1. **State the Problem:** We have a 7.5-kg block resting against a vertical wall with friction coefficients\n static friction $\mu_s = 0.45$, kinetic friction $\mu_k=0.35$, and an applied force $P$ at angle $\theta=40^\circ$. We want to find the smallest force $P$ that keeps the block in equilibrium.\n\n2. **Analyze Forces:** The block weight is $W = mg = 7.5 \times 9.81 = 73.575$ N vertically downward.\n\n3. **Force components:** The force $P$ acts along the rod at $40^\circ$ from horizontal. Resolve $P$ into horizontal ($P \cos 40^\circ$) and vertical components ($P \sin 40^\circ$).\n\n4. **Friction and normal force:** The wall supplies a normal force $N$ horizontally opposing $P \cos 40^\circ$, so $N = P \cos 40^\circ$. The friction force resists sliding vertically and is limited by $f = \mu_s N = \mu_s P \cos 40^\circ$.\n\n5. **Equilibrium conditions:** Vertically, the friction force plus vertical component of $P$ must balance weight:\n$$P \sin 40^\circ + f = W$$\nSubstitute $f$:\n$$P \sin 40^\circ + \mu_s P \cos 40^\circ = W$$\n\n6. **Solve for $P$:**\n$$P (\sin 40^\circ + \mu_s \cos 40^\circ) = W$$\n$$P = \frac{W}{\sin 40^\circ + \mu_s \cos 40^\circ}$$\n\n7. **Calculate numeric value:**\n$\sin 40^\circ \approx 0.6428$, $\cos 40^\circ \approx 0.7660$,\n\n$$P = \frac{73.575}{0.6428 + 0.45 \times 0.7660} = \frac{73.575}{0.6428 + 0.3447} = \frac{73.575}{0.9875} \approx 74.51$$ N\n\n**Final Answer:** The smallest force $P$ to maintain equilibrium is approximately $\boxed{74.5}$ N.