Accelerating wedge with a block at rest on it

[rudransh verma]

Homework Statement

A wedge and a block are moving on a smooth horizontal surface such that the block of mass m is at rest with respect to wedge of mass M. The magnitude of applied force P

Relevant Equations

$F=ma$

I have drawn a fbd and the logic I think is that at rest the block moves down the wedge but when a force P is applied vertical force becomes zero and the horizontal force $F_N\sin \beta$ = P?

 

[TSny]

when a force P is applied vertical force becomes zero

What vertical force becomes zero?

and the horizontal force $F_N\sin \beta$ = P?

This is not correct (except in the limit where the mass M of the wedge goes to zero).

Using a free body diagram for the block and Newton's second law, you should be able to derive an expression for $F_N$ in terms of $mg$ and $\beta$, and also an expression for the acceleration of the block in terms of $g$ and $\beta$.

 

[phinds]

Your problem statement, as is often the case, is incomplete because you seem to have not fully thought through what you are presenting. Why do I say this? Because you have not mentioned friction between the block and the wedge.

I ASSUME that you mean it to be zero but you are forcing me to make an assumption and that makes for a poor problem statement.

If the friction is very high then the block can (1) only move UP the wedge, and that only if the force to the right is quite high, or (2) it can move down the wedge if the force is to the left and high enough.

THINK about your problems before you present them.

 

[rudransh verma]

Your problem statement, as is often the case, is incomplete because you seem to have not fully thought through what you are presenting. Why do I say this? Because you have not mentioned friction between the block and the wedge.

I ASSUME that you mean it to be zero but you are forcing me to make an assumption and that makes for a poor problem statement.

I don’t have to. There is no friction in the problem. This is the complete problem given in the book.

 

[rudransh verma]

What vertical force becomes zero?

The vector sum of vertical normal component and mg is zero.

 

[Lnewqban]

I have drawn a fbd and the logic I think is that at rest the block moves down the wedge but when a force P is applied vertical force becomes zero and the horizontal force $F_N\sin \beta$ = P?

The same logic could make you see that the magnitude of force P is important and why.
Just imagine what should happen for each case of extreme values of P: very small and very high.

 

[phinds]

I don’t have to. There is no friction in the problem. This is the complete problem given in the book.

So you are OK with a bad problem statement. You should not be.

It will not serve you well in the long run.

 

[rudransh verma]

This is not correct (except in the limit where the mass M of the wedge goes to zero).

I am not fully grasping the mathematical eqn of what you want to say but the force P will be greater than the $F_N \sin \beta$. You certainly don’t mean $\lim_{m \rightarrow 0} {F_N \sin \beta }=P$ ?

Using a free body diagram for the block and Newton's second law, you should be able to derive an expression for FN in terms of mg and β, and also an expression for the acceleration of the block in terms of g and β.

$F_N \cos \beta=mg$
$F_N \sin \beta=ma$
After solving, $a=g\tan \beta$
$P=Ma=Mg\tan \beta$ ?

 

[Rikudo]

$P=Ma$ ?

No.
Remember, $F_N$ is an action reaction force. As a result, $P$ is not the only force exerted to the wedge.

 

[Steve4Physics]

This is the complete problem given in the book.

1. The problem-statement in Post #1 does not say what you have to find! You haven't actually told us what the question is! Are your trying to find P in terms of M, m and a for example? Or something else?

2. In this particular problem, the block and wedge are moving together (no relative motion). Would they behave any differently if they were glued together (therefore behaving as a single object of mass of M+m)?

3. It’s worth noting that you have not (IMO) drawn suitable free body diagrams. If you need to analyse the forces on the wedge and the forces on the block, you require two free body diagrams - one for the block and one for wedge.

 

[TSny]

I am not fully grasping the mathematical eqn of what you want to say but the force P will be greater than the $F_N \sin \beta$. You certainly don’t mean $\lim_{m \rightarrow 0} {F_N \sin \beta }=P$ ?

If the wedge's mass $M$ is zero, then it turns out that $P$ has the same value as $F_N \sin \beta$. This is not true for the general case where $M$ is not zero. So, your statement $F_N \sin \beta = P$ in your first post is not valid for the general case.

As others have pointed out, the problem statement is incomplete. It does not state whether or not there is friction between the block and the wedge. Also, there is an incomplete sentence: "The magnitude of the applied force P...".

I'm assuming that you are being asked to find $P$ such that the block does not slide on the wedge for the case where there is no friction between the block and the wedge and no friction between the wedge and the horizontal surface.

$F_N \cos \beta=mg$
$F_N \sin \beta=ma$
After solving, $a=g\tan \beta$

Good

$P=Ma=Mg\tan \beta$ ?

No. See post #9 by @Rikudo

 

[rudransh verma]

As others have pointed out, the problem statement is incomplete. It does not state whether or not there is friction between the block and the wedge. Also, there is an incomplete sentence: "The magnitude of the applied force P...".

There is no mention of friction and so I assume it to be zero. If there will be any friction it will be mentioned in the question.

Yes we need to find P. The magnitude of the applied force P is ?

 

[Delta2]

I think you are making a hidden assumption in your mind that the wedge and the block move with the same constant velocity . Under this assumption it is indeed $P=F_N\sin\beta$ as well also when the mass of wedge limits to zero as @TSny already pointed out.

