope.

The block will stay still if the force pulling it down the slope is not enough to overcome friction.

 

[haruspex]

correct me if I'm wrong

You are.

N is the normal force. (the force that Earth applies)

I assume you mean the normal force from the wedge on the block, as stated in the question.
The force the Earth applies is gravity

 

[haruspex]

Then what will be the direction of friction force.

Friction acts to oppose relative motion of surfaces in contact. Sometimes it is not evident which way that will be. Consider both possibilities.

 

[Akash47]

Is that your question? If it moves, it will slide down the slope. Therefore the friction acts to oppose that motion. It is a vector pointing up the slope.

The block will stay still if the force pulling it down the slope is not enough to overcome friction.

But a horizontal force has been applied in opposite direction to the component of weight which normally pulls down a block(if no other force is applied).So in the normal sense,the friction can work opposite direction of the force $F$.
And I didn't get my full answer by any of your replies cu'z I have wanted to know that how I can relate the terms(mentioned in the question) to form an inequality(so that I can find the range of the angles θ).

 

[BvU]

You haven't got a good grip on the problem yet. Take a closer look at what you think constitutes the normal force and how it comes about.

 

[ZapperZ]

The possible source of confusion here is that there are two "Mg"s, one from the mass of the block, the second from the applied force.f

@Akash47 : Is there a component of the applied force Mg that is perpendicular to the inclined plane, i.e. in the same line as the normal force N? In other words, can you draw and show us the free-body diagram of all the forces perpendicular to the inclined plane?

This is not addressing the entire problem being asked, but it is trying to straighten out the possible source of error of your starting point.

Zz.

 

[Akash47]

The possible source of confusion here is that there are two "Mg"s, one from the mass of the block, the second from the applied force.f

@Akash47 : Is there a component of the applied force Mg that is perpendicular to the inclined plane, i.e. in the same line as the normal force N? In other words, can you draw and show us the free-body diagram of all the forces perpendicular to the inclined plane?

This is not addressing the entire problem being asked, but it is trying to straighten out the possible source of error of your starting point.

Zz.

Thanks for pointing out my error.And now I am not asking the whole solution but just a little bit help and clarification.I am confused that what will be the direction of friction force in this case as there is acting two forces i.e weight and $F$ in opposite direction(please answer this without asking any question).I have read that if the friction force is greater or equal to the opposite force,then the block will remain at rest.But there is acting weight component and $F$ in opposite direction,so how can I relate the terms i.e $Mgcosθ,Mgsinθ$ in an inequality so that I get the angles $θ$ ?

 

[ZapperZ]

Thanks for pointing out my error.And now I am not asking the whole solution but just a little bit help and clarification.I am confused that what will be the direction of friction force in this case as there is acting two forces i.e weight and $F$ in opposite direction(please answer this without asking any question).I have read that if the friction force is greater or equal to the opposite force,then the block will remain at rest.But there is acting weight component and $F$ in opposite direction,so how can I relate the terms i.e $Mgcosθ,Mgsinθ$ in an inequality so that I get the angles $θ$ ?

First of all, you should not be making demands such as "... please answer this without asking any question ... ". We asked questions mainly to either (i) guide you for you to do the problem and/or (ii) if what you stated was unclear or confusing.

Without seeing the free-body diagram, I have no clear idea if you have accounted for ALL the forces acting on the block, and if you've taken into account the right components for each force. Most, if not all, of your problems can be address with such a diagram. So why haven't you spend some time sketching it out, and posting it here? (and yes, that WAS a question!)

Zz.

 

[BvU]

please answer this without asking any question

That will be difficult if there is no question...
You are supposed to find out for yourself about these directions There is a hint in the problem statement: there is apparently a range of $\theta$ for which the block is at rest...

if the friction force is greater or equal to the opposite force

By definition friction is never greater than the opposing force: things don't move in the direction of the friction by themselves

 

[Akash47]

Here is the free body diagram.It is so horrible that I want to say it 'Free soul diagram'(it will take(free) out the soul inside you).

Here (...) means the component of that.But as I don't know the direction of friction force(but know the value),so I couldn't indicate it.And now I can hope that I will get my answers.

 

[ZapperZ]

Here is the free body diagram.It is so horrible that I want to say it 'Free soul diagram'(it will take(free) out the soul inside you).
View attachment 234855
Here (...) means the component of that.But as I don't know the direction of friction force(but know the value),so I couldn't indicate it.And now I can hope that I will get my answers.

See, this is the source of your problem, and we could have easily and clearly addressed this if you had shown this way in the beginning. Your FBD is wrong, and you have not taken into account all the necessary forces.

Here's the first pass of the FBD where I sketch all the actual forces. I decided to simply give them labels here so that we can distinguish what they are. I also do not know the exact direction of the frictional force "f", so I simply make an initial guess. If this direction is wrong, then I'll end up with a negative sign in the end.

