Confusion; one block on top of another block

[x86]

Can anyone help me out with this? Here's the question

1. Homework Statement
A 2.2 kg block sits on top of an even bigger 3.2 kg block. The coefficient of kinetic friction between the two blocks is 0.25. The coefficient of static friction between the 3.2kg block and the floor is 0.32

Find the maximum force that can be applied to the bottom block so the top block doesn't move at all.

Apparently this is the answer, but it doesn't make sense to me.

(2.2kg+3.2kg)*9.8m/s^2 * 0.32 + 2.2kg * 9.8m/s^2 * 0.25 = 22 N

How would I go about doing this? My answer was around 32 N, but apparently its wrong and the above was right. However, it doesn't make sense. Is it correct?

---

Here's how I think it should be done.

6 kg * 9.8m/s^2 * 0.32 = 18.2 N. This is how much force the friction of the bottom block resists.

The top blocks friction is 2.2kg * 9.8m/s^2*0.25 = 5.39 N. So this means that when the bottom block accelerates forward at 2.45 m/s^2 if there were no friction forces the top block would be accelerating at 2.45 m/s^2 relative to the bottom block (backwards).

Meaning that the correct answer should be...

2.45 m/s^2 * 6 kg + 18.2 N which is 33 N. But, apparently this is wrong.

So, am I right or is the other guy right?

 

[nil1996]

about kinetic friction

The coeff. of friction between the two blocks that you have mentioned is kietic.But for the given condition we the coeff of static friction.
static friction is responsible for an object to keep it at rest even if force is applied while kinetic friction is necessary for slowing down an moving object.

So please check the qn

 

[x86]

The coeff. of friction between the two blocks that you have mentioned is kietic.But for the given condition we the coeff of static friction.
static friction is responsible for an object to keep it at rest even if force is applied while kinetic friction is necessary for slowing down an moving object.

So please check the qn

thats what the question said. perhaps they made a typo, and the kinetic friction is really static

 

[barryj]

I think you added 2.2 kg to 3.2 kg and got 6 kg.
Using 6 kg I agree with 33N but correct your addition and see what you get.

 

[x86]

I think you added 2.2 kg to 3.2 kg and got 6 kg.
Using 6 kg I agree with 33N but correct your addition and see what you get.

Ah you're right, it is 5.4 kg. Anyway, thanks for confirming that my method was correct

 

[haruspex]

I agree. To dislodge the top block the acceleration must reach 9.8m/s^2 * 0.25, so the force should be (2.2kg+3.2kg)*9.8m/s^2 * 0.32 + (2.2kg+3.2kg) * 9.8m/s^2 * 0.25

 

[x86]

I agree. To dislodge the top block the acceleration must reach 9.8m/s^2 * 0.25, so the force should be (2.2kg+3.2kg)*9.8m/s^2 * 0.32 + (2.2kg+3.2kg) * 9.8m/s^2 * 0.25

Ah okay thank you. At that force the acceleration should be zero so it would not dislodge the block correct? Due to friction and all

 

[haruspex]

Ah okay thank you. At that force the acceleration should be zero so it would not dislodge the block correct? Due to friction and all

Below that force, the acceleration of the top block will be the same as that of the lower block, so it will stay in place. I'm making two assumptions wrt the OP:
1. That it should have referred to static friction, not kinetic.
2. That "not moving at all" means not moving relative to the lower block. If it really meant not moving within an inertial frame then the answer would just be (2.2kg+3.2kg)*9.8m/s^2 * 0.32.