| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Standard +0.3 This is a straightforward moments problem requiring taking moments about point A to find tension (given answer to verify), then resolving forces horizontally and vertically to find the reaction at A. Standard M2 equilibrium with one geometric insight (right angle), but all steps are routine textbook techniques with no novel problem-solving required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(60T = 15 \times 30\cos\theta\) | M1 | moments about A |
| A1 | ||
| \(60T = 15 \times 30 \times 0.6\) | A1 | \(\cos\theta = 0.6\) |
| \(T = 4.5 \text{ N}\) | A1 | 4 |
| (ii) \(X = T\sin\theta\) | M1 | res. horiz. (or moments) |
| \(X = 3.6 \text{ N}\) | A1 | |
| \(Y + T\cos\theta = 15\) | M1 | res. vert. (3 terms) (or moments) |
| \(Y = 12.3 \text{ N}\) | A1 | |
| \(R = 12.8 \text{ N}\) | A1 | \(\checkmark (\text{their } X^2 + Y^2)\) |
| \(73.7°\) to horizontal | A1 | 6 |
(i) $60T = 15 \times 30\cos\theta$ | M1 | moments about A
| A1 |
$60T = 15 \times 30 \times 0.6$ | A1 | $\cos\theta = 0.6$
$T = 4.5 \text{ N}$ | A1 | 4 | AG
(ii) $X = T\sin\theta$ | M1 | res. horiz. (or moments)
$X = 3.6 \text{ N}$ | A1 |
$Y + T\cos\theta = 15$ | M1 | res. vert. (3 terms) (or moments)
$Y = 12.3 \text{ N}$ | A1 |
$R = 12.8 \text{ N}$ | A1 | $\checkmark (\text{their } X^2 + Y^2)$
$73.7°$ to horizontal | A1 | 6 | or $16.3°$ to vert. $\tan^{-1}(\text{their } Y/X)$ | 10
**or triangle of forces:** Triangle (M1) $R^2 = 15^2 + 4.5^2 - 2x4.5x15x0.6$ (M1A1) $R = 12.8$ (A1) $\sin\theta/4.5 = \sin\theta/12.8$ (M1) $\theta = 16.3°$ to vert. (A1)5\\
\includegraphics[max width=\textwidth, alt={}, center]{35477eb8-59e0-4de6-889c-1f5841f65eec-3_319_650_1219_749}
A uniform $\operatorname { rod } A B$ of length 60 cm and weight 15 N is freely suspended from its end $A$. The end $B$ of the rod is attached to a light inextensible string of length 80 cm whose other end is fixed to a point $C$ which is at the same horizontal level as $A$. The rod is in equilibrium with the string at right angles to the rod (see diagram).\\
(i) Show that the tension in the string is 4.5 N .\\
(ii) Find the magnitude and direction of the force acting on the rod at $A$.
\hfill \mbox{\textit{OCR M2 2005 Q5 [10]}}