But this assumption is only one special case of the problem. In the general case, the block and wedge move not with the same constant velocity but with the same (constant?) acceleration! When the acceleration happens to be zero then they move with the same constant velocity and then it is $P=F_N\sin\beta$

 

[rudransh verma]

Remember, $F_N$ is an action reaction force. As a result, $P$ is not the only force exerted to the wedge.

$F_N\sin \beta=ma$
$P-F_N\sin \beta=Ma$

$P=ma+Ma=(m+M)a$
I get it. P is the force acting on a body of mass $(m+M)$ due to which it accelerates with acceleration $a=g\tan \beta$. Correct amount of force P is causing the two bodies to behave as one like they are glued together. Not less and not more.

Putting the value of a from post#8, we get
$P=(m+M)g\tan \beta$.

But if M=0 then
$P=\frac{F_N\sin \beta}{\cos ^2\beta}$ not $F_N\sin \beta$?

 

[Delta2]

I think you are correct now but You made a small algebraic blunder in your last equation.

if M=0 then from your fourth equation $$P=mg\tan\beta=F_N\cos\beta\frac{\sin\beta}{\cos\beta}=F_N\sin\beta$$

 

[rudransh verma]

You made a small algebraic blunder in your last equation.

What is wrong? I don’t understand.

 

[Delta2]

Well and I can't understand how you end up with that $\cos^2\beta$ in the denominator.

Isn't it (according to your equations in post #8 $$F_N\cos\beta=mg$$ and of course $$\tan\beta=\frac{\sin\beta}{\cos\beta}$$.

Substitute those in the equation $$P=(m+M)g\tan\beta$$ and you get $$P=F_N\sin\beta$$ after some high school algebra and under the condition M=0.

 

[rudransh verma]

Well and I can't understand how you end up with that cos2⁡β in the denominator.

I understand that you are right but $P=mg\tan \beta$
$=\frac{mg \cos \beta \sin \beta}{\cos ^2 \beta}$
$=\frac{F_N\sin \beta}{\cos ^2 \beta}$
Where am I going wrong?

 

[Delta2]

Look more carefully in your equations, it is (under that M=0) $$P=mg\cos\beta\tan\beta$$.

 

[rudransh verma]

Look more carefully in your equations, it is (under that M=0) $$P=mg\cos\beta\tan\beta$$.

$F_N \cos \beta=mg$
But $F_N=mg \cos \beta$?

 

[Mark44]

So you are OK with a bad problem statement.

There is no mention of friction and so I assume it to be zero.

I believe that a frictionless surface is implicit in the problem statement (emphasis added by me). From the OP:

A wedge and a block are moving on a smooth horizontal surface

 

[TSny]

I understand that you are right but $P=mg\tan \beta$
$=\frac{mg \cos \beta \sin \beta}{\cos ^2 \beta}$
$=\frac{F_N\sin \beta}{\cos ^2 \beta}$
Where am I going wrong?

You made the mistake of assuming that $mg \cos \beta = F_N$. The correct relation is $mg = F_N \cos \beta$ as you had in post #8.

(In many inclined plane problems, we do have $mg \cos \beta = F_N$. But not in this problem.)

 

[Delta2]

Seems I blunder too in post #19, I meant to write $$P=F_N\cos\beta\tan\beta$$.

 

[rudransh verma]

You made the mistake of assuming that mgcos⁡β=FN. The correct relation is mg=FNcos⁡β as you had in post #8.

Oh yes! $F_N$ > $mg\cos \beta$.
Thanks everyone!

 

[Delta2]

Yes the equation $F_N=mg\cos\beta$ holds when the block is sliding down the incline while the incline is immovable . Here the situation is a bit different.

 

[haruspex]

I believe that a frictionless surface is implicit in the problem statement (emphasis added by me). From the OP:

Yes, but what is missing is that there is no friction between block and wedge.

 

[Delta2]

Ehm HELP guys, now that I think of it again, if we have a block and wedge moving with the same constant velocity, then by considering a system of reference with y-axis the direction of gravity, we conclude that $$mg=F_N\cos\beta$$ but if we consider a system with y-axis the direction of normal we conclude that $$F_N=mg\cos\beta$$, what's going on here?

 

[haruspex]

with the same constant velocity,

Says who?

 

[Delta2]

Says who?

sorry I don't understand what's wrong with that statement?

 

[haruspex]

sorry I don't understand what's wrong with that statement?

Nothing in post #1 implies the velocity is constant.

 

[Delta2]

Yes ok I made up my own problem because this thing puzzles me...

 

[rudransh verma]

sorry I don't understand what's wrong with that statement?

I think the wedge can move with constant velocity but how can the block move with constant velocity if the net force on it is not zero.

 

[Delta2]

I think the wedge can move with constant velocity but how can the block move with constant velocity if the net force on it is not zero.

Hm, right but if there is friction between block and wedge this can hold true.

 

[haruspex]

Hm, right but if there is friction between block and wedge this can hold true.

Yes, but without information on the coefficient there is no way to answer the question. That only leaves the option of assuming no friction and a non constant velocity.

 

[jbriggs444]

I believe that a frictionless surface is implicit in the problem statement (emphasis added by me). From the OP:

That says that the ground beneath the block is frictionless but says nothing about the wedge/block interface. Since that interface is not mentioned as "smooth" while the ground interface is, I read the problem to be including friction for the wedge/block interface.

If it were a complete problem, context could make interpretation easier.