1. **Problem statement:** Given forces $F_1$, $F_2$, and $F_3$ with the relation $$F_3 = 2F_2 = 3F_1,$$ and distances along the beam segments $AB=5$ cm, $BC=3$ cm, and $CD=4$ cm, we need to compare the torques of these forces about point $B$. 2. **Torque formula:** Torque $\tau$ about a point is given by $$\tau = r F \sin(\theta),$$ where $r$ is the perpendicular distance from the point to the line of action of the force, $F$ is the magnitude of the force, and $\theta$ is the angle between the force and the lever arm. 3. **Calculate distances from point B:** - For $F_1$ and $F_2$ applied at point $C$, distance $r_{BC} = 3$ cm. - For $F_3$ applied at point $D$, distance $r_{BD} = BC + CD = 3 + 4 = 7$ cm. 4. **Angles of forces relative to beam:** - $F_1$ is at $120^\circ$ from the beam at $C$. - $F_2$ is horizontal (along the beam) at $C$, so angle with beam is $0^\circ$. - $F_3$ is vertical downward at $D$, so angle with beam depends on beam orientation but since beam is horizontal, angle is $90^\circ$. 5. **Calculate torque magnitudes:** - Torque of $F_1$ about $B$: $$\tau_1 = r_{BC} F_1 \sin(120^\circ) = 3 F_1 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} F_1.$$ - Torque of $F_2$ about $B$: $$\tau_2 = r_{BC} F_2 \sin(0^\circ) = 3 F_2 \times 0 = 0.$$ - Torque of $F_3$ about $B$: $$\tau_3 = r_{BD} F_3 \sin(90^\circ) = 7 F_3 \times 1 = 7 F_3.$$ 6. **Express $F_2$ and $F_3$ in terms of $F_1$:** - From $F_3 = 3 F_1$, so $F_3 = 3 F_1$. - From $F_3 = 2 F_2$, so $F_2 = \frac{F_3}{2} = \frac{3 F_1}{2} = 1.5 F_1$. 7. **Substitute back to torques:** - $$\tau_1 = \frac{3\sqrt{3}}{2} F_1 \approx 2.598 F_1,$$ - $$\tau_2 = 0,$$ - $$\tau_3 = 7 \times 3 F_1 = 21 F_1.$$ 8. **Compare torques:** - $\tau_1 \approx 2.598 F_1$, - $\tau_3 = 21 F_1$, - $\tau_2 = 0$. Therefore: - Torque of $F_1$ around $B$ is less than torque of $F_3$. - Torque of $F_1$ around $B$ is greater than torque of $F_2$ (which is zero). **Answer:** Option (d) is correct: The torque of $F_1$ around point $B$ is less than the torque of $F_3$.