October 2021 Paper 3 Q10
10
A block \(D\) of weight \(50\,\mathrm{N}\) lies at rest in equilibrium on a fixed rough horizontal surface. A force of magnitude \(15\,\mathrm{N}\) is applied to \(D\) at an angle \(\theta\) to the horizontal (see diagram).
(a) Complete the diagram in the Printed Answer Booklet showing all the forces acting on \(D\). [1]
It is given that \(D\) remains at rest and the coefficient of friction between \(D\) and the surface is 0.2.
(b) Show that \[15\cos\theta - 3\sin\theta \leqslant 10.\] [5]
| Scheme | Marks | AO |
|---|---|---|
 | B1 | 1.2 |
| [1] |
Notes
B1: All three forces added correctly – need not be labelled
Extra forces is B0
| Scheme | Marks | AO |
|---|---|---|
| \(F = 15\cos\theta\) | B1 | 3.3 |
| \(R = 15\sin\theta + 50\) | M1 A1 | 3.3 1.1 |
| \(15\cos\theta \leqslant \frac{1}{5}(15\sin\theta + 50)\) | M1 | 3.4 |
| \(15\cos\theta - 3\sin\theta \leqslant 10\) | A1 | 2.2a |
| [5] |
Notes
B1: Not for just seeing the expression \(15\cos\theta\)
M1: M1 for resolving vertically – allow sign errors and sin/cos confusion (three terms)
Condone inclusion of \(g\) with the 50 for the M mark only
M1: Use of \(F \leqslant \frac{1}{5}R\) with their \(F\) and \(R \neq 50\)
Allow \(F = \frac{1}{5}R\) for the M mark only
A1: AG