Calculating Static Friction Between Two Blocks on a Roughened Surface

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Homework Help Overview

The problem involves calculating the force of static friction between two blocks resting on a roughened surface, with a specific focus on the interaction between the blocks when a horizontal force is applied to the bottom block.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's laws to determine the static friction force, with one participant questioning the use of only the top block's mass in their calculations.

Discussion Status

The discussion includes attempts to clarify the relationship between the forces acting on the blocks and the resulting static friction. Some guidance has been provided regarding the application of Newton's second law specifically to the top block.

Contextual Notes

Participants are navigating the implications of the problem setup, including the effects of the roughened surface and the absence of slipping between the blocks. There is also a reference to an answer key that may influence their reasoning.

BoogieL80
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I was working the following problem:

Two blocks rest on a horizontal frictionless surfce (the top block being 5kg and the bottom block being 10 kg). The surface between the top and bottom blocks is roughened so that there is no slipping between the two blocks. A 30 N force is applied to the bottom block ( a 30 N horizontal force). What is the force of static friction between the top and bottom blocks

The answer key tells me that the answer to the problem is 10 N. I know that fsMAX = coefficednt of static friction * FN. I figured that the normal force for the top block is 49 N and the bottom block is 147 N. I can't figure out what combination of those numbers equal 10 N. Any help would be appreciated.
 
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Hint: Find the acceleration.
 
Re

I know the acceleration is 2m/s, but I'm still a little loss. Do I just take into account the top block and say that since F=ma that fs=ma thus fs = (5kg)(2m/s^2)? If that is the answer, why just use the 5kg top block and not both the 5kg block and 10kg block?
 
That's the answer. To analyze the forces acting on the top block (such as the friction), apply Newton's 2nd law to the top block. (If you apply Newton's 2nd law to both blocks you will not learn anything about the friction between the blocks since, due to the third law, the net friction force on both blocks is zero.)
 
Re

Thank you :smile:
 

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