Having trouble finding the tension of this Newton law problem

[bblair3]

Block A in the figure below has mass mA = 4.5 kg and is sliding down the ramp. Block B has mass mB = 2.2 kg. The coefficient of kinetic friction between block B and the horizontal plane is 0.50. The inclined plane is frictionless and at angle 30°.

part a is to find the tension

the equation I have is Ma(g)sin theta-f(a)=ma a
i also found fa of to be 19.096, which I used \muk mag cos theta

I found the b part which was to find the acceleration: 1.68 m/s^2

using my first equation for tension i keep getting t=13.99
webassign does not like that answer and I can't figure out what I am doing wrong

thanks for any help you guys can give me

 

[gneill]

From the problem description only one of the blocks is on the slope, the other is on a horizontal surface. So I'm not seeing the reason for you're use of the cosine function when you calculate the frictional force.

You should be able to arrive at two expressions relating the tension and acceleration by analyzing the Free Body Diagrams of the two blocks.

 

[bblair3]

if use sin to find Fa it gives me 11.025 which makes the tension equal 3.56 which is also wrong.
maybe I am not writing the equation right.

I am getting Tension to be T=ma g sin theta- fa-ma a

which I am getting 4.5(9.8)sin 30-19.096-(4.5)(9.8) which = 13.99

am I setting the equation wrong??

 

[gneill]

You should be getting two equations for the tension, one for each block, from their FBD's. These equations will each have two unknowns: the tension and the acceleration. From them a single equation for the tension can be assembled (by substituting the acceleration from one into the other).

Can you write the two equations from the FBD's?