What is the Frictional Coefficient for this Homework Statement?

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Homework Help Overview

The discussion revolves around determining the frictional coefficient in a physics problem involving forces acting on blocks. Participants are exploring the relationship between frictional force, normal force, and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for frictional force and question the application of the normal force in the context of the problem. There are attempts to clarify the relationship between various forces and the frictional coefficient.

Discussion Status

The discussion is active, with participants providing insights into the correct application of formulas and the importance of free body diagrams. Some participants suggest checking calculations and revisiting fundamental concepts.

Contextual Notes

There is mention of assumptions regarding the normal force and the conditions for static friction. The discussion also references the need for clarity on the acceleration of the blocks involved.

Priyadarshini
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Homework Statement


WP_20160129_22_40_07_Pro.jpg


Homework Equations


a=meu*g

The Attempt at a Solution


meu= 20/10
= 2
The answer's wrong
 
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Myu*g? What is that suppose to be?
Don't just apply the formula you've learnt,
Identify the normal force
 
Suraj M said:
Myu*g? What is that suppose to be?
Don't just apply the formula you've learnt,
Identify the normal force
The normal force here is the driving force of the vehicle, which is it's mass times the acceleration
 
Priyadarshini said:
The normal force here is the driving force of the vehicle, which is it's mass times the acceleration
You've got the concept right. Check the math again.
 
Exactly so the formula for frictional force is?
Do you thing myu*g represent the frictional force now?
 
Suraj M said:
Exactly so the formula for frictional force is?
Do you thing myu*g represent the frictional force now?
The firctional force is meu*normal force, so the frictional force is meu*M*a
Can I do this:
The driving force- frictional force = (M1+M2) a
friction= [F-(M1+M2)a]
so meu = [F-(M1+M2)a]/M1a
meu= [M1a - (M1+M2)a]/M1a
meu = M2/M1
But then what?
 
Actually, 3 of the answers are correct. Static friction is determined by ##N\mu \geq F ## for no slippage, not ##N\mu = F ##
 
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What is the acceleration of the smaller block?
 
cnh1995 said:
What is the acceleration of the smaller block?
Isn't the acceleration 20m/s^2 for both the blocks?
 
  • #10
Yes. That is what is needed. Write the equation for forces "on" the smaller block. You'll get the value of friction coeff just to avoid slipping. You'll see Chestermiller's reply makes true sense!
 
  • #11
Priyadarshini said:
Isn't the acceleration 20m/s^2 for both the blocks?
Have you drawn a free body diagram showing the forces acting on the smaller mass, or do you feel like you have advanced to the point where you no longer need to use free body diagrams?
 
  • #12
Chestermiller said:
Have you drawn a free body diagram showing the forces acting on the smaller mass, or do you feel like you have advanced to the point where you no longer need to use free body diagrams?[/QUOTE ]
WP_20160130_001.jpg
 
  • #13
Chestermiller said:
Have you drawn a free body diagram showing the forces acting on the smaller mass, or do you feel like you have advanced to the point where you no longer need to use free body diagrams?
cnh1995 said:
Yes. That is what is needed. Write the equation for forces "on" the smaller block. You'll get the value of friction coeff just to avoid slipping. You'll see Chestermiller's reply makes true sense!
WP_20160130_002.jpg

Got it! Thank you!
 

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