Question with masses, friction, and other forces

[mystmyst]

[PLAIN]http://img228.imageshack.us/img228/6816/mechanics1.png

Homework Statement

Horizontal force, which has magnitude F, acts on a system of masses as shown. Both masses are attached to a wheel and rope, both without mass. There is no friction between the wagon and the ground. There is static/kinetic friction between the wagon and the mass above it.

\mu_k < \mu_s

'M' > 'm'

Question: If it's known both masses are moving at the same acceleration, what is the acceleration?

The Attempt at a Solution

Please correct me if I'm mistaken.

Since both bodies are moving with the same acceleration, the static friction, f, between the bodies must be greater than or equal to F.

\vec{F} = M \vec{a} --------- Newton's law...

F - f = (m+'M')a ------ Substituting forces and masses

f = \mu N_m ---- formula

N_m = mg ------ as seen from image

f = \mu mg ------- replace N with mg

F - \mu mg = (m+'M')a ----- replacing f

a = \frac{F - \mu mg}{m+'M'}

And that's my solution. Is there a way to replace F with something else? Is this the correct answer?

Thanks!

 

[mystmyst]

now that I think about it, I think the friction force is in the right direction and not left. Since if there is a force pushing the wagon to the right, the other mass will want to fall off to the left, but the friction force in the other direction is holding it. So I should change the sign from minus to plus.

What do you guys think? agree/disagree?

 

[PhanthomJay]

Neither solution is correct. Since the acceleration of both masses are the same, then you can draw a free body diagram of both masses together. When you do this, all forces in the horizontal direction are internal, except the applied force. Internal forces do not show up in your free body diagram when you isolate the entire system this way.

 

[mystmyst]

Neither solution is correct. Since the acceleration of both masses are the same, then you can draw a free body diagram of both masses together.

Do you mean that if they have the same acceleration it's as if it is one large body instead of two small ones?

What are internal and external forces and when do they and don't they appear in free body diagrams?

Thanks for your help (and patience...)

 

[PhanthomJay]

Do you mean that if they have the same acceleration it's as if it is one large body instead of two small ones?

yes, that is correct.

What are internal and external forces and when do they and don't they appear in free body diagrams?

Thanks for your help (and patience...)

External forces are forces, outside the system you are looking at , which are applied to the system; internal forces are forces within the system, that cancel out due to Newton's 3rd law of action-reaction pairs, and which cannot be found until you break apart the system into free body diagrams of parts of that system, in which case the internal forces become external to your new system.

Now that sounds about as confusing as a Wikipedia definition, so let me try to simplify. When you look at the full system of the wagon, top mass, cable, and pulley, together all as one system, the only forces external to the system, in the vertical direction, are the weight of the system acting downward (the gravity force, or the force of the Earth on the system), and the normal force acting up (the force of the ground on the system); and the only forces external to the system , in the horizontal direction, is the applied force, F (if there was friction between the wagon and ground, that would be an external force also, but there is none in this example). The internal forces...the friction between the wagon and mass, the tension force in the cable, the normal force between the 2 blocks..., do not show up in this diagram. So to find the acceleration in the horizontal direction, use Newton 2 , where the sum of the external horizontal forces (the net external force) is equal to the system's mass time its acceleration in the horizontal direction.

Now suppose you wanted to find the friction force between the wagon and mass. You could draw a free body diagram of the top mass, and identify the forces acting on that new system. In the vertical direction, this would be the weight of the mass, and the normal force of the wagon on the mass, and in the horizontal direction, the forces acting would be the friction force of the wagon on the mass, and the tension force in the cable attached to that mass. No other forces show up in that diagram. You might want to google on 'free body diagrams' if this is not clear.

