Ratio of two masses connected by pulley

[Tasha9000]

Homework Statement

[/B]
Figure 1) Block 1, of mass m1, is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude a and that the coefficient of kinetic friction between block 2 and the plane is μ.

Find the ratio of the masses m1/m2.
Express your answer in terms of some or all of the variables a, μ, and θ, as well as the magnitude of the acceleration due to gravity g.

Homework Equations

Fnet=ma
Ff=muN
W=mg

The Attempt at a Solution

+ /x is direction of acceleration

Forces on m₂

y-axis
N - mg_y=0

x-axis
T - m₂g_x-Ff=m₂a

T - m₂g Cosθ - μm₂Sinθ=m₂a

Forces on m₁

m₁g - T = m₁a

Attempt

I just put all the equations into one and got:

m1/m2=(g(Cosθ - μSinθ))/(g/a)

but it still says it's wrong. It says the final answer doesn't depend on Cosθ or Sinθ

 

[Tasha9000]

Homework Statement

[/B]
Figure 1) Block 1, of mass m1, is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude a and that the coefficient of kinetic friction between block 2 and the plane is μ.

Find the ratio of the masses m1/m2.
Express your answer in terms of some or all of the variables a, μ, and θ, as well as the magnitude of the acceleration due to gravity g.

Homework Equations

Fnet=ma
Ff=muN
W=mg

The Attempt at a Solution

+ /x is direction of acceleration

Forces on m₂

y-axis
N - mg_y=0

x-axis
T - m₂g_x-Ff=m₂a

T - m₂g Cosθ - μm₂Sinθ=m₂a

Forces on m₁

m₁g - T = m₁a

Attempt

I just put all the equations into one and got:

m1/m2=(g(Cosθ - μSinθ))/(g/a)

but it still says it's wrong. It says the final answer doesn't depend on Cosθ or Sinθ

it should be m1/m2=(g(Cosθ - μSinθ))/(g-a)

I also probably messed up the components but I tried both ways

 

[haruspex]

To check whether I have sin and cos the right way round, I consider an extreme case, like $\theta=\pi/2$. Do your equations look right for that case?