# Frictional Coefficient

1. Jan 29, 2016

1. The problem statement, all variables and given/known data

2. Relevant equations
a=meu*g

3. The attempt at a solution
meu= 20/10
= 2

2. Jan 29, 2016

### Suraj M

Myu*g? What is that suppose to be?
Don't just apply the formula you've learnt,
Identify the normal force

3. Jan 29, 2016

The normal force here is the driving force of the vehicle, which is it's mass times the acceleration

4. Jan 29, 2016

### cnh1995

You've got the concept right. Check the math again.

5. Jan 29, 2016

### Suraj M

Exactly so the formula for frictional force is?
Do you thing myu*g represent the frictional force now?

6. Jan 29, 2016

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?

7. Jan 29, 2016

### Staff: Mentor

Actually, 3 of the answers are correct. Static friction is determined by $N\mu \geq F$ for no slippage, not $N\mu = F$

8. Jan 29, 2016

### cnh1995

What is the acceleration of the smaller block?

9. Jan 29, 2016

Isn't the acceleration 20m/s^2 for both the blocks?

10. Jan 29, 2016

### cnh1995

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. Jan 29, 2016

### Staff: Mentor

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. Jan 29, 2016

13. Jan 29, 2016

Got it! Thank you!

14. Jan 29, 2016