Question

An Atwood machine is constructed using two

Answer 1

Answer:

0.47 m/s²

Explanation:

Assuming the string is inelastic, m₃ will accelerate downward at a rate of -a, and m₄ will accelerate upward at a rate of +a.

Draw a four free body diagrams, one for each hanging mass and one for each wheel.

For m₃, there are two forces: weight force m₃g pulling down, and tension force T₃ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₃ − m₃g = m₃(-a)

For m₄, there are two forces: weight force m₄g pulling down, and tension force T₄ pulling up.  Sum of forces in the +y direction:

∑F = ma

T₄ − m₄g = m₄a

For m₁, there are two forces: tension force T₃ pulling down, and tension force T pulling right.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

T₃r₃ − Tr₃ = (m₁r₃²) (a/r₃)

T₃ − T = m₁a

For m₂, there are two forces: tension force T₄ pulling down, and tension force T pulling left.  Sum of the torques in the counterclockwise direction:

∑τ = Iα

Tr₄ − T₄r₄ = (m₂r₄²) (a/r₄)

T − T₄ = m₂a

We now have 4 equations and 4 unknowns.  Let's add the third and fourth equations to eliminate T:

(T₃ − T) + (T − T₄) = m₁a + m₂a

T₃ − T₄ = (m₁ + m₂) a

Now let's subtract the second equation from the first:

(T₃ − m₃g) − (T₄ − m₄g) = m₃(-a) − m₄a

T₃ − m₃g − T₄ + m₄g = -(m₃ + m₄) a

T₃ − T₄ = (m₃ − m₄) g − (m₃ + m₄) a

Setting these two expressions equal:

(m₁ + m₂) a = (m₃ − m₄) g − (m₃ + m₄) a

(m₁ + m₂ + m₃ + m₄) a = (m₃ − m₄) g

a = (m₃ − m₄) g / (m₁ + m₂ + m₃ + m₄)

Plugging in values:

a = (1.64 kg − 1.27 kg) (9.8 m/s²) / (2.5 kg + 2.3 kg + 1.64 kg + 1.27 kg)

a = 0.47 m/s²