Calculating Toolbox Mass in Horizontal Pulley System

[Haniah]

Homework Statement

A group of construction workers are building a house and want to lower down an excess bundle of 1400 kg bricks to ground. They tie one end of a rope to the bundle of bricks, loop it through a pulley, and tie the other end of the rope to a toolbox. If the coefficient of kinetic friction between the toolbox and the floor is 0.690 and the acceleration of the bundle of bricks is 4.56 m/s2, what is the mass of the toolbox?
where A=toolbox
and B=bricks

Homework Equations

T-u_km_Bg=-m_Ba_A
T-m_Ag=m_Aa_A
(m_A-u_km_B)g=-(m_A+m_B)a_A

The Attempt at a Solution

m_A = ((gu_km_B)-(am_B))/(a+g)
I tried to solve by using this rearranged formula but did not get the right answer. I'm not quite sure what I'm doing wrong or what I'm missing. I am also assuming that acceleration (4.56 m/s²) is uniform, is this a correct assumption.

 

[Doc Al]

T-u_km_Bg=-m_Ba_A
T-m_Ag=m_Aa_A

I think you have your A and B reversed. Also, check your signs. (Let "a" be the magnitude of the acceleration.)