Determine the friction force in a stacked block scenario

[jna]

Homework Statement

Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. What is the friction force of block A on block B if the block B has a mass of 24 kg and is accelerating at 0.50 m/s2 to the right relative to the block A?

Homework Equations

see (3)

The Attempt at a Solution

Since the force on block A is 75N and its mass is 30kg, the acceleration it moves with to the left is:
a_A = (75N)/(30kg)
If there was no friction, block B would remain still relative to the background, or, would appear to accelerate at 2.5 m/s^2 to the right with relative to block A. With friction, this acceleration is reduced to 0.5 m/s^2. The friction force can be written as:
F_f = m_B*a_B = (24kg)*((2.5-5)m/s^2)=48 N
The friction force on block B is 48 N and accounts for an acceleration of block B of 2.0 m/s^2 to the left.

The solution above (48N) does not match any of my multiple choices. Could you please check and explain where the attempt made a mistake?

 

[Dewgale]

You forgot to take all the forces into account. Think about this -- given you've got an interaction between two objects, is there any place you may have accidentally broken Newton's 3rd law?

 

[jna]

Hmm, yes I am pretty sure I broke something. So a more careful but still blurry look: Block A exerts force through friction on block B in the direction <--. Obviously, that force exceeded the opposing max friction force of B on A pointing -->, since block B is sliding and accelerating 0.5 m/s^2 -->. Had the max friction force not been exceeded, the force transferred to block B would be 75N, but since we know B is accelerating with 0.5 m/s^2--> relative to A, that friction force is reduced, namely 75N - 24kg*0.5m/s^2 = 63N.
Is that the way to go?

 

[Dewgale]

If the frictional force is the only contact force between the two blocks, is it possible for the frictional force on each block to have different magnitudes?

 

[jna]

I'd say yes.

 

[Dewgale]

I'd say yes.

No, that's a violation of Newton's 3rd law.

The net sum of any force over an entire system must be zero. It's like how in deep space, if I throw something, I shoot off in the other direction. If I apply a force to something, it applies an equal and opposite force back on me. So, if block B applies a friction force $f_F$ to block A, block A must apply an equal and opposite frictional force $-f_F$ back onto block B.

 

[jna]

Ah sorry i actually meant to say the same (misread your question). But can you please help me correct my answer?

 

[Dewgale]

I can't just give you the answer, you need to work through it yourself. I can give you some hints though.

1. There should be no forces in the horizontal direction on block B apart from the frictional force.
   
2. Make sure you read the context around that $0.5 m/s^2$ carefully. It's not quite as clear-cut as it looks.

 

[jna]

It would be helpful if you could point out where I went wrong with the 63N reasoning. I am just unsure what to do with the two hints.

 

[Dewgale]

Well, one issue is that the frictional force is to the left, yet your argument is based around block B accelerating to the right. Clearly this doesn't work out very well! You can also immediately see that it isn't the right answer numerically, since if $f_F = 63 N$, then as it's the only force, it would cause an acceleration of $2.625 m/s^2$ to the left, when you assumed it was $0.5 m/s^2$.

Try solving this problem from the perspective of an observer at rest relative to the ground, who sees block A accelerating with some acceleration $a$, and block B accelerating with some other acceleration, still to the left, but less than $a$.

 

[jna]

I see. Re the first part, fF=63N causing an acceleration of 2.625 m/s^2 was still fine to me because I only assumed a relative reduction of acceleration of B by 0.5 m/s^2 (the 0.5 m/s^2 is specified as "to the right relative to block A). I do understand that the force 63N makes B accelerate to the left but just by 0.5 m/s^2 slower than block A.
Re the second part: my attempt actually sees it from the "background" POV. My argument was not necessarily based around B accelerating to the right, but rather accelerating to the left just 0.5 slower than A (which is given). And I thought to account for that fact I need to subtract 12N (->0.5 m/s^ deceleration) from 75N.
Perhaps I misunderstood you?
(in any case - I do appreciate your time responding and trying to help!)

