# Frictionless pulleys with unknown masses. ATTEMPTED SOLUTION INSIDE!

1. Nov 3, 2008

### soccerdude28

1. The problem statement, all variables and given/known data
Two masses are connected by means of a string which passes over a light frictionless pulley. The coefficient of kinetic friction between the 9kg mass and the table is 0.25. The applied force on the 9kg mass is unknown and the acceleration of the unknown mass, m, is 1m.s^2 down. Calculate:
a. The magnitude of the tension in the string
b. The mass
c. The applied force, Fa, on the 9kg mass

2. Relevant equations
Ff=UkFn
Fnet=ma

3. The attempt at a solution
Alright so i started out with finding frictional force on the 9kg block. Fg=Fn
88.2=Fn
Ff=UkFn
88.2(0.25)
22.05N

Now i found the first equation of just the block with unknown mass
Fnet=ma
Fg-T=ma
9.8m-T=m
T=8.8m

Now second equation of just the 9kg mass
Fnet=ma
T-Fa-Ff=9(1)
T-22.05-Fa=9
T-Fa=31.05

Now the third equation of the full system
Fnet=ma
Fg+Ff+Fa=1(9+m)
9.8m-22.05-Fa=9+m
Fa=-31.05+8.8m

Now from the three equations, i subbed 1 into 2
So: 8.8m-Fa=31.05
Rearrange
8.8m-31.05=Fa

Now i used "Substitution" for the third equation as well as the new one we just found
8.8m-31.05=-31.05+8.8m
and it basically just cancels eachother out, which is why i am confused. What did i do wrong?
Thank you:)​

2. Nov 3, 2008

### soccerdude28

i thought that with an attempted solution, people would be more willing to help, doesnt look like it