1. **Problem Statement:** Determine the value of force $P$ required to push the wedge given that the angle of friction $\phi$ for all surfaces in contact is 15° and the weight $W_B = 0$. The wedge is inclined at an angle $\theta = 60^\circ$ and a vertical force of 400 kN acts downward on the wedge. 2. **Relevant Formulas and Concepts:** - The angle of friction $\phi$ relates the friction force $F$ to the normal force $N$ by $F = N \tan \phi$. - To move the wedge, the applied force $P$ must overcome both the component of the vertical force along the incline and the frictional resistance. - The force equilibrium along the direction of $P$ is: $$ P = W \tan(\theta + \phi) $$ where $W$ is the vertical load (400 kN), $\theta$ is the wedge angle, and $\phi$ is the angle of friction. 3. **Calculations:** - Given: - $W = 400$ kN - $\theta = 60^\circ$ - $\phi = 15^\circ$ - Calculate $\theta + \phi$: $$ 60^\circ + 15^\circ = 75^\circ $$ - Calculate $\tan(75^\circ)$: $$ \tan(75^\circ) \approx 3.732 $$ - Calculate $P$: $$ P = 400 \times 3.732 = 1492.8 \text{ kN} $$ 4. **Interpretation:** The force $P$ required to push the wedge to the left, overcoming friction and the vertical load, is approximately 1492.8 kN. **Final answer:** $$ P \approx 1493 \text{ kN} $$