Tension force of an object on a slope

[Dakkers]

Homework Statement

A box is trying to be pulled by a rope up a hill that is 30 degrees above the horizontal. The coefficient of friction ($$\mu$$) is 0.25. The acceleration is 1.5m/$$s^{2}$$ down the hill. The mass of the box is 2.0 kg. Find the force of tension in the rope.

Homework Equations

Fg = mg
Fn = Fg
Ff = $$\mu$$Fn
Fnet = ma

The Attempt at a Solution

I started by calculating each of the forces that I could.

Fg = mg = (2.0)(9.8) = 19.6 N

Fn = Fg = 19.6 N

Ff = $$\mu$$Fn = (0.25)(19.6) = 4.9 N

I drew a diagram and broke everything into components.

[URL]http://img231.imageshack.us/i/physicsforces.png/[/URL]

I forgot to add that if we do components of each force we have to do the components of acceleration, which becomes either 1.5cos30

or 1.5sin30 [down].

I plugged the components into the Fnet = ma equation but it didn't work, like so:

Fnet = ma (up/down)
Ftsin30 + 19.6sin60 - 19.6 - 4.9sin30 = (2.0)(-1.5sin30)
Ftsin30 - 5.0759... = -1.5
Ftsin30 = 3.5759
Ft = 7.1518

buuut that's not right because, at the least, Ff > Ft since the box is sliding down the hill.

Fnet = ma (left/right)
Ftcos30 - 4.9cos30 - 19.6cos60 - 0 = (2.0)(1.5cos30)
Ftcos 30 + 5.5564... = 2.5980...
Ftcos 30 = -2.95...
Ft = -3.41

once again, wrong.

pleeeease help!​

 

[ideasrule]

I haven't checked your calculations, but this question would be much easier if you broke the force up into a component parallel to the slope, and a component normal to the slope. That way, you'll get two nice equations--one for each component--that are easy to solve.

 

[Dakkers]

not to sound like a complete noob buuut what do you mean by parallel to the slope and normal to the slope?

 

[ideasrule]

You can redefine your coordinates so that the x coordinate represents distance down the slope, and the y coordinate represents distance perpendicular to the slope. I've attached a drawing to try to explain.

 

[Dakkers]

yeah and I made each force into its respective components (as the ugly math says up there) and tried isolating Ft but it was wrong every time. the answer should be about 2.5 (apparently) and I haven't gotten that yet

f

 

[ashishsinghal]

What do you mean by isolate. There will always be other forces involved.
What are the forces acting parallel to the surface of incline?