Three masses are arranged on a table with pulleys, as shown. the pulleys and string are taken to be massless. there is a coefficient of kinetic friction μk = 0.2 between m3 and the table. if m1 =4kg, m3=6kg and m2 =12kg, then the acceleration of the masses is
the figure looks like a table with a box in the middle attached to string and two pulleys on either side with two other boxes hanging off the table.

3.03 m/s²

Explanation:

m₂ > m₁, so m₁ will move up, m₂ will move down, and m₃ will move to the right.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₂ − m₂g = m₂(-a)

T₂ = m₂g − m₂a

Sum of the forces on m₃ in the y direction:

∑F = ma

N − m₃g = 0

N = m₃g

Sum of the forces on m₃ in the x direction:

∑F = ma

T₂ − T₁ − F = m₃a

T₂ − T₁ − Nμ = m₃a

Substituting and solving for acceleration:

(m₂g − m₂a) − (m₁g + m₁a) − (m₃g)μ = m₃a

m₂g − m₂a − m₁g − m₁a − m₃gμ = m₃a

g (m₂ − m₁ − m₃μ) = (m₁ + m₂ + m₃) a

a = g (m₂ − m₁ − m₃μ) / (m₁ + m₂ + m₃)

Given m₁ = 4 kg, m₂ = 12 kg, m₃ = 6 kg, and μ = 0.2:

a = 9.81 (12 − 4 − 6×0.2) / (4 + 12 + 6)

a = 3.03 m/s²

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