Question

A 4 kg block A is placed on the top of an 8 kg block B which rests on a smooth table. A just slips on B when a force of 12 N is applied on A. Then the maximum horizontal force F applied on B to make both A and B move together is?

Answer 1

Hello and welcome to OpenStudy. From what I understand, you need 12N to move 4kg. So to move 4+8 kg is probably just multiplication.

Answer 2

Apparently the answer is 24 N, but I have absolutely no idea how :(

Answer 3

Well my logic works this way: 12N for 4kg, so 3N for each kg. For 8 kg it will be...?

Answer 4

But you're finding the force that needs to be applied on B...

Answer 5

._. You're right. This is challenging my brain too much at 4 am. Do you have any ideas?

Answer 6

I think you need to take the two bodies A and B separately into consideration, and get two separate equations, and solve them..

Answer 7

Oh look what I found! http://lms.ptudep.org/templates/default/images/tests_cloaked/9039.html

Answer 8

It says the answer is 36 N, and the same as my method. Of course, I made the mistake of finding for B only instead of A+B earlier.

Answer 9

Except in the question I asked you, we're finding the force applied on B specifically :(

Answer 10

No we're not. We're looking for the force we apply to B, since B is at the bottom. Applying force to A would only move A, of course. B is WHERE we apply the force, but the magnitude of the force will have to take A's mass into account. Since when we push B, we are applying force to both of them as a system.

Answer 11

Does that make sense?

Answer 12

That's what I thought too, but apparently we need to take the coefficient of friction into consideration as well..

Answer 13

But why would we? Isn't the table smooth? Can you remind me the formula that takes friction into account?

Answer 14

\[mumg\]

Answer 15

Where \[\mu\]

Answer 16

is the coefficient of friction (static or kinetic)

Answer 17

I'm thinking, even if we calculate the friction, it'll be for between A and B where in the end we are moving the whole system on the table. :s

Answer 18

exactly.. you see why I'm confused? :(

Answer 19

Are you sure you need to take friction into account? What is actually preventing you from taking the solution we came up with?

Answer 20

yeah! I'm sure.. They've worked it out in this weird way here and come up with 24 N :(

Answer 21

Oh God then I am so bad to not know this. Sorry man. Good luck though. I'll just crash

Answer 22

No worries, Thank you SO much, you really helped :)

Answer 23

Not sure what I did but you're welcome. Welcome again and enjoy your stay.

Answer 24

Will do :) The stupid book must be wrong. It doesn't make sense..

Answer 25

@Potatoes.ramu
@wolfe8
The answer really is 24 N

Answer 26

How?

Answer 27

Case 1: 12 N force applied on block A
Calculate the combined acceleration of the blocks.
As only friction will contribute to the acceleration of block B, you'll get the max value of friction.

Case 2: 'F' force applied on block B
Let the combined acceleration be 'a'
Now, only friction will contribute to the acceleration of block A;
so you can get the upper value for F such that no slipping occurs.

Answer 28

Omg. That makes perfect sense! Thank you so much!

Answer 29

Your Welcome

Answer 30

I cannot brain this right now. Will revisit someday. Thanks dude.

Answer 31

:)

Answer 32

Now, even I am having problem in thinking straight..Maybe I should take a break..