Determining Coefficient of Friction for Two Boxes with Applied Force

[Dave27]

Homework Statement

Suppose there are 2 boxes over a horizontal surface. One has mass M and the other one has mass m. The one with mass m is over M as shown in the picture and a horizontal force is applied to pull M. There is no friction between M and the surface, but there is friction between m and M. The question is how large the coefficient of friction needs to be to occur no slip at all between m and M if the force applied is 100N. Also, let's say M = 5kg, m = 2kg.

Homework Equations

Some free body diagrams:
For m

For M

Acceleration of the system:
F = (M+m)a
a = F / (M+m)
For m:
The force of friction for it not to slip should be at least equal to F so there's no acceleration.
Ff = u * Fn
Ff = umg
Ff = F_m
umg = ma
a here should be the acceleration of the system

The Attempt at a Solution

[/B]
Plugin in numbers:
u (2kg)(9.8m/s^2) = (2kg)(100N/(5kg+2kg))
u 19.6N = 2kg ( 14.3 m/s^2 )
u 19.6N = 28.6N
u = 1.46 ~

 

[TSny]

Hello.

Neither free body diagram is correct. Reconsider them and see if you can spot any errors.

Keep in mind that in a free body diagram of an object, only the forces that act directly on that object should be included.

Also, don't forget the third law. If block m exerts a force on block M, then block M must exert an equal but opposite force on block m (and vice versa).

 

[BvU]

Hello Dave,

I don't agree with the free body diagram for m: there is no force F acting on it !
Your numerical result is correct, however. Somewhat rare to see such high friction coeffcients, but who knows...
Not that the formal answer to the question should not be in the form $\mu = ...$ but $\mu > ...$

 

[kuruman]

Both your FBDs are not correct.

Top: Force F acts on the bottom mass not on both masses. Given that, in what direction is friction and why?
Bottom: There is no normal force exerted by the table. Mass m pushes down on mass M.

Something to consider: After you have drawn the two FBDs correctly, consider adding all the arrows in them to get the combined FBD for both masses together. You should have a single force F to the right, a single force F_n up, no friction, and weight (m+M)g down. That's not what you get when you add the arrows that you posted.

 

[Dave27]

Hello Dave,

I don't agree with the free body diagram for m: there is no force F acting on it !
Your numerical result is correct, however. Somewhat rare to see such high friction coeffcients, but who knows...
Not that the formal answer to the question should not be in the form $\mu = ...$ but $\mu > ...$

FBD has been one of the most frustrating topics for me so far, I always get them wrong. How did you guys get good at them?!

I was thinking about not posting any FBDs, but you know what, I'm glad I did at the end because otherwise I wouldn't have got an opportunity to learn something.

Somehow my answer was correct (luckily). But you saying "mu > ..." shouldn't it be "mu >= ..." in any case? How is it putting that's equal wrong? Do the forces don't cancel each other out if they're equal? This just confuses me.

Both your FBDs are not correct.

Top: Force F acts on the bottom mass not on both masses. Given that, in what direction is friction and why?
Bottom: There is no normal force exerted by the table. Mass m pushes down on mass M.

Something to consider: After you have drawn the two FBDs correctly, consider adding all the arrows in them to get the combined FBD for both masses together. You should have a single force F to the right, a single force F_n up, no friction, and weight (m+M)g down. That's not what you get when you add the arrows that you posted.

I'm confused, why should friction not be to the left? Or maybe you meant I should consider friction to the right for M because m is on top? and what do you mean there's no normal force exerted by the table? How could that be? isn't it the normal force for M which is just the same as Mg ? Ofc they just cancel each other out but still.

I don't see how is it possible to have just a single force to the right and a single force up, unless you consider the FBD of the 2 boxes then I don't see how should that happen.

Hello.

Neither free body diagram is correct. Reconsider them and see if you can spot any errors.

Keep in mind that in a free body diagram of an object, only the forces that act directly on that object should be included.

Also, don't forget the third law. If block m exerts a force on block M, then block M must exert an equal but opposite force on block m (and vice versa).

Well the book never told me how to do FDBs, I guess that's why I'm so bad at them and I got lucky with the result I guess.

