Kinetic Friction ProblemNeed Help

[mixedtape_15]

Kinetic Friction Problem..Need Help!

Okay...so I really think I'm doing this right..but the CAPA thing won't take my answer so there has to be something I'm doing wrong.
The Question is ...
A 32.0 kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 265 N at an angle of 54.5o above the horizontal. The coefficient of kinetic friction between the box and the surface is 0.330. What is the acceleration of the box?

So I drew my free body diagram and I had the Normal Force and the Weight and cosine of the angle in the y direction and I had the kinetic friction (normal force times kinetic friction coefficient) and weight and sine of the angle in the X direction.

Then I worked it all out and I got
Fy = may
Fy = 0
N - W(cos54.5) = 0
N - mg(cos54.5) = 0
N = 32.0kg (9.81 m/s^2)(cos54.5)
N = 182.3

Fx = max
-fk + T - w(sin54.5) = max
T - (N x uk) - mg(sin54.5) = max
265 - (182.3 x 0.330) - (32kg)(9.81m/s^2)(sin54.5) = 32kg(ax)
then I solved It all and it worked out to be -1.59 m/s^2
but its saying its not right..so I'm confused beyond belief now.
so Please help me .. lol

 

[geoffjb]

Check your trig functions in calculating N.

 

[mixedtape_15]

I've switched the trig functions around and tried that answer but still nothing. and I'm almost positive I have my trig done right anyways...but thanks.

 

[geoffjb]

I've tried this already but it didn't help..besides I'm almost positive that I have my trig done right.

So you're sure N - w \cos{54.5} = 0?

 

[mixedtape_15]

well there is no acceleration in the y direction so I guess it would have to be that.

 

[geoffjb]

Look at the x- and y-components of all the forces, and determine what you have.

\Sigma F_{x} = T_{x} - F_{f}

\Sigma F_{y} = N + T_{y} - mg

 

[geoffjb]

well there is no acceleration in the y direction so I guess it would have to be that.

Why the cosine? Like I said, rethink your trig.

 

[mixedtape_15]

I didn't think you had to incorporate the tension into the Y direction...isn't it just in the X?

 

[geoffjb]

I didn't think you had to incorporate the tension into the Y direction...isn't it just in the X?

Re-read the question, and redraw your free body diagram. The surface is horizontal, but the rope is not.

There is certainly a y-component of the tension, and it alleviates some of the normal force.

 

[mixedtape_15]

okay I'll go do that and hopefully figure this out..thanks

 

[mixedtape_15]

Okay I just redid it and realized my stupid mistake. I thought that the box was on a slope..but then when I reread it I realized that the box was on a straight horizontal and it was the rope that was on the angle. Thanks a bunch! :D