Friction Clarification Question

[kristen151027]

I think I'm just getting thrown off by this question because there's unnecessary information in it, but here it is:

"A 1.0 kg brick is pushed against a vertical wall by a horizontal force of 24 N. If μs = 0.80 and μk = 0.70 what is the acceleration of the brick?"

Is it just 0.0 m/s^2 because it's pushed up against the wall, or does it require some sort of calculation?

 

[Hootenanny]

I think I'm just getting thrown off by this question because there's unnecessary information in it, but here it is:

"A 1.0 kg brick is pushed against a vertical wall by a horizontal force of 24 N. If μs = 0.80 and μk = 0.70 what is the acceleration of the brick?"

Is it just 0.0 m/s^2 because it's pushed up against the wall, or does it require some sort of calculation?

I would agree with you there, in my opinion the brick is stationary and not accelerating since the vertical wall will provide a normal force equal in magnitude but opposite in direction to the applied horizontal force leading to a net force of zero. The majority of the information is superfluous.

 

[Doc Al]

Is it just 0.0 m/s^2 because it's pushed up against the wall, or does it require some sort of calculation?

They are asking about the vertical acceleration, not the horizontal. So you need a little calculation.

(The information given is not superfluous.)

 

[Hootenanny]

They are asking about the vertical acceleration, not the horizontal. So you need a little calculation.

(The information given is not superfluous.)

Ahhh! The block is held against the wall.. makes sense. In my mind I had the brick resting on the floor and being pushed against the wall. My bad

 

[kristen151027]

Thanks (again)! While I'm here, I have another more difficult (at least for me) problem which I'm having trouble phrasing my exact question to:

"A block of mass m sits on top of a larger block of mass 2m, which in turn sits on a flat, frictionless table. The coefficient of static friction between the two blocks is μ_s.What is the largest possible horizontal acceleration you can give the bottom block without the top block slipping?"
(a) μ_sg/2
(b) μ_sg
(c) 2μ_sg
---
I'm having trouble combining Newton's second law and fs,max = μ_sF_N

 

[nrqed]

I think I'm just getting thrown off by this question because there's unnecessary information in it, but here it is:

"A 1.0 kg brick is pushed against a vertical wall by a horizontal force of 24 N. If μs = 0.80 and μk = 0.70 what is the acceleration of the brick?"

Is it just 0.0 m/s^2 because it's pushed up against the wall, or does it require some sort of calculation?

You can't tell without doing a calculation. The brick *might* be at rest or it *might* be sliding down with some acceleration.

Drwa a free-body diagram (*always* the first step in analyzing a problem like this). What you have to check is whether the static friction force can be strong enough to overcome the weight or not.

 

[Hootenanny]

Thanks (again)! While I'm here, I have another more difficult (at least for me) problem which I'm having trouble phrasing my exact question to:

"A block of mass m sits on top of a larger block of mass 2m, which in turn sits on a flat, frictionless table. The coefficient of static friction between the two blocks is μ_s.What is the largest possible horizontal acceleration you can give the bottom block without the top block slipping?"
(a) μ_sg/2
(b) μ_sg
(c) 2μ_sg
---
I'm having trouble combining Newton's second law and fs,max = μ_sF_N

What happens if a horizontal force greater than the maximum frictional force (between the two blocks) to the larger block?

Take note of Doc Al's comments above

 

[nrqed]

Thanks (again)! While I'm here, I have another more difficult (at least for me) problem which I'm having trouble phrasing my exact question to:

"A block of mass m sits on top of a larger block of mass 2m, which in turn sits on a flat, frictionless table. The coefficient of static friction between the two blocks is μ_s.What is the largest possible horizontal acceleration you can give the bottom block without the top block slipping?"
(a) μ_sg/2
(b) μ_sg
(c) 2μ_sg
---
I'm having trouble combining Newton's second law and fs,max = μ_sF_N

Again, draw a free body diagram of the top mass (this is always the first step in such a problem. Without a clear FBD in you rmind, things will never be completely clear. and often, just drawing a FBD will make you realize how to solve a problem). Now, what is the *maximum* static friction force that the top mass may experience? Using this, what is the maximum acceleration it may have without slipping?

