M5 June 2009 Q5
5. Two forces \(\mathbf{F}_1 = (2\mathbf{i} + \mathbf{j})\) N and \(\mathbf{F}_2 = (-2\mathbf{j} - \mathbf{k})\) N act on a rigid body. The force \(\mathbf{F}_1\) acts at the point with position vector \(\mathbf{r}_1 = (3\mathbf{i} + \mathbf{j} + \mathbf{k})\) m and the force \(\mathbf{F}_2\) acts at the point with position vector \(\mathbf{r}_2 = (\mathbf{i} - 2\mathbf{j})\) 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.
(a) Find the magnitude of \(\mathbf{F}_3\). (4)
(b) Find a vector equation of the line of action of \(\mathbf{F}_3\). (8)
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.
(c) Find the magnitude of this couple. (4)
| Scheme | Marks |
|---|---|
| \((2\mathbf{i} + \mathbf{j}) + (-2\mathbf{j} - \mathbf{k}) + \mathbf{F}_3 = \mathbf{0}\) | M1 |
| \(\mathbf{F}_3 = -2\mathbf{i} + \mathbf{j} + \mathbf{k}\) | A1 |
| \(|\mathbf{F}_3| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{6}\) N | M1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \((3\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (2\mathbf{i} + \mathbf{j}) + (\mathbf{i} - 2\mathbf{j}) \times (-2\mathbf{j} - \mathbf{k}) + (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \times (-2\mathbf{i} + \mathbf{j} + \mathbf{k})\) | M1 |
| \((-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + (2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) + \big((y - z)\mathbf{i} + (-2z - x)\mathbf{j} + (x + 2y)\mathbf{k}\big)\) | A3 |
| \(y - z = -1,\ -x - 2z = -3,\ x + 2y = 1\) | DM1 |
| \(x = 1,\ y = 0,\ z = 1\) is a solution | DM1 |
| so, \(\mathbf{r} = (\mathbf{i} + \mathbf{k}) + \lambda(-2\mathbf{i} + \mathbf{j} + \mathbf{k})\) is a vector equn of line of action of \(\mathbf{F}_3\) | M1 A1 |
| (8) |
| Scheme | Marks |
|---|---|
| \((3\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (2\mathbf{i} + \mathbf{j}) + (\mathbf{i} - 2\mathbf{j}) \times (-2\mathbf{j} - \mathbf{k}) = \mathbf{G}\) | M1 |
| \((-\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + (2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = (\mathbf{i} + 3\mathbf{j} - \mathbf{k}) = \mathbf{G}\) | A1 |
| \(|\mathbf{G}| = \sqrt{1^2 + 3^2 + (-1)^2} = \sqrt{11}\) N m | M1 A1 |
| (4) | |
| (16 marks) |