 

[haruspex]

Had the max friction force not been exceeded, the force transferred to block B would be 75N,

Why? Wouldn't that mean there is no net force on A, so it would not move?

Instead of making such leaps of faith, just write down the ΣF=ma equation for each block. Write another equation relating the accelerations.

 

[Dewgale]

One flaw with your approach is that the acceleration of block A isn't $a = \frac{F_a}{M_A} = \frac{75 N}{30 kg}$. Therefore subtracting off $(0.5 m/s^2)\times 24 kg$ won't give you the right answer.

 

[jna]

One flaw with your approach is that the acceleration of block A isn't $a = \frac{F_a}{M_A} = \frac{75 N}{30 kg}$. Therefore subtracting off $(0.5 m/s^2)\times 24 kg$ won't give you the right answer.

Yes, I agree that 75N/30kg was not the right thing to solve this (original attempt). That's why I posted the revised attempt (#3) which does not go into determining the acceleration of the block A at all, just considers forces.

 

[haruspex]

Yes, I agree that 75N/30kg was not the right thing to solve this (original attempt). That's why I posted the revised attempt (#3) which does not go into determining the acceleration of the block A at all, just considers forces.

but you still confused yourself. See post #12.

 

[Dewgale]

Except you're talking about force being transmitted via exceeding the maximum friction. That doesn't really make sense. The frictional force arises in response to $F_A$. We have no idea how much less it is -- so just subtracting 12 off of the applied force doesn't help us, because the net force which causes $a$ is some unknown quantity. You need to actually sit down and solve the equations. I highly recommend trying the approach haruspex suggested -- it'll save you a lot more time.

 

[jna]

Except you're talking about force being transmitted via exceeding the maximum friction. That doesn't really make sense. The frictional force arises in response to $F_A$. We have no idea how much less it is -- so just subtracting 12 off of the applied force doesn't help us, because the net force which causes $a$ is some unknown quantity. You need to actually sit down and solve the equations. I highly recommend trying the approach haruspex suggested -- it'll save you a lot more time.

Let me try (forces acting to the left +, to the right -):
For Block A: (net force A) = (acting force F) - (friction/drag from from block B)
For Block B: (net force B) = (friction force from A)
where (friction force from A) = - (friction/drag from from block B)
--- so:
m_A * a_A = F - F_fr
m_B * a_B = F_fr
and
a_B = a_A - 0.5 m/s^2
Are these equations correct?

 

[Dewgale]

Let me try (forces acting to the left +, to the right -):
For Block A: (net force A) = (acting force F) - (friction/drag from from block B)
For Block B: (net force B) = (friction force from A)
where (friction force from A) = - (friction/drag from from block B)
--- so:
m_A * a_A = F - F_fr
m_B * a_B = F_fr
and
a_B = a_A - 0.5 m/s^2
Are these equations correct?

Yes, those are correct.

 

[haruspex]

where (friction force from A) = - (friction/drag from from block B)

You already dealt with their being opposite when you put a minus sign in the equation for forces on block A.
All your final equations are correct though.

 

[jna]

So, as the solution for F_fr I am getting 80/3 (approx. 26.7) N. Unfortunately, that does not match any of the values in the multiple choice: {A. 99 N, B. 12 N, C. 63 N, D. 75 N}. I am ok with that as long as I know this result is correct - are you getting the same? Thanks for all your help again.

 

[Dewgale]

I am getting the same results yes. Assuming all the information you posted in your original post is correct, that is the correct answer.

 

[jna]

I am getting the same results yes. Assuming all the information you posted in your original post is correct, that is the correct answer.

Thank you!

 

[haruspex]

I am getting the same results yes. Assuming all the information you posted in your original post is correct, that is the correct answer.

Me too.

 

[jna]

Me too.

Thank you!