Solve the Mystery: Why is My Way Wrong?

[suspenc3]

I just recently got my midterm back and I felt this was graded wrong...

I don't have a picture..but just picture a ramp inclined to \phi=30 One mass is sitting on the incline...And the other is dangling from a rope which runs over a pulley

Two masses m_1 & m_2 both have a mass of 2Kg. If m_1 slide up the ramp at a constant velocity, what is the kinetic frictional force on m_1? Assume that the masses of the rope and pulley are negligible, and that the pulley is frictionless.
How I did it...
m_1 x/ T-f_k = m_1a (Eq1)
y/ N-mgsin \phi=0
m_2 x/ T-mgsin \phi=m_2a
y/ T=mgsin \phi (Eq2)
-------------------------------------------------
isolate T in Eq1
T-f_k=m_1a Since a=0
T=f_k
Sub Eq2 into Eq1
mgsin \phi=f_k
2Kg(9.8m/s)sin \phi=f_k
f_k=9.8N
Although i got the right answer he says that it needs to be expressed like this
T-f_k-mgsin \phi=0
f_k=T-mgsin \phi T=mg
f_k=mg-mgsin \phi = 9.8N
Why is my way wrong?

 

[mezarashi]

Could you explain what your variables x, y, and T mean?

You should analyze it through the use of force diagrams. The hanging mass is there only to tell you that the tension in the string, T is equal to mg. From there you free body diagram should focus entirely on the mass on the incline.

 

[marlon]

m_1 x/ T-f_k = m_1a (Eq1)

What is this x ? Is that x=gsin(30°) ?

Besides, in the x-direction (ie along the incline) you have 3 forces
1) T m_2g
2) friction \mu_kN
3) the x-component of gravity : gsin(30°)
I did not put in the right signs of these forces...
I only see two forces in your equation

What is x/T ?

The acceleration of object 1 is indeed 0

You should have m_1a_1=0=m_2g - \mu_kN -mgsin(30°)

y/ N-mgsin \phi=0
m_2 x/ T-mgsin \phi=m_2a

What is m_2 x/T ?

y/ T=mgsin \phi (Eq2)

This is incorrect.

In the y direction (perpendicular to the incline) you have
m_1a_1 = 0 =-mgcos(30°) + N

Where N is the normal force.


But you only need the equation in the x direction to get your answer

marlon