How Can the Bottom Block Receive Maximum Force Without Moving the Top Block?

AI Thread Summary
The discussion revolves around calculating the maximum horizontal force the bottom block can receive without the top block moving, given a frictionless surface. The maximum static frictional force between the blocks is identified as 12N, which is crucial for determining the top block's acceleration. By calculating the acceleration of the 4kg block using the formula F = ma, it is established that the maximum acceleration is 3 m/s². Consequently, the total force required to accelerate both blocks together is 27N. The conversation highlights the simplicity of the problem despite initial confusion, emphasizing the importance of understanding the relationship between force, mass, and acceleration.
Tubs
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If a 4kg block is resting on a 5kg block and the maximum horizontal force exerted on the block without it moving is 12N, what is the maximum horizontal force that the bottom block can receive without the top block moving? The surface between the bottom block and the ground is frictionless.

I can't seem to solve this :frown: any help would be awesome :)
 
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Tubs said:
If a 4kg block is resting on a 5kg block and the maximum horizontal force exerted on the block without it moving is 12N, what is the maximum horizontal force that the bottom block can receive without the top block moving?

The question obviously must mean "what is the maximum force on the bottom block such that the top block will not slide."

From the first part, you have the maximum static frictional force between the two blocks, 12 N.

This static frictional force is the force that must accelerate the top block. With this force, and the given mass, what is the maximum acceleration of the top block?

Once you find that, ask yourself: "Self? if the top block accelerates at this quantity, what total pushing force would be required to get the whole mass accelerating at that quantity?"
 
Just to clarify, is the 12N MAX force applied to the 4kg block?? If it is, think of what force will accelerate the 4kg block forward, while it is positioned on top of the 5kg block. This force has a max value (12N). Associated with this max force is a max acceleration for the 4kg block. If you can determine the max acceleration of the 4kg block, you know the max acceleration of the 4kg/5kg system. See what you can come up with... (1) identify the force that causes the 4kg block's acceleration, and (2) understand how that force, via acceleration, can be used to determine the max possible horizontal force applied to the 5kg block. I will stand by for your response(s).
 
Fnet1 = ma
12 = (4)a
a = 3 m/s^2

Fmax = ma
= 9(3)
= 27N

All good?
 
Looks good. Another example, where all the surfaces are frictionless, would be a 4kg block tethered to the top of the 5kg block via a horizontal string. Assuming the max tension the string could withstand is 12N, the string would be able to accelerate the 4kg block to a max amount of 3m/s/s. And similarly, the max horizontal force applied to the 5kg block would be again 27N.
 
all good :smile:
 
Fermat said:
all good :smile:

The comment on that question was that it was difficult. Compared to these other ones it is simple, I should have seen that answer right away. STUPID STUPID STUPID STUPID :mad:
 
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