# Homework Help: Help with pulley and masses

1. May 30, 2007

### BMWPower06

1. The problem statement, all variables and given/known data
In Figure below, m1 = 11.9 kg and m2 = 4.23 kg.
http://aycu19.webshots.com/image/18298/2003920338738691485_rs.jpg
The coefficient of static friction between m1 and the horizontal surface is 0.526 while the coefficient of kinetic friction is 0.288. If the system is released from rest, what will its acceleration be?

If the system is set in motion with m2 moving downward, what will be the acceleration of the system?
2. Relevant equations
F=MA i assume

3. The attempt at a solution

2. May 30, 2007

### stunner5000pt

ok first of all find the net force (vertical) on m2

what forces are acting on it?

then find the net horizontal force on m1. Firstly assume that they let it go from rest... determine whether or not the system will accelerate. Remember the force on m1 must be bigger than the static fricion force in order for it to move. (accelerate)

If it does then you have calculate the acceleration using the kinetic friction force.

3. May 30, 2007

### PhanthomJay

Noting that the tension in the cable is the same on both sides of the assumed massless frictionles pulley, you should draw a free body diagram (FBD) of each block, identify the forces acting on that block for each case, and use Newton 2 for each block to solve for the unknowns. Are you familiar with FBD's?

4. May 30, 2007

### BMWPower06

yea, i drew the FBD and got the Fn of M1= 116.62 and Fn of M2= 41.454. Now where do i go from here?

5. May 30, 2007

### BMWPower06

Ok so i used this:

and i got -.49m/s^2 but it says the answer is not correct. Anyone know why?

6. May 30, 2007

### BMWPower06

for mew i used the coefficient of Kinetic friction

7. May 31, 2007

### BMWPower06

any1 know whats wrong?

8. May 31, 2007

### PhanthomJay

It looks like you did part 2 first (the moving case) but you slipped on the minus sign. Use downward as the positive direction, try +.49m/s^2 as your answer. But in part 1, take again a look at 'stunner's' hint.

9. May 31, 2007

### stunner5000pt

how did you get this answer??

10. May 31, 2007

### BMWPower06

so i have:
For M2:
Force of m1 on m2 going up and Fmg going down

For M1:
Force of m2 on m1 going right and Ff going left.

Do i use Ff=(.526)*(11.9)*(9.8)
then where do i go from here?

11. May 31, 2007

### stunner5000pt

that 'force of m1 on m2' is called the tension because there is a string connecting the tow

can you write out what you said in the form of equations ? Maybe then you will be able to see what to solve for. Equations should be of the form
Fnet = sum of all of the forces in that direction