A lift of mass 16000 kg is connected to a rope that passes over a drum of 4 m diameter, and a balance mass of 6000 kg is attached to the other end of the rope. The drum has a mass of 1250 kg and a radius of gyration of 450 mm , and is driven through a \( 32: 1 \) reduction gear by a motor with an efficiency of \( 85 \% \). The rotating parts of the motor have a mass of 110 kg and a radius of gyration of 125 mm . The frictional torque in the motor is 30 Nm . Calculate the motor torque for an upward lift acceleration of \( 0,2 \mathrm{~m} / \mathrm{s}^{2} \). Please draw the free body diagram for you to understand the instruction.

To calculate the motor torque needed for the upward lift acceleration, we shall use the information provided about the lift, the counterbalance mass, and the dynamic forces involved. First, we need to determine the net force acting on the lift. The gravitational force acting on the lift is given by: \[ F_g = m_{\text{lift}} \cdot g = 16000 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 156960 \, \text{N} \] The gravitational force acting on the counterbalance weight is: \[ F_{\text{counter}} = m_{\text{counter}} \cdot g = 6000 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 58860 \, \text{N} \] The net force required to accelerate the lift upwards at \(0.2 \, \text{m/s}^2\) can be calculated using Newton's second law (\(F = m \cdot a\)): \[ F_{\text{net}} = m_{\text{lift}} \cdot a = 16000 \, \text{kg} \cdot 0.2 \, \text{m/s}^2 = 3200 \, \text{N} \] Now, to find the total upward force exerted by the motor, we needs to account for both the weight of the lift and the required net force for acceleration: \[ F_{\text{motor}} = F_g - F_{\text{counter}} + F_{\text{net}} \] \[ F_{\text{motor}} = 156960 \, \text{N} - 58860 \, \text{N} + 3200 \, \text{N} = 97900 \, \text{N} \] Next, we must convert this force to torque at the drum: The torque can be calculated as: \[ \tau_{\text{drum}} = F_{\text{motor}} \cdot r_{\text{drum}} = F_{\text{motor}} \cdot \left(\frac{d_{\text{drum}}}{2}\right) = 97900 \, \text{N} \cdot 2 \, \text{m} = 195800 \, \text{Nm} \] Considering the gear reduction of \(32:1\), we can calculate the motor torque: \[ \tau_{\text{motor}} = \tau_{\text{drum}} \cdot \text{Reduction Ratio} \] \[ \tau_{\text{motor}} = \frac{195800 \, \text{Nm}}{32} \approx 6125 \, \text{Nm} \] Finally, we must also account for the frictional torque experienced in the motor. The total motor torque needed is: \[ \tau_{\text{total motor}} = \tau_{\text{motor}} + \text{Friction Torque} = 6125 \, \text{Nm} + 30 \, \text{Nm} = 6155 \, \text{Nm} \] Thus, the required motor torque for an upward lift acceleration of \(0.2 \, \text{m/s}^2\) is approximately **6155 Nm**.