Friction Question (Maybe more physics than trig)

[Gruns]

Hello!
For the life of me I can't figure out what I'm missing with this problem:
A 6.26-kg block is placed on top of a 14.5-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.689. What is the maximum horizontal force that can be applied before the 6.26-kg block begins to slip relative to the 14.5-kg block, if the force is applied to (a) the more massive block and (b) the less massive block?

I'm able to solve for (a) with no problem. (In this case, 140 N)
I have tried every conceivable answer for (b) with no luck at all.

Here's what I know:
Weight of block A: 61.348 N, frictional force=42.27
Weight of block B: 142.1 N, frictional force=97.91
Weight of blocks A+B: 203.448 N, frictional force 140.18
The weight of both blocks times the coefficient of friction gives us the answer to (a).
Logicially I expect the answer for (b) to be very small.
According to all of my Free Body Diagrams, it should be 97.91-42.27=55.64, but this is not correct. (I actually expcet it to be much smaller than this even.)
As guesses I've tried the friction force of just block A, just block B, and the same as (a) but these are incorrect, as I expected.
Clearly I am missing something, but I have no idea what it is.
Appreciate any help anyone has!
Thanks

 

[Deleted member 4993]

Hello!
For the life of me I can't figure out what I'm missing with this problem:
A 6.26-kg block is placed on top of a 14.5-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.689. What is the maximum horizontal force that can be applied before the 6.26-kg block begins to slip relative to the 14.5-kg block, if the force is applied to (a) the more massive block and (b) the less massive block?

try 42.27 N
- that is the only relevant frictional force for movement of A relative to B.

 

[Gruns]

Hello and thank you for your reply!
42.27 is the frictional force of block A alone. As I said, I've tried this and it was incorrect. I've pretty much tried every possible known force, as well as all of the known forces minus the smaller force, if that makes sense. Iin my crude experimentaion of placing a small block on top of a larger block, no matter how little force I apply to the top(smaller) block, the bottom block does not move. Granted, I can't replicate a frictionless tabletop, but still, it seems that the force applied to the top is going to be very, very small. (I tried 0.001 as a guess, and it was wrong, too.)
Still searching for an answer!
Gruns

 

[Deleted member 4993]

Hello!
For the life of me I can't figure out what I'm missing with this problem:
A 6.26-kg block is placed on top of a 14.5-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.689. What is the maximum horizontal force that can be applied before the 6.26-kg block begins to slip relative to the 14.5-kg block, if the force is applied to (a) the more massive block and (b) the less massive block?

If this is the problem - with frictionless table - then both the blocks will start to move with slightest pull (on either of the blocks). There cannot be any relative motion between the blocks.

 

[wjm11]

A 6.26-kg block is placed on top of a 14.5-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.689. What is the maximum horizontal force that can be applied before the 6.26-kg block begins to slip relative to the 14.5-kg block, if the force is applied to (a) the more massive block and (b) the less massive block?

I'm able to solve for (a) with no problem. (In this case, 140 N)
I have tried every conceivable answer for (b) with no luck at all.

Here's what I know:
Weight of block A: 61.348 N, frictional force=42.27
Weight of block B: 142.1 N, frictional force=97.91
Weight of blocks A+B: 203.448 N, frictional force 140.18
The weight of both blocks times the coefficient of friction gives us the answer to (a).
Logicially I expect the answer for (b) to be very small.
According to all of my Free Body Diagrams, it should be 97.91-42.27=55.64, but this is not correct.

Hello, Gruns,

As Subhotosh has pointed out, there is only one frictional force that may be observed in this problem. It is between the two blocks.

Since your masses are given with three significant digits, be sure to use at least that for g: 9.81m/s^2. This results in a weight of 61.4 N for the top block and a friction force of 42.3 N developing between the blocks.

Force applied to top block:

A force applied to the top block exceeding 42.3 N will cause it to slide. (Perhaps an incorrect number of significant digits is causing problems if you are answering online/on computer.)

***

Force applied to bottom block:

Since 42.3 N can be applied to the top block before it will slip, this would result in an acceleration of

a = F/m = 42.3 N/ 6.26 kg = 6.759 m/s^2

If we apply force to the bottom block, the top block can only be accelerated through friction between the blocks. This means that if we attempt to accelerate both blocks at a higher rate than 6.759 m/s^2, the top block will slip because their will not be sufficient friction force to accomplish this.

What force would be required to accelerate both blocks at this rate?

F = ma = (6.26 kg + 14.5 kg)( 6.759 m/s^2) = 140.32 N => 140 N

This means that if 140 N or less is applied to the bottom block, the two blocks will accelerate together. If 140 N is exceeded, the top block will slip.

Hope that helps.

 

[Gruns]

Thanks again for the replies, but again, no luck.
140 N was indeed the answer I got for the bottom block(part A) and it was correct.
Unfortunately, 42.3 was INCORRECT for part B, and I don't know why.
Yes, it was answered via computer and the deadline is now past, so no worries trying to figure it out now, other than for my own personal knowledge. In a week or so the professor should unlock the homework again and I may be able to see the 'correct" answer.
Thanks again!
Gruns

 

[Deleted member 4993]

This problem was poorly worded at best. Please post the "correct" answer when you get it - so that we can understand what did the questioneer had in mind!!!