FBD+Friction question

1. Dec 6, 2015

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 7, 2015

JeremyG

Let's start by getting your free body diagrams right first.

For the FBD of the two masses in contact, remember that if you have a friction force acting on m1 by m2, by Newton's third law (N3L), there should be an equal and opposite friction force acting on m2 by m1. What other forces do you have to consider N3L for?

For your own understanding, try labelling the forces with labels that include the nature of the force (e.g. friction, gravitational etc.) as well as the two bodies involved. An example would be something like what I have above: "Friction force on M1 by M2". Then as long as you have both m1 and m2 in your labels, by N3L, you must make sure that it also appears in the FBD of the other mass.

Show us your corrected FBD and we'll continue with the other parts from there,

3. Dec 7, 2015

Hey JeremyG,

I corrected the FBD with both on it, i'm not sure if this is what it should look like. I thought about N3L when the Fa on m1 from m2, there should be a -Fa m2 from m1, I think it's Fn but I'm not certain.

It looks wrong because I think where Fn2 is I would have an opposite for on

I edited it, I believe this FBD should be the correct one

Last edited: Dec 7, 2015
4. Dec 7, 2015

JeremyG

Hi, your FBD is still slightly incorrect. I will point out your mistakes more clearly this time since you have thought about it for some time. Also note that it is more helpful for yourself and your instructor to draw the FBD separately for m1 and m2. One FBD for one body in a system.

1) Fa does not have a reaction force in this FBD, because Fa is a force BY an external body ON m2. We call this an "external force". The reaction force of this Fa will have the label "force BY m2 ON the external body". But we will not be drawing in this force because in this FBD we are only considering m1 and m2. Every other body is considered an external body and not in our system. Look at the body after the "ON" to figure out which FBD the force is in.
NB. This is why you don't draw in the reaction force for your weight as well.

2) What about Ffr? In my first post, I labelled it as "friction force BY m2 ON m1". So your Ffr on m1 is correct. BUT: Notice that both bodies are inside this label, so its reaction force MUST be inside the FBD of m2 as well! What would be the label of the force then? Use N3L.

3) Do the same inference for FN2 in your diagram. I assume you were drawing it as "Normal contact force BY m2 ON m1", but what about its reaction force?

Do you find that labelling it as above helps you figure out which forces should be in which FBD, as well as which forces should have reaction forces appearing in the corresponding diagram?

Continue the good work! We are making progress!

5. Dec 21, 2015

I found drawing the two separate FBD to clear things up for me.

I'm still sort of confused why N3L doesn't apply in this since FFa on M1 from M2 would cause a reaction on M2 from M1. Is that the reason we have Fn M2onM1?

The friction aspect of the FBD I do understand on M1 and it makes strange sense it should be on M2, but I don't have a clear grasp of why that is. I feel like Ffr M2onM1 is the reaction force but the why for the direction of the force is rather vauge.

P.s Sorry about the delay, I had a rough week with a big energy test and planetary forces.

6. Dec 21, 2015

JeremyG

N3L does apply for Fa, but it is an external force and its origin is from some external body, so we only see the effect on M2. As far as M1 is concerned, Fa is non-existent. So where does the pushing force on M1 come from you might ask? Well, from your FBD of M1, it becomes apparent that this pushing force comes from the Normal contact force of M2 on M1.

Also, the frictional force of M1 on M2 should act downwards, not upwards. N3L tells us that if the frictional force of M2 on M1 is upwards (this upward force keeps M1 from falling), its reaction force on M2 should act downwards, opposite in direction.

7. Dec 21, 2015

Staff: Mentor

8. Dec 23, 2015

thank you Chestermiller for the FBDs, they look different from what i'm accustomed to, but they display everything in such clarity.

would it be correct to state that

assuming
Y is + upwards
X is + to the left

FnetY M1=FFR+(-M1g)=aM1=0
Fnety M2=FN+(-M2g)+(-FFR)=aM2=0

Since we know μ, we can say μ*FFR=FN M1
Since N3L we can conclude that
FN M1=-FN M2
which would be the contact force of M2 on M1?
I'm uncertain if that would be the minimum contact force requires for M1 not to fall, or if there's a different way I should look at it.

9. Dec 23, 2015

Staff: Mentor

Your vertical force balance on M1 is correct if "a" in your equation is the vertical acceleration ay1. Your vertical force balance on M2 is incorrect. It should read: $N_{table}-F_{FR}-M_2g=0$. Can you see that from the free body diagram?

In terms of the normal contact force between the two bodies (of magnitude FN), you can see that FN is drawn with opposite directions in the two free body diagrams, indicating that Newton's 3rd law applies. This is equivalent to your equation FN M1= -FN M2. So, FN in the free body diagram for M1 is the contact force exerted by M2 on M1, and FN in the free body diagram for M2 is the contact force exerted by M1 on M2.

That is not the minimum contact force required for M1 not to fall. Your vertical force balance on M1, although correct as it stands, needs to be developed further, by taking into account the relationship between the normal contact force FN and the vertical friction force FFR for the situation in which M1 is just about to slip (i.e., fall). This relationship involves the coefficient of static friction. What is this relationship? If you substitute this relationship into you vertical force balance for M1, what do you obtain? But, you're not through yet.

The question doesn't ask for the minimum contact force required for M1 not to fall. The question asks you what is the minimum value of the applied force Fa required for M1 not to fall. That will be different than the contact force between M1 and M2. You will not be able to answer any of these questions until you start by writing down a force balance on M1 in the horizontal direction. So let's see what you come up with for that. ( After you get that correct, there will still be more work to be done.)

Chet

Last edited: Dec 23, 2015