Are Individual Normals Irrelevant in Solving a Pulley and Wedge System?

[Differentiate it]

You won't need to worry about that to solve the problem, but you can figure that out after if you like once you have the proper set of equations.

So what you're telling me is that the 2 normal forces exerted are not equal? Why?

 

[erobz]

So what you're telling me is that the 2 normal forces exerted are not equal? Why?

For two reasons. a portion of the pistons own weight is contributing to the lower normal and its accelerating down the incline, which has an opposite effect, but not necessarily exactly offsetting for all accelerations.

 

[Differentiate it]

For two reasons. a portion of the pistons own weight is contributing to the lower normal and its accelerating down the incline, which has an opposite effect, but not necessarily exactly offsetting for all accelerations.

Oh, so basically, one of the normals(acting up the incline) counters Mgsinθ and the other(acting down the incline) is the one contributing to the acceleration of M. Is that correct?

 

[erobz]

Trust m

Oh, so basically, one of the normals(acting up the incline) counters Mgsinθ and the other(acting down the incline) is the one contributing to the acceleration of M. Is that correct?

The thing is you don't need two Normals to solve this problem, but you must have at least 1 normal. It should be less confusing that way. Just assume the the pistion is just a tiny bit loose ( like a real one would be)

 

[Differentiate it]

Trust m

The thing is you don't need two Normals to solve this problem, but you must have at least 1 normal.

Alright, but last question, could you just confirm what i said in my previous reply?

Oh, so basically, one of the normals(acting up the incline) counters Mgsinθ and the other(acting down the incline) is the one contributing to the acceleration of M. Is that correct?

 

[erobz]

Alright, but last question, could you just confirm what i said in my previous reply?

I don't think that's it. One of the Normals gets a portion of the small masses own weight added to it while the top doesn't.

 

[Differentiate it]

All I can say is under the assumption of two normals ( meaning you forcefully squeezed that pistion in there) that is how the equations make it seem.

Ok alright

 

[erobz]

Ok alright

No see my edit. I've changed my mind since it seems like you are talking about it countering the sleeves weight. I didn't catch that at first

 

[erobz]

How about just forget about the two normals for now and pretend its just a little loose in the slot ( like a real one). Put the normal on the lower side of the slot and solve the resulting equations.

 

[Differentiate it]

Trust m

The thing is you don't need two Normals to solve this problem, but you must have at least 1 normal. It should be less confusing that way. Just assume the the pistion is just a tiny bit loose ( like a real one would be)

Last question:
Could you please elaborate why the normal that is facing downwards is greater?

 

[erobz]

Could you please elaborate why the normal that is facing downwards is greater?

The incline is the reason. Its supporting some of its own weight because of the inclination.

If it were squeezed in there horizontal ( $\theta = 0$ ) the Normal forces ( what ever they may be from the "squeezing" - are equal $N_l = N_u$ ), if you take it and rotate it lettting $\theta = \frac{\pi}{2}$ the upper normal remains $N_u$ and the lower normal would be $N_l+mg$

Anyhow, if this is all overly confusing you don't have to assume two Normals to solve this, but you better have at least 1.

 

[haruspex]

Last question:
Could you please elaborate why the normal that is facing downwards is greater?

Consider the horizontal case. The only forces on the small mass with a horizontal component are those normal forces. If they were equal and opposite then the small mass would not move horizontally, yet the container clearly will accelerate to the right. Hence the left-hand normal force must be the greater.
Now, it is impossible to say exactly what the two normal forces are. If we make the fit a bit snugger then both normal forces increase by the same amount. These increases balance, so do not affect the motion. So you can safely assume that one of the normal forces is zero. Which one? Doesn’t matter. If you pick the wrong one the other will come out negative, but it makes no difference to the algebra and the result.

 

[Orodruin]

The point is that the individual normals are irrelevant to the solution of the problem. They are both a force along the incline between the two masses so only their sum is relevant and you can treat just the sum. It doesn’t matter if the mass m is squeeze fitted or not as long as there is no internal friction.