Question

A device known as Atwood's machine consists of two masses hanging from the ends of a vertical rope that passes over a pulley. Assume the rope and pulley are massless and there is no friction in the pulley. Mass m1 is greater than mass m2. Find expressions for the magnitude of their acceleration, ????,a, and the tension in the rope, T. Express your answers in terms of the masses and ????, the acceleration due to gravity.

Answer 1

Answer:

$a= \frac{(m_{1} -m_{2} )*g}{m_{1} +m_{2} }$

$T= \frac{m_{2}g(1+m_{1} -m_{2} ) }{m_{1}+m_{2}  }$

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

M1 free body diagram : Look at the attached graphic

∑F = m₁*a

W₁ -T= m₁*a

W₁ - m₁*a = T  Equation 1

M2 free body diagram :Look at the attached graphic

∑F = m₂*a

T-W₂= m₂*a

W₂ + m₂*a = T  Equation 2

Equation 1  = Equation 2

W₁ - m₁*a = W₂ + m₂*a

W₁ - W₂ =  m₁*a + m₂*a

m₁*g -m₂*g = a* (m₁ + m₂)

a = (m₁*g -m₂*g) / (m₁ + m₂)

$a= \frac{(m_{1} -m_{2} )*g}{m_{1} +m_{2} }$

Calculation of the tension in the rope (T)

We replace a in the equation 2

W₂ + m₂*a = T

W₂ + m₂*g*(m₁ -m₂) /  (m₁ + m₂) = T

m₂*g + m₂*g*(m₁ -m₂) /  (m₁ + m₂) = T

m₂*g( (1+ (m₁ -m₂)) /  (m₁ + m₂) = T

m₂*g (1+m₁ -m₂) /  (m₁ + m₂) = T

$T= \frac{m_{2}g(1+m_{1} -m_{2} ) }{m_{1}+m_{2}  }$

Answer 2

The expression for the magnitude of the acceleration of the masses in Atwood's machine is $a=\frac{(M_1 - M_2)g}{M_1+M_2}$.

The expression for the tension of the rope in Atwood's machine is

$T=\frac{2M_1M_2\,g}{M_1+M_2}$

Newton's Second Law of Motion

This problem can be analysed using a free-body diagram.

According to Newton's Second law of motion;

$\sum F=ma$

Therefore, from the freebody diagram of first mass, we can write;

$T-W=-M_1 \,a\\\\\implies T=M_1g-M_1a$

Also, from the freebody diagram of second mass, we can write;

$T-W=M_2 \,a\\\\\implies T=M_2g+M_2a$

Equating the RHS of both the expressions, we get;

$M_1g-M_1a=M_2g+M_2a\\\\M_1a+M_2a=M_1g-M_2g\\\\\implies a=\frac{(M_1 - M_2)g}{M_1+M_2}$

Now, substituting the value of 'a' in any of the above net force expressions, we get;

$T=M_2 g + M_2\frac{(M_1 - M_2)g}{M_1+M_2}=\frac{M_1M_2 g+M_2M_2g+M_1M_2 g-M_2M_2g}{M_1+M_2}$

$\implies T=\frac{2M_1M_2\,g}{M_1+M_2}$

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