Well then, let me try that again.

For the whole system (the 2 boxes) On the horizontal plane there's the Force applied to the right which is 100N. On the vertical there's the force of gravity pointing down which is (M+m)g and the normal force is up which should be the same as the force of gravity because there's no acceleration on the vertical.

For the box M, on the horizontal there's the force applied to the right. On the vertical there's the force of grativity pointing down which is Mg, the force of contact with m which is mg and is also pointing down, and pointing up there's a reaction which should be the sum of both of these forces.

For the box m, on the horizontal there's the force of friction to the left because M accelerates to the right and rubs m. On the vertical there's the force of gravity pointing down which is mg and an equal reaction force pointing up.

Well, Idk why something tells me I'm still wrong. I'll await your response and thanks in advance.

 

[Marc Rindermann]

what you need to consider is that there is only one external force $F$ acting on $M$ to the right. this this force $F$ is accelerating $(M + m)$ to the right at $a = \frac{F}{M + m}$

If there is no slip then $m$ is accelerated with the same $a_m = a$. But what force is it that accelerates $m$ to the right?
What would be the effect on $a_m$ if there was another force acting on $m$ to the left? Could still $a_m = a$?

 

[Dave27]

Yes I was thinking this on the way home I think is like this:

 

[Dave27]

uh? am I still wrong?

 

[BvU]

I think you are doing just fine now.

Just one suggestion: do not use the same symbol (here $F_g$) for different forces if you can avoid it. In this case just writing $mg$ instead of $F_g=mg$ is already enough. That way you also avoid that what in one diagram is called $F_g$, is designated $F_c$ in another

 

[Dave27]

Thank you very much sir!

 

[kuruman]

To enhance the suggestions by @BvU regarding the FBDs in posting #7:

Do not insert equations like F_g = mg. Equalities of this sort will emerge later when you apply Newton's 2nd Law. Just use symbols with action-reaction counterparts showing the same symbols but with arrows in opposite directions. The FBDs are your guide for knowing what to put on the left side of F_net=ma for the constituents of the net force in the vertical and horizontal directions.

In the middle FBD you show an additional force F_c down which is the force exerted by the top mass on the bottom. If that's the action, it has no reaction counterpart in the top FBD, instead you show F_n. This illustrates the point I made above. Despite my criticism, I echo the observation by @BvU that you are doing just fine. It takes a bit of practice to sort things out.

 

[Dave27]

To enhance the suggestions by @BvU regarding the FBDs in posting #7:

Do not insert equations like F_g = mg. Equalities of this sort will emerge later when you apply Newton's 2nd Law. Just use symbols with action-reaction counterparts showing the same symbols but with arrows in opposite directions. The FBDs are your guide for knowing what to put on the left side of F_net=ma for the constituents of the net force in the vertical and horizontal directions.

In the middle FBD you show an additional force F_c down which is the force exerted by the top mass on the bottom. If that's the action, it has no reaction counterpart in the top FBD, instead you show F_n. This illustrates the point I made above. Despite my criticism, I echo the observation by @BvU that you are doing just fine. It takes a bit of practice to sort things out.

What should I put instead of Fn then? In my mind Fn should be just Fc + Mg, and Fc is just mg and that added is the same thing as Fg for the top FBB. Perhaps I'm not labeling the forces properly

 

[BvU]

Good question. I appreciate your asking this: sometimes it's easier to comment on an answer than it is to come up with something better. I thought about it and came with this:

and I have to admit it lacks elegance. And consistence ("Gravity" in the top FBD = "Gravity on m" in the bottom one, so I should have used "Gravity on m" in the top one too ). Open for suggestions for improvement !
But, more important: do you agree with the idea that an FBD should contain a minimum of interpretation ? So: no adding up, but two separate forces.

Having said all that: once you become a bit more experienced it is common to use designations like "mg"

 

[TSny]

Are there two gravitational forces acting on M?

 

[kuruman]

I started answering, but then I revised my answer because of post #13 by @BvU. The FBDs posted are not quite right. The action-reaction counterparts between the masses are not labeled properly. "Normal from M" should be "F from M on m" and "Gravity on m" (bottom drawing) should be "F from m on M".