M5 June 2015 Q3
3. A rigid body is in equilibrium under the action of three forces \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\)
\(\mathbf{F}_1\) and \(\mathbf{F}_2\) act at the points with position vectors \(\mathbf{r}_1\) and \(\mathbf{r}_2\) respectively, where
\(\mathbf{F}_1 = (2\mathbf{j} + \mathbf{k})\) N \(\qquad \mathbf{r}_1 = (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\) m
\(\mathbf{F}_2 = (-2\mathbf{i} - \mathbf{j})\) N \(\qquad \mathbf{r}_2 = (-\mathbf{i} - \mathbf{j} + \mathbf{k})\) m
| Scheme | Marks |
|---|---|
| \((2\mathbf{j} + \mathbf{k}) + (-2\mathbf{i} - \mathbf{j}) + \mathbf{F}_3 = \mathbf{0} \Rightarrow \mathbf{F}_3 = (2\mathbf{i} - \mathbf{j} - \mathbf{k})\) N | M1 A1 |
| Magnitude \(= \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{6}\) N | M1 A1 |
| (4) |
Notes
First M1 for \(\Sigma\mathbf{F}_i = \mathbf{0}\)
First A1 for \(\mathbf{F}_3 = (2\mathbf{i} - \mathbf{j} - \mathbf{k})\)
Second M1 for \(|\mathbf{F}_3| = \sqrt{2^2 + (-1)^2 + (-1)^2}\)
Second A1 for \(\sqrt{6}\) or 2 sf or better.
| Scheme | Marks |
|---|---|
| \((\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \times (2\mathbf{j} + \mathbf{k}) + (-\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (-2\mathbf{i} - \mathbf{j}) + (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \times (2\mathbf{i} - \mathbf{j} - \mathbf{k})\) | M1 |
| \(= -2\mathbf{i} - \mathbf{j} + 2\mathbf{k} + \mathbf{i} - 2\mathbf{j} - \mathbf{k} + (-y + z)\mathbf{i} + (2z + x)\mathbf{j} + (-x - 2y)\mathbf{k}\) | A3 |
| \(-1 - y + z = 0\) \(-3 + 2z + x = 0\) \(1 - x - 2y = 0\) | M1 A1 |
| \(x = 1,\ y = 0,\ z = 1\) is a solution | M1 |
| \(\mathbf{r} = (\mathbf{i} + \mathbf{k}) + t(2\mathbf{i} - \mathbf{j} - \mathbf{k})\) | A1 |
| (8) | |
| (12 marks) |
Notes
First M1 for consistent \(\Sigma\,\mathbf{r} \times \mathbf{F}\) or \(\Sigma\,\mathbf{F} \times \mathbf{r}\) using their \(\mathbf{F}_3\)
First A3 for correct vector products (for either of above) −1 for each incorrect product
Second M1 for equating all 3 components to zero
Fourth A1 for 3 correct equations
Third M1 for trying to find a point and getting an equation in correct form (or any other complete method)
Fifth A1 for answer (non-unique)
N.B. They could take moments about another point e.g. \(\mathbf{r}_1\) or \(\mathbf{r}_2\)
3(b) Alt (Concurrency Principle)
| \(\mathbf{r} = (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) + \lambda(2\mathbf{j} + \mathbf{k})\) \(\mathbf{r} = (-\mathbf{i} - \mathbf{j} + \mathbf{k}) + \mu(-2\mathbf{i} - \mathbf{j})\) | M1 A3 |
| \(1 + 0 = -1 - 2\mu\) \(2 + 2\lambda = -1 - \mu\) \(2 + \lambda = 1\) | M1 A1 |
| \(\Rightarrow \lambda = \mu = -1\) so point has pv \((\mathbf{i} + \mathbf{k})\) | M1 |
| \(\mathbf{r} = (\mathbf{i} + \mathbf{k}) + t(2\mathbf{i} - \mathbf{j} - \mathbf{k})\) | A1 |
First M1 for finding equations of lines of action (and later equating)
First A3 for correct equations −1 each error
Second M1 for equating all 3 components
Fourth A1 for 3 correct equations
Third M1 for trying to find a point and getting an equation in correct form (or any other complete method)
Fifth A1 for answer (non-unique)