Now, you should always draw a second one where you resolve the forces along the directions that you wish to solve it. In this case, I decide to have my cartesian coordinate along the inclined plane. This is my second FBD:

I'm not going to solve this for you. All I want to know if you can tell (i) the difference between this FBD and yours, and (ii) why they are different.

Zz.

 

[BvU]

Do not despair and realize we really want to help you, not just drive you crazy with qusestions upon questions

In your diagram I miss the original forces: $Mg$ from gravity, the external force with the confusing $Mg$, the Normal force and the friction force. All are real forces.

Drat, Zap zapped faster he really wants to help too! Must be a good feeling !

 

[Akash47]

The difference of The FBD of yours and mine is that you indicate both $Fsinθ$ and $Wsinθ$ downwards which I indicate upwards.You have assumed that $f$ and $F$ is in the same direction.Now how can I relate the terms?Is it like that:
$Fcosθ+f ≥ Wsinθ$(but it can lead us that $Fcosθ ≥ Wsinθ$ and then the block will move towards the $F$ which breaks the condition).
That's what I tried to tell you from the beginning.Then how will I set up an inequality to find the range?
Please don't make the argument further and directly answer my question(I am not asking full solution).I am an olympiad student and so you can understand how important time is to me.

 

[BvU]

The difference of The FBD of yours and mine is that you indicate both Fsinθ and Wsinθ downwards which I indicate upwards.You have assumed that f and F is in the same direction

Not so. Your components do not add up to W and F, you miss N and the friction.

Please don't make the argument further

This is not an argument, it is fundamental error on your part.

I am an olympiad student and so you can understand how important time is to me.

So you do not want to waste it. Calm down and re-read the help you've received so far. When you are ready, we can continue.

 

[Akash47]

Calm down and re-read the help you've received so far.

You are returning me the same thing.There's hardly time for me to calm down!And I think,if you want,you can save my time easily by answering the questions or pointing out the errors without making other discussions.

 

[ZapperZ]

The difference of The FBD of yours and mine is that you indicate both $Fsinθ$ and $Wsinθ$ downwards which I indicate upwards.You have assumed that $f$ and $F$ is in the same direction.Now how can I relate the terms?Is it like that:
$Fcosθ+f ≥ Wsinθ$(but it can lead us that $Fcosθ ≥ Wsinθ$ and then the block will move towards the $F$ which breaks the condition).
That's what I tried to tell you from the beginning.Then how will I set up an inequality to find the range?
Please don't make the argument further and directly answer my question(I am not asking full solution).I am an olympiad student and so you can understand how important time is to me.

1. Why would F sin θ be in the +y direction? Look at the direction of the axes! If it helps, tilt your head to the left! Is there any component of F in the +y direction?

2. I only have to assume the direction of f, which I had stated clearly in my post. I do not have to assume the direction of F cos θ. It is GIVEN in the problem.

3. I don't understand the inequality. The problem clearly stated that the mass is not moving. This means that this is a static problem! The sum of all the components of the forces is zero!

I have a horrible feeling that I may have to also write down the sum of the forces here because you don't seem to understand what to do for a problem like this.

Zz.

 

[ZapperZ]

BTW, I want you to look AGAIN at your FBD and tell me if you do not see something obviously wrong with the y-component of your diagram. There is only ONE force acting in that direction, meaning that there's will be a net acceleration in that direction. Your block, according to your FBD, is accelerating perpendicular to the inclined plane!

Does this make sense to you?

Zz.

 

[Akash47]

I have just show the upward forces of weight and $F$ but forgot to show the downward same forces.And there is component of $F$ in the +y as well as in the -y direction which crosses each other.But you are saying that sum of all the forces along X axis(as well as in Y axis) will be zero and that ends up in a equation and giving a single value of θ which is $θ=tan^{-1}\frac {1+μ}{1-μ}$.The questions asks for the range of angles(that means there are infinitely many angles).So should I make the use of: 0< θ<90 for showing the range?
Are you feeling horrible now?

 

[ZapperZ]

I have just show the upward forces of weight and $F$ but forgot to show the downward same forces.

Then your FBD is incomplete, and it makes it wrong. It means that if this is your starting point, everything that comes from here will also be wrong.

And there is component of $F$ in the +y as well as in the -y direction which crosses each other.

This makes zero sense, and it is wrong. F only has a -y direction. You showed nothing in the -y direction in your FBD. If you think that there is a +y direction, please show a sketch with JUST F, and prove what you just said.

But you are saying that sum of all the forces along X axis(as well as in Y axis) will be zero and that ends up in a equation and giving a single value of θ which is $θ=tan^{-1}\frac {1+μ}{1-μ}$.The questions asks for the range of angles(that means there are infinitely many angles).So should I make the use of: 0< θ<90 for showing the range?
Are you feeling horrible now?

