M5 June 2013 Q4
4. Three forces \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) act on a rigid body. The forces \(\mathbf{F}_1\) and \(\mathbf{F}_2\) act through the points with position vectors \(\mathbf{r}_1\) and \(\mathbf{r}_2\) respectively.
\(\mathbf{r}_1 = (-2\mathbf{i} + 3\mathbf{j})\) m, \(\mathbf{F}_1 = (3\mathbf{i} - 2\mathbf{j} + \mathbf{k})\) N
\(\mathbf{r}_2 = (3\mathbf{i} + 2\mathbf{k})\) m, \(\mathbf{F}_2 = (-2\mathbf{i} + \mathbf{j} - \mathbf{k})\) N
Given that the system \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) is in equilibrium,
The force \(\mathbf{F}_3\) is replaced by a force \(\mathbf{F}_4\) acting through the point with position vector \((\mathbf{i} - 2\mathbf{j} + 3\mathbf{k})\) m. The system \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_4\) is equivalent to a single force \((3\mathbf{i} + \mathbf{j} + \mathbf{k})\) N acting through the point with position vector \((\mathbf{i} + \mathbf{j} + \mathbf{k})\) m together with a couple.
| Scheme | Marks |
|---|---|
| \((3\mathbf{i} - 2\mathbf{j} + \mathbf{k}) + (-2\mathbf{i} + \mathbf{j} - \mathbf{k}) + \mathbf{F}_3 = \mathbf{0}\) | M1 |
| \(\mathbf{F}_3 = (-\mathbf{i} + \mathbf{j})\) (N) | A1 |
| (2) |
Notes
M1 for \(\sum \mathbf{F}_i = \mathbf{0}\)
A1 for \((-\mathbf{i} + \mathbf{j})\)
| Scheme | Marks |
|---|---|
| \((-2\mathbf{i} + 3\mathbf{j}) \times (3\mathbf{i} - 2\mathbf{j} + \mathbf{k}) + (3\mathbf{i} + 2\mathbf{k}) \times (-2\mathbf{i} + \mathbf{j} - \mathbf{k}) + \mathbf{r} \times (-\mathbf{i} + \mathbf{j}) = \mathbf{0}\) | M1 |
| \((3\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}) + (-2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) + (-z\mathbf{i} - z\mathbf{j} + (x + y)\mathbf{k}) = \mathbf{0}\) | A2,1,0 |
| \(1 - z = 0,\ 1 - z = 0,\ -2 + x + y = 0\) | M1 |
| \(\mathbf{r} = (2\mathbf{i} + \mathbf{k}) + t(-\mathbf{i} + \mathbf{j})\) is a solution | A1 |
| OR Use Concurrency Principle | |
| (5) |
Notes
In (b) condone consistent use of F x r.
First M1 for \(M(O)\) or \(M(P_1)\) or \(M(P_2)\) (must be using correct forces)
First A1 and Second A1 -1each product.
Second M1 for changing equation to \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{F}_3\)
Third A1 for any correct equation.
OR: Use Concurrency Principle
First M1 for \(\mathbf{r}_1 + s\mathbf{F}_1 = \mathbf{r}_2 + t\mathbf{F}_2\)
First A1 for \(s = 1\) or \(t = 1\)
Second A1 for \(\mathbf{i} + \mathbf{j} + \mathbf{k}\)
Second M1 for changing equation to \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\)
Third A1 for any correct equation.
| Scheme | Marks |
|---|---|
| \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_4 = (3\mathbf{i} + \mathbf{j} + \mathbf{k}) \Rightarrow \mathbf{F}_4 = (2\mathbf{i} + 2\mathbf{j} + \mathbf{k})\) | M1 A1 |
| \(\mathbf{r}_1 \times \mathbf{F}_1 + \mathbf{r}_2 \times \mathbf{F}_2 + (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) \times (2\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = (\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (3\mathbf{i} + \mathbf{j} + \mathbf{k})\) | M1 |
| \((3\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}) + (-2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) + (-8\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}) = (2\mathbf{j} - 2\mathbf{k}) + \mathbf{G}\) | A2,1,0 ft |
| \(\mathbf{G} = (-7\mathbf{i} + 4\mathbf{j} + 6\mathbf{k})\) | A1 |
| \(|\mathbf{G}| = \sqrt{(-7)^2 + 4^2 + 6^2} = \sqrt{101}\) (Nm) | M1 A1 |
| (8) | |
| (15 marks) |
Notes
First M1 \(\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_4 = (3\mathbf{i} + \mathbf{j} + \mathbf{k})\)
First A1 for \(\mathbf{F}_4 = (2\mathbf{i} + 2\mathbf{j} + \mathbf{k})\)
Second M1 for \((\mathbf{r}_1 \times \mathbf{F}_1) + (\mathbf{r}_2 \times \mathbf{F}_2) + (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) \times (\text{their } \mathbf{F}_4) = (\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (3\mathbf{i} + \mathbf{j} + \mathbf{k})\)
Second A1ft and Third A1ft for correct equation with products evaluated (-1 ee)
Fourth A1 for \(\mathbf{G} = (-7\mathbf{i} + 4\mathbf{j} + 6\mathbf{k})\)
Third M1 for \(|\mathbf{G}| = \sqrt{(-7)^2 + 4^2 + 6^2}\)
Fifth A1 for \(\sqrt{101}\) oe (2 or more SF)
(In the printed mark scheme the ends of the second scheme line and of the Second M1 note, and the mark “A2,1,0 ft on a…”, are cut off at the edge of the table; the final bracket \((3\mathbf{i} + \mathbf{j} + \mathbf{k})\) is completed here from the question.)