Two Mass In Contact On Slope

[myxomatosii]

Homework Statement

Two packages at UPS start sliding down the 20° ramp shown in Figure P8.25. Package A has a mass of 5.0 kg and a *coefficient of friction of 0.20. Package B has a mass of 10 kg and a coefficient of friction of 0.15. How long does it take package A to reach the bottom?

I call the larger mass m₂ and the smaller mass m₁.

http://img166.imageshack.us/img166/1899/p825.gif

Homework Equations

F=ma
f_k=μ_kn

The Attempt at a Solution

I have drawn two free body diagrams.

m₁

up: n₁
down:m₁gCos(θ)
left:F_{2 on 1} , m₁gSin(θ)
right: f_k1

m₂

up: n₂
down: m₂gCos(θ)
left: m₂gSin(θ)
right:F_{1 on 2} , f_k2

From this I wrote.

Σ μ θ

ΣF_x1=f_k1-F_{2 on 1}-m₁gSin(θ) = m₁a_x1

ΣF_y1=n₁-m₁gCos(θ)

therefore

n₁=m₁gCos(θ)

ΣF_x2=F_{1 on 2}+f_k2-m₂gSin(θ)=m₂a_x2

ΣF_y2=n₂-m₂gCos(θ)

therefore

n₂=m₂gCos(θ)





Now where to go from here? I am not sure but I believe I need to find the acceleration of the system since both are bound together by this law. By finding that I would then know a_x1 and a_x2.

I'm just not sure how... two different μ_k confuse me, how can I find the total acceleration of the system if both are dragged by different coefficients?

 

[myxomatosii]

Σ μ θ

Perhaps if I draw a third free body diagram representing the total mass being oppossed by two forces of friction. f_k1 and f_k2?

f_k1=μ_k1m₁gCos(θ)

f_k2=μ_k2m₂gCos(θ)

Using total mass to find the force pressing against those two forces..?

w₁₂=m₁₂gCos(θ)

Maybe? Its my idea at the moment.

It would allow me to find a=ΣF/m₁₂

 

[myxomatosii]

Σ μ θ

Perhaps if I draw a third free body diagram representing the total mass being oppossed by two forces of friction. f_k1 and f_k2?

f_k1=μ_k1m₁gCos(θ)

f_k2=μ_k2m₂gCos(θ)

Using total mass to find the force pressing against those two forces..?

w₁₂=m₁₂gCos(θ)

Maybe? Its my idea at the moment.

It would allow me to find a=ΣF/m₁₂

Using the method in the quote I got.

f_k1=μ_k1m₁gCos(θ)=9.21N

f_k2=μ_k2m₂gCos(θ)=13.81N

Oops below, it was Sin not Cos! (Had it wrong in the post above)

w₁₂=m₁₂gSin(θ)=50.277

So would that mean ΣF₁₂=50.277N-13.81N-9.21N=27.575N

a₁₂=F/m₁₂=1.817m/s²?

 

[myxomatosii]

Using the method in the quote I got.

f_k1=μ_k1m₁gCos(θ)=9.21N

f_k2=μ_k2m₂gCos(θ)=13.81N

Oops below, it was Sin not Cos! (Had it wrong in the post above)

w₁₂=m₁₂gSin(θ)=50.277

So would that mean ΣF₁₂=50.277N-13.81N-9.21N=27.575N

a₁₂=F/m₁₂=1.817m/s²?

And..

F_{2 on 1}=f_k1-m₁gSin(θ)-m₁a_x1

F_{1 on 2}=m₂a_x2-f_k2+m₂gSin(θ)


Is that completely off base?

 

[myxomatosii]

If the acceleration I found for the total system was correct...

a₁₂=a₁=a₂=1.817m/s²

then

v_f=(2aΔx)^(1/2) <-- sqrt

v_f=4.263m/s

Δt=v_f/a₁₂

Δt=2.346s?

 

[myxomatosii]

The system didn't accept 2.346s.

I just checked all my work and worked through it all again and got the same answer.

So is something wrong with my original free-body diagram, logic or both?

I think I need to factor in F_{1 on 2} and F_{2 on 1} to find the acceleration but I can't think of how I would do that..