Friction Clarification Question

[Hootenanny]

Wow...finally! Thank you sooo much! This summer work for AP Physics is killing me. I'm a good student (I got a 5 on the AP test in Chemistry this past year), but it's kind of weird not to have a teacher...and the textbook is on the computer, so it's kind of a pain.

It was my pleasure, yes I am not fond of 'remote learning' as it were; I feel isolated without someone to discuss my thoughts with. As for summer work ... ... I have vowed to do nothing that resembles work this summer; perhaps ill advised - but only time will tell

Again, it was a pleasure helping you.

 

[kristen151027]

I just found something in that last step of the problem...
If "2m" were substituted into both m's: \mu_{s}mg = ma
then wouldn't it be equal to (μ_s)(g)
?

 

[Doc Al]

Okay so if the maximum force is the frictional force, then the acceleration comes from:

F = ma
\mu_{s}mg = ma
and
m = 2m (for the bigger block)
so the answer is:

(μ_s)(g)/2 ?

I don't understand what you're doing here.

You correctly found the maximum acceleration of the top block. Well, since the bottom and top blocks move together (no slipping, remember?) what must be the acceleration of the bottom block? Hint: This requires NO calculation, only understanding.

Now, if you wanted to find out what force you would have to apply to the bottom block in order to produce the given acceleration--that's a different question. To answer that, apply Newton's 2nd law to the two-block system.

 

[kristen151027]

The acceleration of the bottom block is the same as the top block.

You correctly found the maximum acceleration of the top block.

Where was the maximum acceleration of the top block stated?

 

[Hootenanny]

Ohh, so if you solve for "a", the masses cancel out and the answer is (μ_s)(g).?

You correctly found the maximum acceleration of the top block. Well, since the bottom and top blocks move together (no slipping, remember?) what must be the acceleration of the bottom block?

Of course! I feel ashamed now. My apologies yet again.

 

[kristen151027]

That's okay! So my answer of (μ_s)(g) is correct?

 

[Hootenanny]

That's okay! So my answer of (μ_s)(g) is correct?

Yeah, don't know what I was thinking last night

 

[kristen151027]

Awesome, thanks!