1. **Problem statement:** A block weighing 39.0 N is pressed against a vertical wall by a force $F$ acting at an angle $\theta=30.0^\circ$ to the horizontal. The coefficient of kinetic friction between the block and the wall is $\mu_k=0.25$. We need to find the magnitude of $F$ when the block moves at constant velocity \n\n2. **Key concepts:** Since the block moves at constant velocity (either sliding up or sliding down), the net force in both vertical and horizontal directions is zero. The friction force acts opposite the direction of sliding and is $f_k=\mu_k N$, where $N$ is the normal force (horizontal component of $F$). \n\n3. **Set forces in components:** Let the positive $x$-direction be horizontal into the wall, positive $y$-direction vertically upward. \n- Horizontal force balance:\n$$ \sum F_x = 0 \implies N = F \cos \theta $$\n- Vertical force balance for sliding up: \n$$ \sum F_y = 0 \implies F \sin \theta = W + f_k $$\n- Vertical force balance for sliding down: \n$$ \sum F_y = 0 \implies F \sin \theta + f_k = W $$\n\n4. **Calculate normal force:**\n$$ N = F \cos \theta $$\n\n5. **Friction force:**\n$$ f_k = \mu_k N = \mu_k F \cos \theta $$\n\n6. **(a) Block sliding up:** The friction force opposes the motion (downward), so vertical balance is: \n$$ F \sin \theta = W + \mu_k F \cos \theta $$\nRearranged for $F$:\n$$ F (\sin \theta - \mu_k \cos \theta) = W $$\n$$ F = \frac{W}{\sin \theta - \mu_k \cos \theta} $$\nSubstitute values:\n$$ F = \frac{39.0}{\sin 30^\circ - 0.25 \times \cos 30^\circ} $$\nCalculate:\n$$ \sin 30^\circ = 0.5, \quad \cos 30^\circ \approx 0.866 $$\n$$ F = \frac{39.0}{0.5 - 0.25 \times 0.866} = \frac{39.0}{0.5 - 0.2165} = \frac{39.0}{0.2835} \approx 137.54 \text{ N} $$\n\n7. **(b) Block sliding down:** Friction force acts upward (opposes downward motion), so vertical balance is: \n$$ F \sin \theta + \mu_k F \cos \theta = W $$\nFactor out $F$:\n$$ F (\sin \theta + \mu_k \cos \theta) = W $$\n$$ F = \frac{W}{\sin \theta + \mu_k \cos \theta} $$\nSubstitute values:\n$$ F = \frac{39.0}{0.5 + 0.25 \times 0.866} = \frac{39.0}{0.5 + 0.2165} = \frac{39.0}{0.7165} \approx 54.44 \text{ N} $$\n\n**Final answers:**\n- (a) Force magnitude to slide block up: $F \approx 137.5$ N\n- (b) Force magnitude to slide block down: $F \approx 54.4$ N