Let's break down part (b) of the problem into two steps and provide explanations after each: --- ## Step 1: Draw a Free-Body Diagram and Set Up Equilibrium Conditions **Explanation:** Let’s identify all forces acting on the beam: - Force by the bear (\( F_{bear} = 800\,N \)), located at \( x = 1.05\,m \) from the wall. - Weight of the beam (\( F_{beam} = 200\,N \)), acting at the center (\( L/2 = 3.25\,m \)). - Weight of the basket (\( F_{basket} = 80\,N \)), at the end (\( L = 6.5\,m \)). - Tension in the wire (\( T \)), at the end, at an angle \( \theta = 60^\circ \). - Reaction forces at the wall: horizontal (\( H \)) and vertical (\( V \)) components. **Set up rotational equilibrium** (about the left end of the beam): \[ \sum \tau = \] - The counterclockwise moments (from \( T \)) must balance all clockwise moments (from the weights): \[ T \sin\theta \cdot L = F_{bear} x + F_{beam} \frac{L}{2} + F_{basket} L \] --- ## Step 2: Solve for the Tension \( T \) in the Wire **Explanation:** Plug in the values: - \( L = 6.5\,m \) - \( x = 1.05\,m \) - \( \sin 60^\circ = .866 \) - \( F_{bear} = 800\,N \) - \( F_{beam} = 200\,N \) - \( F_{basket} = 80\,N \) \[ T \cdot .866 \cdot 6.5 = 800 \times 1.05 + 200 \times 3.25 + 80 \times 6.5 \] \[ T \cdot 5.629 = 840 + 650 + 520 \] \[ T \cdot 5.629 = 201 \] \[ T = \frac{201}{5.629} \approx 357.2\,N \] --- ## **Final Answer:** **The tension in the support wire when the bear is at \( x = 1.05\,m \) is approximately \( \boxed{357\,N} \).**