4. (a) Two particles of masses 6 kg and 7 kg are connected by a light inextensible string passing over a smooth light fixed pulley which is fixed to the ceiling of a lift - Leaving Cert Applied Maths - Question 4 - 2013

When the lift is accelerating, we have to adjust the forces accordingly. The effective acceleration of the system becomes: $a = g - \frac{g}{8} = \frac{7g}{8}$

Now, writing the equations for each mass:

For the 7 kg mass:

=> T = 7g - 7 \frac{g}{8} = \frac{9g}{8}$$ And for the 6 kg mass: $$T - 6g = 6a \\ => T - 6g = 6 \frac{g}{8} \\ => T = 6g + 6 \frac{g}{8} = \frac{48g + 6g}{8} = \frac{54g}{8}$$ The tension T can therefore be expressed as: $$T = \frac{189g}{26} \approx 71.24 ext{ N}$$

To calculate the tension in the string, we can set up the forces acting on the masses. The forces can be expressed as:

For mass $m_1$: $T - m_1g = m_1a$

For mass $m_2$: $m_2g - T = m_2a$

From these equations, we can express the system as:

Adding both equations: $T - m_1g + m_2g - T = (m_1 + m_2)a$

This simplifies to: $m_2g - m_1g = (m_1 + m_2)a$

Now isolating for T gives us: $T = \frac{m_2g - m_1g + m_p(g)}{2}$ Thus: $T = \frac{m_1m_2g}{m_1 + m_2 + 4m_p}$