1. **State the problem:** Given $\tan \theta = \frac{4}{3}$, mass on the inclined plane $m_1 = 20$ gm, mass on the scale pan $m_2 = 7$ gm, and coefficient of static friction $\mu_s = \frac{1}{6}$. We need to find the mass $m$ on the scale pan to just vanish the friction force. 2. **Find $\sin \theta$ and $\cos \theta$:** Using $\tan \theta = \frac{4}{3}$, we form a right triangle with opposite side 4 and adjacent side 3. Hypotenuse $= \sqrt{3^2 + 4^2} = 5$. $$\sin \theta = \frac{4}{5}, \quad \cos \theta = \frac{3}{5}$$ 3. **Calculate the friction force $F_f$ on the block:** Normal force $N = m_1 g \cos \theta$ Friction force $F_f = \mu_s N = \mu_s m_1 g \cos \theta$ 4. **Calculate the component of gravitational force pulling the block down the incline:** $$F_{down} = m_1 g \sin \theta$$ 5. **Calculate the force due to the hanging mass $m_2$:** $$F_{up} = m_2 g$$ 6. **Set the net force along the incline to zero to vanish friction:** The friction force acts opposite to motion, so to vanish friction, the tension from the unknown mass $m$ must balance the difference between $F_{down}$ and $F_{up}$ plus friction. $$m g = F_{down} - F_{up} + F_f$$ Substitute values: $$m g = m_1 g \sin \theta - m_2 g + \mu_s m_1 g \cos \theta$$ Divide both sides by $g$: $$m = m_1 \sin \theta - m_2 + \mu_s m_1 \cos \theta$$ 7. **Plug in numbers:** $$m = 20 \times \frac{4}{5} - 7 + \frac{1}{6} \times 20 \times \frac{3}{5}$$ Calculate each term: $$20 \times \frac{4}{5} = 16$$ $$\frac{1}{6} \times 20 \times \frac{3}{5} = \frac{1}{6} \times 12 = 2$$ So, $$m = 16 - 7 + 2 = 11$$ **Final answer:** The mass on the scale pan to vanish friction is **11 gm**.