Frictionless pulleys with unknown masses. ATTEMPTED SOLUTION INSIDE!

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[left]

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:)

soccerdude28

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

PhanthomJay

Without a picture, it appears your problem is missing some information, if I am understanding it correctly. It looks like you have a 9kg block on the rough table, connected by a string which wraps around a pulley at the edge of the table and connects to a hanging block of unknown mass m, which accelerates at 1m/s^2 downward. There is also an applied force Fa on the 9kg block, acting to the left. Is that correct? If so, the problem has an infinite number of solutions; I can make m as large as I want, and adjust Fa acordingly to give the proper acceleration of 1m/s^2. That third equation of the system (which I don't encourage using) is extraneous, and doesn't buy you anything. Please clarify the question.