| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: reduction to single force |
| Difficulty | Challenging +1.3 This M5 question requires computing moments of forces about a point and applying the condition for reduction to a couple. Part (a) involves finding F₃ from the zero resultant condition, then calculating the couple using cross products—systematic but multi-step. Part (b) requires finding a new line of action given a different couple, involving setting up and solving vector equations. While requiring solid 3D vector mechanics and careful algebraic manipulation, the techniques are standard for M5 with no novel conceptual insight needed. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.04a Calculate moments: about a point |
Three forces $\mathbf{F}_1 = (3i - j + k)$ N, $\mathbf{F}_2 = (2i - k)$ N, and $\mathbf{F}_3$ act on a rigid body.
The force $\mathbf{F}_1$ acts through the point with position vector $(i + 2j + k)$ m, the force $\mathbf{F}_2$ acts through the point with position vector $(i - 2j)$ m and the force $\mathbf{F}_3$ acts through the point with position vector $(i + j + k)$ m.
Given that the system $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ reduces to a couple $\mathbf{G}$,
\begin{enumerate}[label=(\alph*)]
\item find $\mathbf{G}$.
[6]
\end{enumerate}
The line of action of $\mathbf{F}_3$ is changed so that the system $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ now reduces to a couple $(6i + 8j + 2k)$ N m.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find an equation of the new line of action of $\mathbf{F}_3$, giving your answer in the form $\mathbf{r} = \mathbf{a} + t\mathbf{b}$, where $\mathbf{a}$ and $\mathbf{b}$ are constant vectors.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q2 [11]}}