Finding acceleration and tension on an incline

[soysauce]

Homework Statement

Hi everyone, I hope you're enjoying your weekends!
I've been trying this question for a while not having any luck, I'm hoping someone can tell me where I'm going wrong

The question is:
Two boxes are connected by a rope that passes through a pulley on the corner of a incline. Box A is 2.5 kg and box B is 5.5 kg. The coefficient of kinetic friction on the incline is 0.54. Box A is on the incline and box B hangs over. The angle of the incline is 25.4º. Find the acceleration and the tension.

The answers given are 3.8 m/s2 [down] and 33 N [up]

Homework Equations

I got these equations from here:https://www.physicsforums.com/showthread.php?t=27346 and I understand why I would use them

F_t = m₁ *a + m₁gcosθ + μm₁gcosθ and
F_t = m₂(g-a)

The Attempt at a Solution

m1 = 2.5 kg, m2 = 5.5 kg, μ = 0.54, θ = 25.4°

I know I can equate the two tension equations to get:

m₂(g-a) = m₁ *a + m₁gcosθ + μm₁gcosθ

But when I plug in the known values and solve:

5.5(9.8-a) = 2.5*a + 2.5*9.8*cos(25.4) + 0.54*2.5*9.8cos(25.4)

I get a value of a = 2.4

Could someone tell me where I'm going wrong here?
Thanks so much in advance!

 

[Doc Al]

F_t = m₁ *a + m₁gcosθ + μm₁gcosθ and
F_t = m₂(g-a)

In that first equation, one of those cosines should be a sine.

But please don't use someone else's derived equations. Derive them yourself, using Newton's 2nd law. Then you'll understand where they come from.

 

[soysauce]

In that first equation, one of those cosines should be a sine.

But please don't use someone else's derived equations. Derive them yourself, using Newton's 2nd law. Then you'll understand where they come from.

Whoops, that's true!
Thanks so much, I'll definitely be sure to in the future.