What is the Frictional Coefficient for this Homework Statement?

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SUMMARY

The frictional coefficient (meu) can be calculated using the formula meu = M2/M1, where M1 and M2 represent the masses involved in the system. The normal force is defined as the driving force of the vehicle, calculated as mass times acceleration (M*a). To determine the frictional force, the equation F - (M1 + M2)a = friction can be utilized, leading to the conclusion that static friction is determined by the inequality N*mu ≥ F for no slippage. A free body diagram (FBD) is essential for visualizing the forces acting on the smaller mass to accurately calculate the friction coefficient.

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  • Understanding of Newton's laws of motion
  • Familiarity with free body diagrams (FBD)
  • Knowledge of static friction concepts
  • Basic algebra for manipulating equations
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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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