# Three masses are connected as shown, with weightless and inextensible ropes over a frictionless pulley. The planes where ( m_1 ) and ( m_2 ) lie are horizontal and frictionless. When the system is released, ( m_3 ) moves downward. Find the acceleration of

## Ali_Kaya_Akin’s Question

Ali_Kaya_Akin shared the following problem:

Three masses are connected as shown, with weightless and inextensible ropes over a frictionless pulley. The planes where ( m_1 ) and ( m_2 ) lie are horizontal and frictionless. When the system is released, ( m_3 ) moves downward. Find the acceleration of the system in terms of the gravitational acceleration ( g ).

The choices are:
A) ( \frac{2}{5}g )
B) ( \frac{1}{5}g )
C) ( \frac{1}{2}g )
D) ( \frac{1}{4}g )
E) ( \frac{1}{3}g )

To determine the acceleration of the system, we must apply Newton’s second law to each of the masses.

1. Equation of motion for ( m_3 ):

• The downward force due to gravity on ( m_3 ) is ( m_3 g ).

• The tension in the rope is ( T ).

• Given that ( m_3 ) moves downward with acceleration ( a ), we can write:

m_3 g - T = m_3 a \quad \text{(Equation 1)}
2. Equation of motion for ( m_1 ) and ( m_2 ):

• For ( m_1 ), the force ( T_1 ) due to the tether is balanced since both ( m_1 ) and ( m_2 ) are moving horizontally without friction:

T_1 = m_1 a \quad \text{(Equation 2)}
• For ( m_2 ), the situation is similar:

T_2 = m_2 a \quad \text{(Equation 3)}
3. Combining the forces:

• The total mass being accelerated is the sum of all the masses:

\text{Total mass} = m_1 + m_2 + m_3
• The net force on the system is the unbalanced force caused by ( m_3 ):

m_3 g = (m_1 + m_2 + m_3) a
4. Solving for ( a ):

a = \frac{m_3 g}{m_1 + m_2 + m_3}

Substituting the given values ( m_1 = 1 , \text{kg} ), ( m_2 = 2 , \text{kg} ), and ( m_3 = 3 , \text{kg} ):

a = \frac{3g}{1 + 2 + 3} = \frac{3g}{6} = \frac{1}{2}g

Therefore, the correct answer is ( \boxed{\frac{1}{2}g} ).