| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Topic | Moments |
| Type | Coplanar forces in equilibrium |
| Difficulty | Challenging +1.2 This M5 question requires computing moments about a point, finding equilibrium conditions, and understanding couples—all standard Further Maths mechanics techniques. Part (a) is straightforward force balance, part (b) involves cross products and solving for a line of action (procedural but multi-step), and part (c) requires understanding that a couple's moment is independent of point. While it's a substantial 16-mark question requiring careful vector manipulation, the concepts are core M5 syllabus without requiring novel insight. |
| Spec | 1.10g Problem solving with vectors: in geometry3.04b Equilibrium: zero resultant moment and force4.04g Vector product: a x b perpendicular vector |
Two forces $\mathbf{F}_1 = (2i + j)$ N and $\mathbf{F}_2 = (-2j - k)$ N act on a rigid body. The force $\mathbf{F}_1$ acts at the point with position vector $\mathbf{r}_1 = (3i + j + k)$ m and the force $\mathbf{F}_2$ acts at the point with position vector $\mathbf{r}_2 = (i - 2j)$ m. A third force $\mathbf{F}_3$ acts on the body such that $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ are in equilibrium.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of $\mathbf{F}_3$.
[4]
\end{enumerate}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find a vector equation of the line of action of $\mathbf{F}_3$.
[8]
\end{enumerate}
The force $\mathbf{F}_3$ is replaced by a fourth force $\mathbf{F}_4$, acting through the origin $O$, such that $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_4$ are equivalent to a couple.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the magnitude of this couple.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q5 [16]}}