Friction and magnitude between two blocks

AI Thread Summary
A block of mass 4.0 kg is placed on top of a 6.0 kg block, requiring a horizontal force of at least 19 N to make the top block slip. To find the maximum force that can be applied to the lower block while keeping both blocks moving together, the coefficient of friction must be calculated. The normal force acting on the top block is equal to its weight, which is 4 kg multiplied by gravity. The resulting acceleration of the blocks when the maximum force is applied is determined to be 2 m/s². Understanding the forces at play, including drawing a free body diagram, is crucial for solving the problem effectively.
trogdor5
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Homework Statement


A block of mass 4.0 kg is put on top of a block of mass M = 6.0 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 19 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.
(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.
1 N

(b) Find the magnitude of the resulting acceleration of the blocks.
2 m/s2

The Attempt at a Solution



I'm really not sure how I would go about solving this. What I did was this:
19N - mu*4*9.8 = 0
To try and find the coefficient of friction between the two blocks but I'm not sure if that's right and even if it is, how I would go about finding the rest of the solution to the problem. Please help!
 
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trogdor5 said:

Homework Statement


A block of mass 4.0 kg is put on top of a block of mass M = 6.0 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 19 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.
(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.
1 N

(b) Find the magnitude of the resulting acceleration of the blocks.
2 m/s2

The Attempt at a Solution



I'm really not sure how I would go about solving this. What I did was this:
19N - mu*4*9.8 = 0
To try and find the coefficient of friction between the two blocks but I'm not sure if that's right and even if it is, how I would go about finding the rest of the solution to the problem. Please help!

You have enough information in the problem to find mu. 19 N (fsmax) and the normal force on the 4 kg block (in this case supplied by the 6 kg block)

So if you apply a small horizontal force to the 6 kg block, what force will keep the 4 kg block moving with it? (a hint: What would happen if the surfaces between the two blocks were frictionless?)

If you pull the 6 kg block hard enough, the 4 kg block slips because...?
 
I'm honestly lost. Any way you can write out the steps a bit and I'll follow (not much explanation is necessary, as I kind of understand).
 
trogdor5 said:
I'm honestly lost. Any way you can write out the steps a bit and I'll follow (not much explanation is necessary, as I kind of understand).

I don't think we are supposed to do that except if we get more info from you.

Can you draw all the forces acting on the 4 kg mass if you were pulling with a horizontal force of 2N on the 6 kg block? In other words, can you draw a free body diagram of the situation I have presented? and do you know what the acceleration of the 4 kg block would be?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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