Calculating Force in a Frictionless System with Inclined Mass and Wheel

[BoldKnight399]

A 15 kg mass is attached to a chord that is wrapped around a wheel with a radius 10.2 cm. The acceleration of the mass down the frictionless incline is .84 m/s^2. Assume the axle of the wheel to be frictionless. Find the force in the rope. The acceleration due to gravity is 9.81 m/s^2. Answer in units of N.

Ok so I don't truly understand how to approach this problem. Since it is part of my homework having to deal with torque, I feel that torque should have something to do with it, but I can't figure out how to relate it. If I don't use torque, can I just say that:
F=ma?
and if so, I understand that my mass is 15 kg. But my acceleration, how would I take that into account since it is on an incline of 38 degrees (sorry, it has a picture but it will not cut and paste for some reason.)

If anyone has any suggestions of how to approach this problem or how to find the acceleration, that would be great!

 

[kuruman]

Draw a free body diagram and use Newton's Second Law. As far as the acceleration of the mass is concerned, it makes no difference what's at the other end of the rope to maintain the tension that you are looking for. To finish the problem, you will need the angle of the incline which is not mentioned in the statement of the problem.

 

[BoldKnight399]

Ok so I tried to draw a force diagram. I have that the angle is 38 degrees. So drawing the force diagram, I found that the weight of the block was split, going towards the left and pointed straight down. Now I was thinking that the tension and force of the string would be
Fnet=Fgravity-Fstring
so
m*a=cos(38)m*g-Fstring
(15)(.84)=cos(38)(15)(9.81)-Fstring
12.6=115.9557-Fstring

does that make sense or is Newton turning in the grave right now?

 

[kuruman]

The equation makes sense only if 38^o is the angle of the incline with respect to the vertical, not the horizontal. If the angle is (conventionally) defined with respect to the horizontal, you need to change the cosine to a sine.