In a butcher shop, a horizontal steel bar of mass \(4.00 \mathrm{~kg}\) and length \(1.20 \mathrm{~m}\) is supported by two vertical wires attached to its ends. The butcher hangs a sausage of mass \(2.40 \mathrm{~kg}\) from a hook that is at a distance of \(0.20 \mathrm{~m}\) from the left end of the bar. What are the tensions in the two wires?

The forces acting on the steel bar are: 1. The weight of the bar: \(W_b = m_b \cdot g = 4.00 \,\text{kg} \cdot 9.81 \,\text{m/s}^2\) 2. The weight of the sausage: \(W_s = m_s \cdot g = 2.40 \,\text{kg} \cdot 9.81 \,\text{m/s}^2\) 3. The tension in the left wire: \(T_1\) 4. The tension in the right wire: \(T_2\) The torques caused by these forces are: 1. The torque due to the weight of the bar: \(\tau_b = W_b \cdot d_b\), where \(d_b\) is the distance from the pivot point to the center of mass of the bar. 2. The torque due to the weight of the sausage: \(\tau_s = W_s \cdot d_s\), where \(d_s=0.20 \,\text{m}\) is given. 3. The torque due to the tension in the left wire: \(\tau_1 = T_1 \cdot d_1\), where \(d_1 = 0\) since the left wire is the pivot point. 4. The torque due to the tension in the right wire: \(\tau_2 = T_2 \cdot d_2\), where \(d_2\) is the distance from the pivot point to the right wire attachment.

Calculate the weight of the bar \(W_b\) and the weight of the sausage \(W_s\): \(W_b = 4.00 \,\text{kg} \cdot 9.81 \,\text{m/s}^2 = 39.24 \,\text{N}\) \(W_s = 2.40 \,\text{kg} \cdot 9.81 \,\text{m/s}^2 = 23.544 \,\text{N}\)

To calculate the torques, we need the distances to the center of mass of the bar (\(d_b\)) and the distance from the left wire to the right wire (\(d_2\)): \(d_b = \dfrac{1.20 \,\text{m}}{2} = 0.60 \,\text{m}\) \(d_2 = 1.20 \,\text{m}\)

We have two equilibrium equations: one for the sum of forces, and another one for the sum of torques: 1. Sum of forces equation: \(T_1 - T_2 = W_s\) 2. Sum of torques equation: \(\tau_b + \tau_s = \tau_2\) Calculating the torques: \(\tau_b = W_b \cdot d_b = 39.24 \,\text{N} \cdot 0.60 \,\text{m} = 23.544 \,\text{Nm}\) \(\tau_s = W_s \cdot d_s = 23.544 \,\text{N} \cdot 0.20 \,\text{m} = 4.7 \,\text{Nm}\) \(\tau_2 = T_2 \cdot d_2\) Substitute the torques into the sum of torques equation: \(23.544 \,\text{Nm} + 4.7 \,\text{Nm} = T_2 \cdot 1.20\, \text{m}\)

Solve the sum of torques equation for \(T_2\): \(T_2 = \dfrac{23.544 \,\text{Nm} + 4.7 \,\text{Nm}}{1.20 \,\text{m}} = 23.536 \,\text{N}\) Now, we can solve the sum of forces equation for \(T_1\): \(T_1 - T_2 = W_s\) \(T_1 = T_2 + W_s = 23.536 \,\text{N} + 23.544 \,\text{N} = 47.08 \,\text{N}\) Thus, the tension in the left wire (\(T_1\)) is \(47.08\, \text{N}\) and the tension in the right wire (\(T_2\)) is \(23.536\, \text{N}\).