M5 June 2011 Q4
4. Two forces \(\mathbf{F}_1 = (3\mathbf{j} + \mathbf{k})\) N and \(\mathbf{F}_2 = (4\mathbf{i} + \mathbf{j} - \mathbf{k})\) N act on a rigid body. The force \(\mathbf{F}_1\) acts at the point with position vector \((2\mathbf{i} - \mathbf{j} + 3\mathbf{k})\) m and the force \(\mathbf{F}_2\) acts at the point with position vector \((-3\mathbf{i} + 2\mathbf{k})\) m. The two forces are equivalent to a single force \(\mathbf{R}\) acting at the point with position vector \((\mathbf{i} + 2\mathbf{j} + \mathbf{k})\) m together with a couple of moment \(\mathbf{G}\).
Find,
A third force \(\mathbf{F}_3\) is now added to the system. The force \(\mathbf{F}_3\) acts at the point with position vector \((2\mathbf{i} - \mathbf{k})\) m and the three forces \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) are equivalent to a couple.
| Scheme | Marks |
|---|---|
| \(\mathbf{R} = (3\mathbf{j} + \mathbf{k}) + (4\mathbf{i} + \mathbf{j} - \mathbf{k})\) | M1 |
| \(= (4\mathbf{i} + 4\mathbf{j})\) (N) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \((\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \times (4\mathbf{i} + 4\mathbf{j}) + \mathbf{G} = (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) \times (3\mathbf{j} + \mathbf{k}) + (-3\mathbf{i} + 2\mathbf{k}) \times (4\mathbf{i} + \mathbf{j} - \mathbf{k})\) | M1 |
| \((-4\mathbf{i} + 4\mathbf{j} - 4\mathbf{k}) + \mathbf{G} = (-10\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}) + (-2\mathbf{i} + 5\mathbf{j} - 3\mathbf{k})\) | A2 |
| \(\mathbf{G} = (-8\mathbf{i} - \mathbf{j} + 7\mathbf{k})\) (N m) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\mathbf{F}_3 = -\mathbf{R} = (-4\mathbf{i} - 4\mathbf{j})\) | B1 |
| \(\mathbf{G} = (2\mathbf{i} - \mathbf{k}) \times (-4\mathbf{i} - 4\mathbf{j}) + (-12\mathbf{i} + 3\mathbf{j} + 3\mathbf{k})\) | M1 A1 |
| \(= (-16\mathbf{i} + 7\mathbf{j} - 5\mathbf{k})\) | A1 |
| \(|\mathbf{G}| = \sqrt{(-16)^2 + 7^2 + (-5)^2}\) | M1 |
| \(= \sqrt{330}\) (N m) | A1 |
| (6) | |
| (12 marks) |