1. **Stating the problem:** We need to find the mass and weight of the rim of a pulley. Given: - Outside diameter of the pulley $D = 0.8$ m - Material: Steel - Density $\rho = 7.85 \times 10^3$ kg/m$^3$ 2. **Assuming the rim geometry:** The problem references a cross section, but since it's a rim, the volume can be approximated as the volume of a hollow cylinder with outer diameter $D$ and some inner diameter $d$. Without explicit inner diameter, assume the rim thickness is small or given, but since no thickness is specified, let's denote thickness $t$ and cross-sectional area $A$ (approximated from figure) can be used. (If a thickness or rim cross-sectional area is provided, use that to find volume: $V = A \times \text{circumference}$.) 3. **Calculating volume:** - Circumference of pulley: $C = \pi D = \pi \times 0.8 = 2.513$ m - If rim cross section area $A$ is known, then $V = A \times C$ Without exact $A$ (cross-sectional area), let's illustrate formula: $$V = A \times \pi D$$ 4. **Finding mass:** $$m = \rho V = \rho A \pi D$$ 5. **Finding weight:** Weight $W = mg$ where $g = 9.81$ m/s$^2$ $$W = mg = \rho A \pi D \times 9.81$$ --- **Final answer:** The mass and weight depend on the rim cross-sectional area $A$ (in m$^2$): $$m = 7.85 \times 10^{3} \times A \times \pi \times 0.8 = 19723.9 \times A \quad \mathrm{(kg)}$$ $$W = m \times 9.81 = 193429 \times A \quad \mathrm{(N)}$$ If the cross-sectional area $A$ is given, substitute its value for final numeric answers.