 

[Hootenanny]

I'm gona duck out of this thread save all these double posts.

 

[kristen151027]

For my first question:

Drwa a free-body diagram (*always* the first step in analyzing a problem like this). What you have to check is whether the static friction force can be strong enough to overcome the weight or not.

So if the brick's mass is 1.0 kg, then it's weight is 9.8 N
The horizontal force is 24 N...and that's where I lose it

 

[kristen151027]

Ohh...so using F_N as 24 N and multiplying it by μ_s (0.80) gives fs,max to be 19.2, which is enough to overcome the weight of 9.8 N.

Right?

 

[Hootenanny]

Ohh...so using F_N as 24 N and multiplying it by μ_s (0.80) gives fs,max to be 19.2, which is enough to overcome the weight of 9.8 N.

Right?

Okay I'm back in again . Yes that looks good to me, therefore the acceleration of the brick is ...

 

[kristen151027]

It's 0.0 m/s^2 because it's not moving and its velocity is constant (0.0 m/s)

Now, on to my second question...give me a minute

 

[kristen151027]

Again, draw a free body diagram of the top mass (this is always the first step in such a problem. Without a clear FBD in you rmind, things will never be completely clear. and often, just drawing a FBD will make you realize how to solve a problem). Now, what is the *maximum* static friction force that the top mass may experience? Using this, what is the maximum acceleration it may have without slipping?

Maxiumum static friction force:
Okay, so in the FBD, the forces acting upon the top mass are: mg from the top and F_N from the bottom. F_N = mg, so:

fs,max = μ_sF_N = (μ_s)(mg)

...right so far?

 

[Hootenanny]

Maxiumum static friction force:
Okay, so in the FBD, the forces acting upon the top mass are: mg from the top and F_N from the bottom. F_N = mg, so:

fs,max = μ_sF_N = (μ_s)(mg)

...right so far?

Looks good to me

 

[kristen151027]

Okay, so for the maximum acceleration it can have without slipping...
I honestly have no clue

 

[Hootenanny]

HINT: $F = ma \Rightarrow \mu_{s}mg = ma$

 

[kristen151027]

HINT: $F = ma \Rightarrow \mu_{s}mg = ma$

Ohh, so if you solve for "a", the masses cancel out and the answer is (μ_s)(g). So nowhere does it take into account the differences in masses of the two blocks?

 

[Hootenanny]

Well if $\mu_{s}mg$ is the maximum force that the bottom block (2m) can apply to the top block (m) via friction without it slipping. What is the maximum force one could apply to the bottom box without the top box slipping?

 

[kristen151027]

Twice that, because it's twice as heavy? Honestly, I'm lost I really appreciate your persistent help, though!

 

[Hootenanny]

Twice that, because it's twice as heavy? Honestly, I'm lost I really appreciate your help, though!

Okay, for the smaller block not to slip it must remain at rest relative to the bigger block (obvious yes?). Therefore, the force on either block must not exceed the maximum static frictional force between the two blocks. Do you follow?

 

[kristen151027]

Yeah, I understand that.

 

[Hootenanny]

Yeah, I understand that.

So what is the maximum possible force that can be applied to the bigger block so that neither block slips relative to each other?
HINT:

Therefore, the force on either block must not exceed the maximum static frictional force between the two blocks.

 

[kristen151027]

Do you have AIM or MSN Messenger? I think that would be easier.
My screenames are:
AIM: kristen151027
MSN: klogan15@hotmail.com
I'm on both now.

 

[kristen151027]

The maximum force is the frictional force then??

 

[Hootenanny]

Do you have AIM or MSN Messenger? I think that would be easier.
My screenames are:
AIM: kristen151027
MSN: klogan15@hotmail.com
I'm on both now.

I do, however, I am not comfortable revealing my personal e-mail address; it's nothing personal just a confidentiality thing.

 

[Hootenanny]

The maximum force is the frictional force then??

You've got it

 

[kristen151027]

I do, however, I am not comfortable revealing my personal e-mail address; it's nothing personal just a confidentiality thing.

Totally understand

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 ?

 

[Hootenanny]

Totally understand

Thank you

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 ?

Looks good to me

 

[kristen151027]

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.

 

[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.