 

[mystmyst]

Thanks again so much. I'm understanding this much better. (Now I understand I totally forgot about the tension T)

But I still have a few questions:

1) People here have mentioned that those internal forces get canceled out. How do they get canceled out? Clearly they don't get canceled out if we use them in the free body diagrams. (I think the answer has to do with the bodies with different acceleration)

2) It says that F = 2T. I'm assuming they came to this conclusion by looking at the forces acting on the pulley. But the thing I don't get is that it says the bodies are moving at the same acceleration rate. Shouldn't that mean that F > 2T? As in, I thought F=2T only when the wagon isn't moving.

Thanks again!

 

[PhanthomJay]

Thanks again so much. I'm understanding this much better. (Now I understand I totally forgot about the tension T)

But I still have a few questions:

1) People here have mentioned that those internal forces get canceled out. How do they get canceled out? Clearly they don't get canceled out if we use them in the free body diagrams. (I think the answer has to do with the bodies with different acceleration)

When you look at the 2 blocks together as a system, those internal forces are still there, but they exist in force pairs, equal and opposite, which, as you can see when you 'operate' on the system and look at each block separately, act on different parts of the system. They don't show up in the free body of the 2 masses together, because they act opposite each other within that system. If the acceleration of the 2 masses are different, then when you look at the free body of the smaller block, the max kinetic friction available (u_kN) , is the friction force, and solve for the small block acceleration accordingly. (Note that you could still look at the 2 blocks together as a system, but when you solve for the acceleration of the system, you get the acceleration of the center of mass of the system, but don't worry about that for now, it is a bit confusing, but I thought I'd point it out).

2) It says that F = 2T. I'm assuming they came to this conclusion by looking at the forces acting on the pulley. But the thing I don't get is that it says the bodies are moving at the same acceleration rate. Shouldn't that mean that F > 2T? As in, I thought F=2T only when the wagon isn't moving.

Thanks again!

Good point, but the problem says the pulley has no mass, so since F -2T = ma, m of the pulley is 0, so F =2T. This would not be the case if the pulley had mass.

 

[mystmyst]

When you look at the 2 blocks together as a system, those internal forces are still there, but they exist in force pairs, equal and opposite, which, as you can see when you 'operate' on the system and look at each block separately, act on different parts of the system. They don't show up in the free body of the 2 masses together, because they act opposite each other within that system. If the acceleration of the 2 masses are different, then when you look at the free body of the smaller block, the max kinetic friction available (u_kN) , is the friction force, and solve for the small block acceleration accordingly. (Note that you could still look at the 2 blocks together as a system, but when you solve for the acceleration of the system, you get the acceleration of the center of mass of the system, but don't worry about that for now, it is a bit confusing, but I thought I'd point it out).

Ok, thanks.

Good point, but the problem says the pulley has no mass, so since F -2T = ma, m of the pulley is 0, so F =2T. This would not be the case if the pulley had mass.

Thank you so much! Now it makes so much sense!

 

[mystmyst]

So question:

Is the tension T acting on both masses facing the same direction to the right?

Also, when we look at the pulley, we see force F to the right and two tension forces to the left. We can conclude that F=2T since it's given m=0, right? (F-2T=ma, m=0...)

And lastly, is it correct to say the only external force acting on the masses horizontally is the tension from the rope? (maybe tension is internal, I have no idea, but the point I'm trying to make is that it's the only force acting on the masses horizontally) I originally thought it was F but now I think F has nothing to do with the masses. F acts on the pulley which acts on the rope attached to the masses.

 

[PhanthomJay]

So question:

Is the tension T acting on both masses facing the same direction to the right?

most definitely, tension forces always pull away from the objects on which they act

Also, when we look at the pulley, we see force F to the right and two tension forces to the left. We can conclude that F=2T since it's given m=0, right? (F-2T=ma, m=0...)

Yes , as long as the pulley is frictionless as well. For massless and frictionless pulleys, the tension is always the same in the cord that is wrapped around the pulley.

