JAMB Physics 2018 Paper

Concept:The moment (or torque) of a force about a point is the turning effect it produces about that point.It is calculated as the product of the force and the perpendicular distance from the point to the line of action of the force.Moment of a force $= F \times d_{\perp}$Explanation:In this question, the force is $F = 5\,N$ and the point about which the moment is taken is $X$.From the diagram, the line of action of the $5\,N$ force acting at $Y$ passes directly through $X$.Therefore, the perpendicular distance from $X$ to the line of action of the force is $d_{\perp} = 0\,m$.The rod length $XYZ = 2\,m$ is not the perpendicular distance, so it does not contribute to the moment.When the line of action of a force passes through the pivot, the force produces no turning effect about that pivot.Substituting into the moment formula:$$M = 5\,N \times 0\,m = 0\,Nm$$Hence, the moment of the force about point $X$ is zero.Answer:A. $0\,Nm$