1. Problem 1: A body of weight 800 gm.wt is on a smooth inclined plane at angle $\theta$ where $\sin \theta = \frac{6}{10}$. A horizontal force $F$ keeps it in equilibrium. Find $F$ and the reaction $R$ of the plane. 2. Step 1: Convert given data. Weight $W = 800$ gm.wt. $\sin \theta = \frac{6}{10} = 0.6$. From $\sin^2 \theta + \cos^2 \theta = 1$, $$\cos \theta = \sqrt{1 - 0.6^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8.$$ 3. Step 2: Resolve forces. Weight acts vertically downward. Force $F$ acts horizontally. Reaction $R$ acts perpendicular to the plane. 4. Step 3: Set coordinate axes along and perpendicular to the plane. Component of weight along plane: $W \sin \theta = 800 \times 0.6 = 480$ gm.wt. Component of weight perpendicular to plane: $W \cos \theta = 800 \times 0.8 = 640$ gm.wt. 5. Step 4: Express horizontal force $F$ in components along and perpendicular to the plane. Horizontal force makes angle $\theta$ with the plane normal. Component of $F$ along plane: $F \cos \theta$. Component of $F$ perpendicular to plane: $F \sin \theta$. 6. Step 5: Equilibrium conditions. Along plane: $F \cos \theta = W \sin \theta = 480$. Perpendicular to plane: $R = W \cos \theta + F \sin \theta = 640 + F \times 0.6$. 7. Step 6: Solve for $F$. $$F \cos \theta = 480 \implies F = \frac{480}{0.8} = 600 \text{ gm.wt}.$$ 8. Step 7: Calculate reaction $R$. $$R = 640 + 600 \times 0.6 = 640 + 360 = 1000 \text{ gm.wt}.$$ --- 9. Problem 2: A smooth ball of weight $90\sqrt{2}$ Newton rests on a smooth wall and is suspended by a string fixed at a point on the ball surface and at a point on the wall vertically above the contact point. Length of string is twice the radius $r$ of the ball. Find the wall reaction $R_w$ and tension $T$ in the string. 10. Step 1: Let radius of ball be $r$. String length $= 2r$. The string is fixed at the ball surface and vertically above the contact point on the wall. 11. Step 2: Geometry. The ball touches the wall at one point. The string forms a triangle with the radius and vertical line. The string length $2r$ is the hypotenuse of a right triangle with one side $r$ (radius) and the other side unknown. 12. Step 3: Find angle $\alpha$ between string and vertical wall. Using Pythagoras: $$ (2r)^2 = r^2 + x^2 \implies 4r^2 = r^2 + x^2 \implies x^2 = 3r^2 \implies x = r\sqrt{3}.$$ Angle between string and vertical is $\theta = \arctan \frac{x}{r} = \arctan \sqrt{3} = 60^\circ$. 13. Step 4: Forces on ball. Weight $W = 90\sqrt{2}$ N acts vertically down. Tension $T$ acts along string at $60^\circ$ to vertical. Wall reaction $R_w$ acts horizontally. 14. Step 5: Equilibrium conditions. Horizontal: $R_w = T \sin 60^\circ = T \times \frac{\sqrt{3}}{2}$. Vertical: $T \cos 60^\circ = W = 90\sqrt{2}$. 15. Step 6: Solve for $T$. $$T = \frac{W}{\cos 60^\circ} = \frac{90\sqrt{2}}{\frac{1}{2}} = 180\sqrt{2} \text{ N}.$$ 16. Step 7: Calculate $R_w$. $$R_w = T \times \frac{\sqrt{3}}{2} = 180\sqrt{2} \times \frac{\sqrt{3}}{2} = 90 \sqrt{6} \text{ N}.$$ Final answers: - Problem 1: $F = 600$ gm.wt, $R = 1000$ gm.wt. - Problem 2: $T = 180\sqrt{2}$ N, $R_w = 90\sqrt{6}$ N.