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