The magnitude of tension force on each strings are 148.8 N and 136.8 N respectively.
Given data:
The mass of blocks are [tex]m_{1}=12.0 \;\rm kg[/tex] and [tex]m_{2}=19.0 \;\rm kg[/tex].
The mass of solid cylinder is, M = 9.20 kg.
The radius of solid cylinder is, r = 0.200 m.
In the given problem, there are two cases:
- For the first block, there are two forces: m₁g pulling down and tension T₁ pulling up.
-
For the second block, there are two forces: m₂g pulling down and tension T₂ pulling up.
For the pulley, there are two torques: T₁r pulling counterclockwise and T₂r pulling clockwise.
m₁ moves up, so sum the forces in the +y direction. Then apply the equilibrium of force condition as,
∑F = ma
T₁ − m₁g = m₁a
Since, m₂ moves down, so sum the forces in the -y direction:
∑F = ma
m₂g − T₂ = m₂a
The pulley rotates clockwise, so sum the torques in the clockwise direction:
∑τ = Iα
T₂r − T₁r = (½ Mr²) (a/r)
T₂ − T₁ = ½ Ma
If we add the first two equations together:
T₁ − m₁g + m₂g − T₂ = m₁a + m₂a
T₁ − T₂ + (m₂ − m₁) g = (m₁ + m₂) a
Substitute:
½ Ma + (m₂ − m₁) g = (m₁ + m₂) a
(m₂ − m₁) g = (m₁ + m₂ − ½ M) a
a = (m₂ − m₁) g / (m₁ + m₂ − ½ M)
Substitute the values:
a = (19.0 kg − 12.0 kg) (9.8 m/s²) / (12.0 kg + 19.0 kg − ½ (9.20 kg))
a = 2.60 m/s²
Plug this into the first two equations to find the two tensions.
T₁ = m₁g + m₁a = 148.8 N
T₂ = m₂g − m₂a = 136.8 N
Thus, we can conclude that the magnitude of tension force on each strings are 148.8 N and 136.8 N respectively.
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