What is the maximum horizontal force

[pkossak]

Two masses M1=3.80kg and M2=8.10kg are stacked on top of each other as shown in the figure. The static coefficient of friction between M1 and M2 is μs=0.400. There is no friction between M2 and the surface below it. What is the maximum horizontal force that can be applied to M1 without M1 sliding relative to M2?

http://www.msu.edu/~kossakze/physics2.gif

This one I'm not really sure how to start. I thought that if I took M1 (37.278 N) * 0.4, than that would give me the frictional force pushing in the opposite direction, or the max force without M1 sliding. It's not asking whether M2 would slide or not, so shouldn't that have no effect? Fairly perplexed here.

 

[Doc Al]

You are on the right track. The only horizontal force on M2 is the static friction from M1. You know the maximum value of that friction force, so you can now calculate the maximum acceleration of M2. Now use the fact that both blocks move together to determine force F.

 

[pkossak]

Friction force can only be 3.8 kg*9.81 m/s^2*.4 = 14.9112 N, correct? I just don't understand how that is not also maximum amount of horizontal force that can be applied without M1 slipping.

 

[Doc Al]

Friction force can only be 3.8 kg*9.81 m/s^2*.4 = 14.9112 N, correct?

Correct.

I just don't understand how that is not also maximum amount of horizontal force that can be applied without M1 slipping.

That friction force is the maximum available between the blocks. To have the friction force equal that maximum, you have to apply a force F great enough to accelerate the blocks so that the friction needs to be at maximum to prevent slipping. Figure out the acceleration that you must attain, then use Newton's law to figure out F.

If you applied a force F equal to the maximum friction force it would turn out that the actual friction force would be reduced to some sub-maximal value. In other words, that wouldn't be the maximum force you could apply.

 

[pkossak]

yeah that was pretty dumb of me. Thank you! 14.9112, friction force, divided by the coefficient of state friction equals 21.09 N.

 

[Doc Al]

... friction force, divided by the coefficient of state friction equals...

Huh? You didn't get the answer by doing that!

 

[pkossak]

really? It says the answer is 21.9 N. Wouldn't that make sense to be the maximum? I guess maybe not, as that would be the answer without friction, but is it just a coincidence that this is the correct answer?

 

[Doc Al]

I'm not questioning the answer; I'm questioning how you arrived at the answer.

You did not get 21.09 by dividing 14.91 by 0.4!

 

[pkossak]

oh my, your right. I can't believe how much of a coincidence this was. I was using the coefficient of static friction from another problem (.707) on accident, and I got the right answer here!

 

[pkossak]

So that set me back a little bit. Would the answer by chance be

(3.8kg*9.81m/s^2) - (3.8kg*9.81m/s^2*0.4)?

I have a feeling this is wrong too because I didn't use M1 at all. I don't know why I'm having so much trouble here

 

[Doc Al]

Please see my comments in posts #2 & #4. Start by finding the maximum acceleration.

 

[mukundpa]

M1 is not slipping relative to M2 but both M1 and M2 are accelerating with the same acceleration (no relative motion). The friction between the two (backward on M1 and Forward on M2) (a single force cannot be produced, every action has equal and opposite reaction) will try to stop M1 and accelerate M2 in forward direction. If the two blocks are moving together may be considered as a single system and acceleration of M1 and M2 is given by
a = F/(M1+M2). Here friction force's' are internal forces.
Now apply Newton second law for separate blocks to gat friction.

Magnitude of friction force(s) depends on the acceleration of M1+M2 and is max. when acceleration is max. i.e. F is max. After that slipping between them will be there and acceleration of M1 will increase but acceleration of M2 will remain the same.