Question

A 2 kg and a 4 kg hang freely at opposite ends of a light inextensible string which passes over a small and light pulley
fixed to a rigid support. Calculate the acceleration of the system

Answer 1

Answer on Question #51063, Physics, Mechanics | Kinematics | Dynamics

A $2\mathrm{kg}$ and a $4\mathrm{kg}$ hang freely at opposite ends of a light inextensible string which passes over a small and light pulley fixed to a rigid support. Calculate the acceleration of the system.

Solution

Fig.1

According to Newton's second law

$$\left\{ \begin{array}{l} \ddot {a} m _ {2} = m _ {2} \ddot {g} + \vec {T} \\ \ddot {a} m _ {1} = \vec {T} + m _ {1} \ddot {g} \end{array} \right.$$

than

$$\left\{ \begin{array}{l} a m _ {2} = m _ {2} g - T \\ a m _ {1} = T - m _ {1} g \end{array} \right.$$

From Eq.(2)

$$a = g \frac {m _ {2} - m _ {1}}{m _ {2} + m _ {1}} = 9. 8 m / s ^ {2} \cdot \frac {4 k g - 2 k g}{4 k g + 2 k g} = 3. 2 7 m / s ^ {2}$$

Answer: $a = g\frac{m_2 - m_1}{m_2 + m_1} = 3.27m / s^2$

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