An Atwood's machine consists of blocks of masses m1=12.0kg and m2=19.0kg attached by a cord running over a pulley. The pulley is a solid cylinder with mass M=9.20kg and radius r=0.200m the block of mass m2 is allowed to drop, and the core turns the pulley without slipping. If the pulley axis is frictionless, find the linear acceleration of the system and the tension on each string.

The magnitude of tension force on each strings are 148.8 N and 136.8 N respectively.

Given data:

The mass of blocks are  and  .

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 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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2.60 m/s²

149 N and 137 N

Explanation:

Draw free body diagrams for the blocks and the pulley.

For the first block, there are two forces: m₁g pulling down and T₁ pulling up.

For the second block, there are two forces: m₂g pulling down and 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:

∑F = ma

T₁ − m₁g = m₁a

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)

Plug in 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

Rounded to three significant figures, the tensions are 149 N and 137 N.

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

wavelength divides in half and speed stays the same.