| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Moderate -0.3 This is a straightforward two-part statics problem requiring moment equilibrium about the hinge and force resolution. The setup is standard (uniform rod with hinge and perpendicular force), requiring only basic moment calculation (taking moments about A eliminates the reaction) and Pythagoras for the resultant force. Slightly easier than average due to the perpendicular force simplifying the moment arm and the routine nature of both parts. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(16L\cos\theta = 4 \times 2L\) → \(\theta = 60°\) or \(\pi/3°\) or \(1.05°\) | M1, A1 [2] | Moments about A, accept L = 1 |
| (ii) \(X = 4\sin60°\) and \(Y = 16 - 4\cos60°\) = \(\sqrt{(4\sin60°)^2 + (16 - 4\cos60°)^2}\) = \(14.4\) N; \(\alpha = 76.1°\) | B1, M1, A1ft, B1 [4] | \(\tan\alpha = (16 - 4\cos60°)/(4\sin60°)\); It cv(X,Y). \(\alpha = 76.1°\); R = 14.4 N |
(i) $16L\cos\theta = 4 \times 2L$ → $\theta = 60°$ or $\pi/3°$ or $1.05°$ | M1, A1 [2] | Moments about A, accept L = 1
(ii) $X = 4\sin60°$ and $Y = 16 - 4\cos60°$ = $\sqrt{(4\sin60°)^2 + (16 - 4\cos60°)^2}$ = $14.4$ N; $\alpha = 76.1°$ | B1, M1, A1ft, B1 [4] | $\tan\alpha = (16 - 4\cos60°)/(4\sin60°)$; It cv(X,Y). $\alpha = 76.1°$; R = 14.4 N1\\
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A uniform $\operatorname { rod } A B$ of weight 16 N is freely hinged at $A$ to a fixed point. A force of magnitude 4 N acting perpendicular to the rod is applied at $B$ (see diagram). Given that the rod is in equilibrium,\\
(i) calculate the angle the rod makes with the horizontal,\\
(ii) find the magnitude and direction of the force exerted on the rod at $A$.
\hfill \mbox{\textit{CAIE M2 2011 Q1 [6]}}