M5 June 2005 Q3
3. A system of forces acting on a rigid body consists of two forces \(\mathbf{F}_1\) and \(\mathbf{F}_2\) acting at a point \(A\) of the body, together with a couple of moment \(\mathbf{G}\). \(\mathbf{F}_1 = (\mathbf{i} + 2\mathbf{j} - \mathbf{k})\) N and \(\mathbf{F}_2 = (-2\mathbf{i} + \mathbf{j} + 3\mathbf{k})\) N. The position vector of the point \(A\) is \((\mathbf{i} + \mathbf{j} + \mathbf{k})\) m and \(\mathbf{G} = (7\mathbf{i} - 3\mathbf{j} + 8\mathbf{k})\) Nm.
Given that the system is equivalent to a single force \(\mathbf{R}\),
(a) find \(\mathbf{R}\), (2)
(b) find a vector equation for the line of action of \(\mathbf{R}\). (7)
| Scheme | Marks |
|---|---|
| \(\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2\) | M1 |
| \(\mathbf{R} = (-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})\) N | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| M(\(O\)): \(\begin{pmatrix}1\\1\\1\end{pmatrix}\times\begin{pmatrix}-1\\3\\2\end{pmatrix} + \begin{pmatrix}7\\-3\\8\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}\times\begin{pmatrix}-1\\3\\2\end{pmatrix}\) | M1 A2ft on \(\mathbf{R}\) |
| \(\begin{pmatrix}-1\\-3\\4\end{pmatrix} + \begin{pmatrix}7\\-3\\8\end{pmatrix} = \begin{pmatrix}2y-3z\\-z-2x\\3x+y\end{pmatrix}\) | A1 A1 |
| \(\Rightarrow\ \begin{pmatrix}6\\-6\\12\end{pmatrix} = \begin{pmatrix}2y-3z\\-z-2x\\3x+y\end{pmatrix}\) | |
| Take \(z = 0\), one solution is \(x = 3,\ y = 3,\ z = 0\) | M1 |
| \(\therefore\ \mathbf{r} = \begin{pmatrix}3\\3\\0\end{pmatrix} + \lambda\begin{pmatrix}-1\\3\\2\end{pmatrix}\) is an equation of the line of action of \(\mathbf{R}\) | A1 |
| (7) | |
| (9 marks) |
Notes
The scheme links the M1 for finding a point on the line to the first M1 of (b).