| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2004 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | 3D force systems: reduction to single force |
| Difficulty | Challenging +1.2 This is a standard M5 three-dimensional moments question requiring systematic application of equilibrium conditions (sum of forces = 0) and moment calculations using cross products. While it involves 3D vectors and multiple steps, the method is routine for Further Maths students: find F₃ from force equilibrium, then calculate the couple magnitude using r × F. No novel insight required, just careful vector arithmetic. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
\begin{enumerate}
\item Three forces $\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }$ and $\mathbf { F } _ { 3 }$ act on a rigid body. $\mathbf { F } _ { 1 } = ( 12 \mathbf { i } - 4 \mathbf { j } + 6 \mathbf { k } ) \mathrm { N }$ and acts at the point with position vector $( 2 \mathbf { i } - 3 \mathbf { j } ) \mathrm { m } , \mathbf { F } _ { 2 } = ( - 3 \mathbf { j } + 2 \mathbf { k } ) \mathrm { N }$ and acts at the point with position vector $( \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }$. The force $\mathbf { F } _ { 3 }$ acts at the point with position vector $( 2 \mathbf { i } - \mathbf { k } ) \mathrm { m }$.
\end{enumerate}
Given that this set of forces is equivalent to a couple, find\\
(a) $\mathbf { F } _ { 3 }$,\\
(b) the magnitude of the couple.\\
\hfill \mbox{\textit{Edexcel M5 2004 Q1 [7]}}