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
Answer:
To determine the acceleration of the system, we must apply Newton’s second law to each of the masses.
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Equation of motion for m_3 :
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The downward force due to gravity on m_3 is m_3 g .
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The tension in the rope is T .
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Given that m_3 moves downward with acceleration a , we can write:
m_3 g - T = m_3 a \quad \text{(Equation 1)}
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Equation of motion for m_1 and m_2 :
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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)}
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Combining the forces:
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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
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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} .