And lastly, is it correct to say the only external force acting on the masses horizontally is the tension from the rope? (maybe tension is internal, I have no idea, but the point I'm trying to make is that it's the only force acting on the masses horizontally) I originally thought it was F but now I think F has nothing to do with the masses. F acts on the pulley which acts on the rope attached to the masses.

Yes, I believe what you are saying is correct, but to be sure, what you have done here is to isolate the 2 masses from the pulley, in which case, the tension forces that were internal to the pulley-and-2 masses system, become external forces when you isolate the 2 masses as you have done. In which case, yes, 2T is the only horizontal force acting. If you were to look instead at the full pulley-and-2 masses system, then F is the only horizontal force acting. You get the same results. The tougher part in this problem would be to determine the magnitude and direction of the friction force between the 2 masses, which you would have to find by isolating one of the masses separate from the rest of the system.
You've done nice work so far in trying to get a handle on this!

 

[mystmyst]

If you were to look instead at the full pulley-and-2 masses system, then F is the only horizontal force acting.

Why is the tension not a horizontal force also? I would think:

F-2T=(m+M)a

Hm, but I guess that can't be right since F=2T...I'm gettin confused again ):

You've done nice work so far in trying to get a handle on this!

Thanks! And you've done an amazing job explaining and being patient!

 

[mystmyst]

Actually, since the tension force is in the same direction force F, it would be

F+2T=(m+M)a

and that makes a bit more sense but I know 2T isn't supposed to be there...
just a bit more explanation till I fully understand this q.

 

[PhanthomJay]

Actually, since the tension force is in the same direction force F, it would be

F+2T=(m+M)a

and that makes a bit more sense but I know 2T isn't supposed to be there...
just a bit more explanation till I fully understand this q.

Let's get right to the point:

When you look at the entire system , as shown in your sketch, the only external horizontal force acting is F. Everything else in the horizontal direction...friction force between mass and wagon, and cord tension forces ... are internal forces, and do not show up in your diagram. Thus, applying Newton's 2nd law,

F_net = F = (m + M)a

and with F and m and M as given quantities, then a = F/(m + M).

This is the acceleration of the wagon/mass system, and also the accelerationof the mass or wagon alone, since it is given that they all move together.

Now to find the tension force in the cables, draw a free body diagram of the pulley, and you get F =2T, as discussed, thus T = F/2.

Now draw a free body diagram of the mass and wagon together, isolating them from the pulley. Now your internal Tension forces become external to this new system, and thus, again per Newton 2

F_net = (m + M)(a)
2T = (m + M)(a), or
a = 2T/(m + M)

now since T =F/2, then
a= {2(F/2)}/(m + M)
a = F/(m + M)

which is a check on your first result...the same.

If you wanted to find the friction acting between the mass and wagon, then isolate the top mass in a free body diagram...the friction and tension forces become external to that mass, and per Newton 2

F_Net = ma
T + f_friction = ma

now you can solve for the friction force, substituting as req'd.

Now to check this result ., isolate the wagon only:

F_net = Ma
T - f_friction= Ma
and when you solve for the friction force with appropriate substitutions, the result will be the same.


Note that you do not know the direction of the friction force, but you do know that in whatever direction it acts on the mass, it will act in the opposite direction on the wagon.

 

[mystmyst]

Let's get right to the point:

When you look at the entire system , as shown in your sketch, the only external horizontal force acting is F. Everything else in the horizontal direction...friction force between mass and wagon, and cord tension forces ... are internal forces, and do not show up in your diagram.

Ah, that's great! I think that clears up everything.

So to be sure I understand,

F=ma is in situations where F's are external forces. And when you isolate masses internal forces become external.

 

[PhanthomJay]

Ah, that's great! I think that clears up everything.

So to be sure I understand,

F=ma is in situations where F's are external forces. And when you isolate masses internal forces become external.

Remember it's F_net = ma, that is, the net force, (or sum of all forces) = mass x acceleration. And yes, you have the correct understanding of the external/internal forces, and how they are applied to free bodies, nice job!