Question about finding static friction coefficient

AI Thread Summary
The discussion revolves around calculating the coefficient of static friction between two stacked boxes, given that a force of 600 N is applied to the top box while the bottom box has a static friction coefficient of 0.3 with the floor. The required force to move both boxes is determined to be 540 N, indicating that the static friction between the boxes must be sufficient to prevent the top box from slipping off. Participants clarify that the problem focuses on static friction, not kinetic friction, as the boxes move together. The key point is that the coefficient of friction between the two boxes must be high enough to transmit the necessary force without slipping. Understanding these dynamics is crucial for solving the problem accurately.
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Homework Statement



Two boxes each weighing 900 N are stacked on a horizontal floor. Box A is on the bottom, box B is on the top. The coefficient of static friction between box a and the floor is 0.3. A worker applies a force of 600 N to box b parallel to the floor, and the stack moves together along the floor. Which of the following could be the coefficient of static friction between box a and box b?

1) 0.7
2) 0.5
3) 0.4
4) 0.3
5) 0.1

Homework Equations



Ffriction = mu * Fnormal

The Attempt at a Solution


I have no idea how to solve for the coefficient of friction between the two boxes. If it was between the floor and the combination of the two boxes, I would have treated the boxes as one and then find the resulting coefficient of friction.

Could anyone please help me get started and solve this problem?

Thanks a lot!
 
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Although 600N is applied not all of that is needed to move the boxes along the ground. Calculate what the required force is. How does that force get transmitted from the man to the lower box?
 
CWatters said:
not all of that is needed to move the boxes along the ground.

How is not all of that needed?
 
The problem says that the coefficient of friction with the ground is 0.3.

0.3 * 1800 = 540N

Only 540N of the 600N is required to move the boxes.

How is that transmitted to the lower box?
 
Why would it give you coefficient of static friction between box a and the floor? Shouldn't it be coefficient of kinetic friction?

Weird..
 
540N is what is required to move the boxes and that's what has to be transmitted through the interface between box A and B. The coefficient of friction there must be high enough to allow that. If it wasn't the top box would move and the bottom one would stay put.
 
Biker said:
Why would it give you coefficient of static friction between box a and the floor? Shouldn't it be coefficient of kinetic friction?

Weird..
The question asks which answers could be right. For that, we must allow all valid values for the kinetic friction. What is the range of valid values?
(But you do not actually need to worry about that. Just consider the point at which static friction between lower box and ground is overcome, but static friction between the boxes is not.)
 
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