1. **State the problem:** A body of weight $W$ gm wt is on a rough horizontal plane. A horizontal force of 100 gm wt makes it just about to move. When the plane is tilted at $45^\circ$ and a force of magnitude $150\sqrt{2}$ gm wt acts up the slope, the body is again about to slide. Find the coefficient of static friction $\mu$ between the body and the plane. 2. **Analyze the first scenario (horizontal plane):** - Weight $W$ acts vertically downward. - Normal reaction $N = W$ (since plane is horizontal). - Force $F_1 = 100$ gm wt acts horizontally. - The body is just about to move, so friction force $f_s = F_1 = 100$ gm wt. - Friction force $f_s = \mu N = \mu W$. Thus, from the first scenario: $$\mu W = 100 \implies \mu = \frac{100}{W}$$ 3. **Analyze the second scenario (inclined plane at $45^\circ$):** - The plane is inclined at $\theta = 45^\circ$. - Weight components: - Perpendicular to plane: $W \cos 45^\circ = \frac{W}{\sqrt{2}}$ - Parallel to plane (down slope): $W \sin 45^\circ = \frac{W}{\sqrt{2}}$ - Normal reaction $N = W \cos 45^\circ = \frac{W}{\sqrt{2}}$ - Force $F_2 = 150\sqrt{2}$ gm wt acts up the slope. - The body is about to slide up the slope, so friction acts down the slope. 4. **Set up equilibrium along the slope:** - Forces up slope: $F_2 = 150\sqrt{2}$ - Forces down slope: component of weight $\frac{W}{\sqrt{2}}$ plus friction $f_s = \mu N = \mu \frac{W}{\sqrt{2}}$ Since the body is about to move up, the net force up equals net force down: $$F_2 = \frac{W}{\sqrt{2}} + \mu \frac{W}{\sqrt{2}} = \frac{W}{\sqrt{2}}(1 + \mu)$$ Substitute $F_2$: $$150\sqrt{2} = \frac{W}{\sqrt{2}}(1 + \mu)$$ Multiply both sides by $\sqrt{2}$: $$150 \times 2 = W (1 + \mu) \implies 300 = W (1 + \mu)$$ 5. **Use the value of $\mu$ from step 2:** $$\mu = \frac{100}{W}$$ Substitute into equation: $$300 = W \left(1 + \frac{100}{W}\right) = W + 100$$ Solve for $W$: $$W = 300 - 100 = 200$$ 6. **Find $\mu$ using $W=200$:** $$\mu = \frac{100}{200} = 0.5$$ **Final answer:** The coefficient of static friction $\boxed{0.5}$.