But I never said that this is the END of the solution! I've been trying to do this ONE STEP AT A TIME, since you have issues with almost every step of the problem! I've been trying to get you to do one correct thing after another. You still can't seem to get the FBD right, and you're already thinking about the final answer?

But you are right. I feel "horrible" right now. So horrible, in fact, that I'm going to bow out of this thread and consider what I've done to be a total waste of my time.

Zz.

 

[BvU]

Are you feeling horrible now?

No because if I did I might be inclined to spoil it for you by putting down the answer. So there's still hope.

However, I now wonder what you did in part (i) of the exercise ... there you need N and F_f too

 

[haruspex]

You have assumed that f and F is in the same direction.

Not sure whether ZapperZ was assuming that. As I wrote, both possibilities should be considered. It is OK to show a particular direction in the diagram if we bear in mind that the sign of f is uncertain. But if you are uncomfortable with that, draw both diagrams.

but it can lead us that Fcosθ≥Wsinθ

Not sure what you mean by that.

then the block will move towards the F which breaks the condition).

And indeed in general that is possible. If we ignore the given fact that F=Mg, Fand μ could be small while θ is large. Whether it is possible given that F=Mg needs to be determined.

 

[jbriggs444]

I have just show the upward forces of weight and FFF but forgot to show the downward same forces

In a free body diagram one draws the forces acting on one body only (and clearly identifies that body). If you consider that weight has both an upward and a downward component than it would seem likely that you are invoking Newton's third law and including both third-law partner forces, even though they do not act on the same object. That would be erroneous.

Edit: Alternately, it could suggest that you are pre-invoking Newton's second law before writing down a force balance equation. i.e. trying to find a second-law counter-force for each individual applied force. That defeats the purpose of a free body diagram. The key purpose being to help you build the force balance equations.

 

[haruspex]

you are saying that sum of all the forces along X axis(as well as in Y axis) will be zero and that ends up in a equation and giving a single value of θ

No, it does not lead to that. To get that, you would be assuming that F_{static friction}=μ_sN, but that there is an error in that equation. What should the equation be?

 

[Akash47]

Maybe the error in the equation is that I should write $F_{static,max}=μF_n$ instead of $F_{static}=μF_n$.
As you said that both the direction of friction should be considered,then I could use the fact like that:
First, assume that $F_s$(friction) is in the direction of the applied force $F$(when $Fcosθ>Wsinθ$).Then considering the fact that block is not moving,we get$:$
$F_{s,max} ≥ Fcosθ-Wsinθ$
Secondly,$F_{s,max} ≥ Wsinθ-Fcosθ$ (when $Wsinθ>Fcosθ$)
And of course,there is still a case when $Fcosθ=Wsinθ$ and in this case, the solution is: θ=45.
I hope,considering these cases end up the solution.

 

[haruspex]

Yes, that is all correct, except I do not understand this line:

there is still a case when Fcosθ=Wsinθ

In what sense is this a special case? Isn't it just encompassed by the range of values for which the system is static?

 

[Akash47]

Oh! It's not a special case.When $Fcosθ=Wsinθ$,then there is no friction force working.How will you explain it?

 

[haruspex]

Oh! It's not a special case.When $Fcosθ=Wsinθ$,then there is no friction force working.How will you explain it?

What's to explain? If the net of the other forces tending to make the surfaces slide against each other is zero then there is no frictional force.

 

[Akash47]

I was saying that there is a case $θ=45^°$ when also the block will not move and there will be no friction.And the range of the values of θ is$:tan^{-1}\frac{1-μ}{1+μ} ≤θ≤ tan^{-1}\frac{1+μ}{1-μ}$.And to get the value θ=45,you have to put μ=0 in either the lower bound or the upper bound.But can a static friction co-efficient of a surface ever
be zero(at least not in this problem where μ is clearly mentioned)?Then what is your answer to this?
Though the question asks for the range of values in terms of μ (which don't really need the value θ=45),I think the full solution also need the value $θ=45^°$.

 

[jbriggs444]

I was saying that there is a case θ=45 when also the block will not move and there will be no friction.And the range of the values of θ is$: \frac{1-μ}{1+μ} ≤θ≤ \frac{1+μ}{1-μ}$.And to get the value θ=45,you have to put μ=0 in either the lower bound or the upper bound.But can a static friction co-efficient of a surface ever
be zero(at least not in this problem where μ is clearly mentioned)?Then what is your answer to this?
Though the question asks for the range of values in terms of μ (which don't really need the value θ=45),I think the full solution also need the value θ=45.

It is clear that $\theta$ = 45 degrees falls within the solution set regardless of the value of $\mu$

You are going to need to write down some equations so that we can see where you are going wrong. It is pretty clear that those formulas for $\theta$ are not correct. For one thing, they give a result that is unitless, but you are using degrees, so $\theta$ should not be unitless.