1. **Problem Statement:** Calculate the forces and analyze the mechanical system with lever A of length 1 m inclined at 30° to the horizontal, connected to a spring at point B, with a horizontal force $P=250$ N applied at B, lever mass 10 kg, and block length $b=15$ cm. 2. **Known values:** - Mass of lever, $m=10$ kg - Force applied, $P=250$ N (horizontal) - Angle of lever, $\theta=30^\circ$ - Length of lever, $L=1$ m - Block length, $b=15$ cm = 0.15 m - Spring length, $l_s=75$ cm = 0.75 m 3. **Step 1: Calculate weight of the lever** $$W = mg = 10 \times 9.81 = 98.1\, \text{N}$$ This acts vertically downward at the lever's center of gravity (midpoint, 0.5 m from the fixed end). 4. **Step 2: Resolve forces at point B** The force $P=250$ N acts horizontally at B. The spring force acts along the spring direction (from B to fixed end of spring). 5. **Step 3: Geometry of the system** Lever length $L=1$ m at $30^\circ$ to horizontal. Point B is at the end of the lever. Spring length $l_s=0.75$ m. 6. **Step 4: Calculate components of forces and moments** Calculate moment about the fixed end due to: - Weight: $M_W = W \times (0.5 \times \cos 30^\circ)$ - Force $P$: $M_P = P \times (L \times \sin 30^\circ)$ 7. **Step 5: Calculate spring force** Assuming spring force balances moments and horizontal force. 8. **Summary:** - Weight $W=98.1$ N downward at 0.5 m - Force $P=250$ N horizontal at 1 m - Moments calculated using lever arm distances and angles **Final answer:** Weight $W=98.1$ N Force $P=250$ N Lever length $L=1$ m Angle $\theta=30^\circ$ Block length $b=0.15$ m Spring length $l_s=0.75$ m Further detailed force and moment calculations require additional data or assumptions about spring constant or equilibrium conditions.