## 5a
| Moments about $A$: $6g(\times 1) \times \sin 70 = T \sin 30 \times 2$ | M1A2 |
| $T = 55.3$(N) or 55 (N) | A1 (4) |
| **Notes:** First M1 for a complete method to find $T$, with usual rules, correct no. of terms, allow sin/cos confusion, dim correct (missing $g$ is an A error) and allow incorrect angles; First A2 for a correct equation (A1A0 for one error); Third A1 for 55 (N) or 55.3 (N) |

## 5b
| Resolve horizontally: $T \cos 50 = R(= 35.5)$ | B1 ft |
| Resolve vertically: $T \sin 50 = 6g \pm F$ | M1A1 ft |
| $\|F\| = 16.5$ (16.473...) | |
| Use $F = \mu R$: $\mu = \frac{6g - T \sin 50}{T \cos 50}$ (with their values) | M1 |
| $= 0.464$ or 0.46 | A1 (5) |
| **Notes:** First B1 ft for resolving horizontally ($T$ does not need to be substituted); First M1 for resolving vertically with usual rules, must be using 40⁰ or 50⁰; First A1 ft for a correct equation ($T$ does not need to be substituted); Second M1, independent, for use of $F = \mu R$, must have found an $F$ and an $R$; Second A1 for 0.46 or 0.464; N.B. They may resolve in other directions e.g. along the rod or perpendicular to the rod or take moments e.g. about $B$ or $C$ |

## 5c
| Use of tan and their components: $\tan^{-1}\left(\frac{35.5}{16.5}\right)$ | M1 |
| $= 65.1°$ or 65⁰ to the upward vertical or equivalent (24.9⁰ or 25⁰ above the horizontal) | A1 (2) [11] |
| **Notes:** First M1 for a complete method to find the angle (must have found the two components) with either the horizontal or vertical; First A1 for 65⁰ or 65.1⁰ to the upward vertical oe (A0 for just an angle) or the angle marked on a clear diagram with an arrow. |

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