Physics
posted by Josh on .
A 390 kg motor boat, moving down a frictionless ramp, is connected over a massless pulley to a 193 kg counterweight that rests on a horizontal surface. The ramp is inclined by 28.6°, and the coefficient of kinetic friction between the couterweight and the horizontal surface is 0.507.
What is the tension in the rope connecting the boat and the counterweight?
Formula:
Xaxis: m1gcos0=mTa
m2gcos9=mTa
YAxis: mgsin0 = 0
So I'm really stuck here. I tried to figure out the relative formula's but I think I'm doing it wrong. I just need help getting started. Thanks!

It is uncertain to me where the weight is, vertical or not to the pulley. It matters.
why are these questions labeled physics? Do you have a math teacher teaching physics? 
I do not know why you have cos9 in the second formula, nor what mTa is supposed to mean. I suppose the number 0 is supposed to be theta (28.6 degrees.
There is no hyphen in coefficient.
I do appreciate your showing your work. Almost no one does that here anymore, and I wish they would.
You need to write two equations of motion: one for the direction of the boat down the ramp, and one for the counterweight in the horizontal direction. Rope tension T will be a second variable you must solve for with the two equations.
390*g sin28.6  T = 390 a
T  193*g*(0.507) = 193 a
You can eliminate T by adding the two equations. Once you have a, solve for T using either equation.