Multiple Pulleys and Masses with Friction

[Wara]

Multiple Pulleys and Masses with Friction (Picture Included)

Homework Statement

http://content.screencast.com/users/Waraa/folders/Snagit/media/2d699ba1-32e0-4d9c-b79f-1a1d85dccd12/03.19.2012-15.49.46.png

I have solved the question but I am still iffy as to whether my solution is correct or not.

Homework Equations

Let m₁ = 10kg.
Let m₂ = 20kg.

(1) m₁a₁ = T₁ - μm₁g
(2) m₂a_a = m₂g - T₂
(3) 2T₁ = T₂
(4) 2a₂ = a₁

The Attempt at a Solution

T₂ = m₂g
T₂ = (20kg)(9.8m/s²)
T₂ = 196N

T₁ = $\frac{T_{2} + μm_{1}g}{2}$
T₁ = $\frac{196N + (0.32)(10kg)(9.8m/s^{2})}{2}$
T₁ = 113.68N

Sub (3) and (4) into (2)
m₂a₂ = m₂g - T₂
m₂($\frac{a_{1}}{2}$) = m₂g - 2T₁

Called this new equation (5).
(5) m₂($\frac{a_{1}}{2}$) = m₂g - 2T₁


(1) m₁a₁ = T₁ - μm₁g
(5) m₂($\frac{a_{1}}{2}$) = m₂g - 2T₁

(1) + (5)
m₁a₁ + m₂($\frac{a_{1}}{2}$) = T₁ - μm₁g + m₂g - 2T₁
a₁(m₁ + m₂($\frac{1}{2}$) = -μm₁g + m₂g - T₁
a₁(10kg + 20kg($\frac{1}{2}$) = -(0.32)(10kg)(9.8m/s²)+(20kg)(9.8m/s²)-113.68N
a₁ = 2.5448 m/s²

from (4)
2a₂ = a₁
a₂ = $\frac{a_{1}}{2}$
a₂ = $\frac{2.548m/s^{2}}{2}$
a₂ = 1.274 m/s²

Therefore, T₁ = 113.68N, T₂ = 196N, a₁ = 2.548 m/s², and a₂ = 1.274 m/s².

My question is if my solution is correct. Please check and reply back with suggestions or corrections. Thank you so much.

 

[Wara]

Please help me with this question. have i done this question correctly?