Multiple Pulleys and Masses with Friction

Wara

1. The problem statement, all variables and given/known data

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

2. Relevant equations
Let m₁ = 10kg.
Let m₂ = 20kg.

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

3. The attempt at a solution
T₂ = m₂g
T₂ = (20kg)(9.8m/s²)
T₂ = 196N

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

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

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

(1) m₁a₁ = T₁ - μm₁g
(5) m₂([itex]\frac{a_{1}}{2}[/itex]) = m₂g - 2T₁

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

from (4)
2a₂ = a₁
a₂ = [itex]\frac{a_{1}}{2}[/itex]
a₂ = [itex]\frac{2.548m/s^{2}}{2}[/itex]
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?

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