Forces: Pulley and Slope Constraining the Motion of two Masses

[Mongster]

Homework Statement

In the figure shown below, two blocks are connected by a massless, inextensible string over a frictionless pulley. The mass of Block A is 10kg and the mass of Block B is unknown.
Block A is on the incline with Q=37deg. The coefficient of kinetic friction between Block A and incline is Uk=0.30. The system of blocks has an acceleration of 0.80m/s^2 with Block B moving downwards.

Calculate mass of Block B.

My ans: 7.025kg, which isn’t correct.

Relevant Equations

F=ma
Ff = Uk x FN

 

[haruspex]

"Block B moving downwards."
Your equations are all as though block B is moving upwards.

 

[Lnewqban]

Mb suddenly appeared when you combined equations (1) and (2).
Revise equations prior combining them.

Concentrate on the summation of forces aligned with the slope.
If there is an acceleration up hill, there must be a greater force pulling in that direction.

Forces perpendicular to the slope are balanced, there is no movement in that direction.

 

[haruspex]

Mb suddenly appeared when you combined equations (1) and (2).

That all looked right to me. There are m_Ba and m_Bg in (1), and they reappear with correct signs after combining with (2).
I'm fairly certain the error is that the equations assume m_B is rising instead of falling. Each assumption will lead to a solution but yield different values.

 

[Mongster]

I changed the sign convention for the pulley in accordance to the direction of motion. However I still could not obtain the answer :(

 

[Mongster]

Update: Found a couple of mistakes in my second attempt at this Qn

1. My component force for Ma.g for Block A has been mixed up. Basically M.agSin37 & M.agCos37 are on the wrong sides.

2. Forgot to input Frictional Force (Fr) in Eqn1 + Eqn2.

Thank you all for the valuable input!

 

[haruspex]

Update: Found a couple of mistakes in my second attempt at this Qn

1. My component force for Ma.g for Block A has been mixed up. Basically M.agSin37 & M.agCos37 are on the wrong sides.

2. Forgot to input Frictional Force (Fr) in Eqn1 + Eqn2.

Thank you all for the valuable input!

So do you have the right answer now?
Btw, in case you did not recognise it, 37 degrees is a favourite angle with problem setters because it is the smallest angle in a 3-4-5 triangle. Knowing this, you can write down the trig functions immediately.

 

[Mongster]

Apologies for the late response. Yes the answer is now correct. Thank you for the heads up, I never realized that but now it makes sense why--- with regards to their love for the angle 37deg.