Maximum Force for Block on Bracket Friction?

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Homework Help Overview

The problem involves a 10kg block resting on a 5kg bracket on a frictionless surface, with static and kinetic friction coefficients provided. The original poster seeks to determine the maximum force that can be applied to the block without it sliding off the bracket, as well as the acceleration of the bracket.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster calculates the maximum static friction force and questions whether this value can be equated to the applied force. Some participants suggest applying Newton's second law to both the block and the system as a whole to clarify the relationship between the forces and the resulting motion.

Discussion Status

The discussion is ongoing, with participants exploring the implications of static friction and acceleration. There is a divergence in understanding whether the maximum static friction force directly translates to the maximum applied force, with some participants challenging this assumption and advocating for a more thorough application of Newton's laws.

Contextual Notes

Participants are navigating the complexities of static friction in a dynamic scenario, questioning the assumptions about equilibrium and acceleration in the context of the problem.

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Homework Statement


A 10kg block is resting on a 5kg bracket, which rests on a frictionless surface. The coef. of static and kinetic friction between the block and bracket are .4 and .3, respectively. Find out a) the max force that can be applied to the block without the block sliding on the bracket and b) the acceleration of the 5kg bracket.

http://img114.imageshack.us/img114/1958/56qe9.jpg

Homework Equations



Fnet = m * a

The Attempt at a Solution



This seems like a very simple question but I'm just having an issue with it (plus the book calls it a "challenging" problem so it just seems too easy).

Finding a) was pretty simple, just do 10 * .4 * 9.81 (Normal Force * \muk). I came up with 39.24N as the answer.

To find b I thought it would just have to apply Newton's second law.

39.24N = (5kg + 10kg) * a
a = 2.62m/s

But since this seemed too easy so I thought that I was wrong. Since the bracket exerts a frictional force on the block, do I have to include the force of the block on the bracket as given by Newton's third law? That seemed logical to me but then it would obviously negate the force of the string which would mean no acceleration, and it seems extremely obvious that there would be acceleration.

Any help would be great, thanks.
 
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Proximity said:
Finding a) was pretty simple, just do 10 * .4 * 9.81 (Normal Force * \muk). I came up with 39.24N as the answer.
What you found is the maximum value of static friction, which is not the same thing as the applied force F. (If F simply equaled the max friction force then the block would be in equilibrium.)

So it's not that simple. Apply Newton's 2nd law to the block and to the system as a whole.
 
That is what F equals though, isn't it? The question asks the max force that can be applied to the block so it doesn't slide. The max force that the static friction can apply is 39.24N, and so the max force on the string that can be applied is also 39.24N.
 
Proximity said:
That is what F equals though, isn't it?
No.
The question asks the max force that can be applied to the block so it doesn't slide. The max force that the static friction can apply is 39.24N, and so the max force on the string that can be applied is also 39.24N.
That would be true if it wasn't accelerating. But it is. Apply Newton's 2nd law.
 

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