Let's solve the problem step-by-step: ### **1. Given Data** - Mass of lamp, \( m = 80 \) kg - Diameter of rod AB, \( d_{AB} = 10 \) mm - Diameter of rod BC, \( d_{BC} = 8 \) mm - Angle at point B = \( 60^\circ \) (between the rods) - Gravitational acceleration, \( g = 9.81 \) m/s² ### **2. Forces Acting on the Lamp** The lamp is suspended and is in equilibrium. The forces are: - Weight of the lamp, \( W = m \cdot g \) - Tension in rod AB, \( T_{AB} \) - Tension in rod BC, \( T_{BC} \) #### Calculate the weight: \[ W = 80 \times 9.81 = 784.8 \text{ N} \] ### **3. Free Body Diagram & Equilibrium Equations** At point B (where the lamp is hanging), the forces must balance. Let's resolve the forces: - Both rods form an angle of 60°. - Assume symmetry, so the angles with the vertical are each \( \theta = 30^\circ \). #### **Vertical Force Balance** \[ T_{AB} \sin(30^\circ) + T_{BC} \sin(30^\circ) = W \] Since \(\sin(30^\circ) = 0.5\): \[ 0.5T_{AB} + 0.5T_{BC} = 784.8 \] \[ T_{AB} + T_{BC} = 1569.6 \tag{1} \] #### **Horizontal Force Balance** \[ T_{AB} \cos(30^\circ) = T_{BC} \cos(30^\circ) \] Since \(\cos(30^\circ) = 0.866\): \[ 0.866T_{AB} = 0.866T_{BC} \] \[ T_{AB} = T_{BC} \] #### **Substitute into (1):** \[ T_{AB} + T_{AB} = 1569.6 \] \[ 2T_{AB} = 1569.6 \] \[ T_{AB} = 784.8\ \text{N}, \quad T_{BC} = 784.8\ \text{N} \] --- ### **4. Find the Normal Stress in Each Rod** The normal (axial) stress, \(\sigma = \frac{T}{A}\) #### **For Rod AB:** - Diameter \( d_{AB} = 10 \) mm \( = 0.01 \) m - Area \( A_{AB} = \pi \left(\frac{d_{AB}}{2}\right)^2 \) \[ A_{AB} = \pi \left(\frac{0.01}{2}\right)^2 = \pi (0.005)^2 = 7.854 \times 10^{-5} \text{ m}^2 \] \[ \sigma_{AB} = \frac{T_{AB}}{A_{AB}} = \frac{784.8}{7.854 \times 10^{-5}} = 9995 \text{ kPa} = 10.0 \text{ MPa} \] #### **For Rod BC:** - Diameter \( d_{BC} = 8 \) mm \( = 0.008 \) m - Area \( A_{BC} = \pi \left(\frac{d_{BC}}{2}\right)^2 \) \[ A_{BC} = \pi (0.004)^2 = \pi \times 1.6 \times 10^{-5} = 5.027 \times 10^{-5} \text{ m}^2 \] \[ \sigma_{BC} = \frac{T_{BC}}{A_{BC}} = \frac{784.8}{5.027 \times 10^{-5}} = 15608 \text{ kPa} = 15.6 \text{ MPa} \] --- ## **Final Answers** - **Normal stress in AB (\(\sigma_{AB}\)):** **10.0 MPa** - **Normal stress in BC (\(\sigma_{BC}\)):** **15.6 MPa** If you need any additional explanation, feel free to ask!

Certainly! Here's the translation of the question into English: --- **Question:** A lamp with a mass of 80 kg is hanging as shown below, supported by rods AB and BC. If the diameters of the rods are 10 mm for AB and 8 mm for BC, determine the normal (axial) stress in each rod. --- Let me know if you'd like any further assistance!

The normal (axial) stress in each rod is calculated by dividing the tension force by the cross-sectional area, reflecting the internal force per unit area experienced by the material.