| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring moments about the hinge, Pythagoras to find the angle, and resolution of forces. The setup is straightforward with clear geometry, and the method is routine for M2 level—take moments to find rod force, then resolve horizontally and vertically. Slightly easier than average due to the systematic approach and clear diagram. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance of rod from hinge is \(\dfrac{2.4}{2.5}(0.7)\) or \(0.7\cos 16.26°\) \((= 0.672)\) | B1 | May be implied in moment equation |
| For taking moments about the hinge (3 term equation) | M1 | |
| \(0.672F = 68 \times 1.2 + 750 \times 2.4\) | A1 ft | |
| Force is \(2800\) N | A1 | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X = 784\) (ft for \(0.28F\)) | B1 ft | |
| For resolving vertically (4 term equation) | M1 | |
| \(Y = 1870\) (ft for \(0.96F - 818\)) | A1 ft | 3 marks total |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance of rod from hinge is $\dfrac{2.4}{2.5}(0.7)$ or $0.7\cos 16.26°$ $(= 0.672)$ | B1 | May be implied in moment equation |
| For taking moments about the hinge (3 term equation) | M1 | |
| $0.672F = 68 \times 1.2 + 750 \times 2.4$ | A1 ft | |
| Force is $2800$ N | A1 | **4 marks total** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X = 784$ (ft for $0.28F$) | B1 ft | |
| For resolving vertically (4 term equation) | M1 | |
| $Y = 1870$ (ft for $0.96F - 818$) | A1 ft | **3 marks total** |
> **SR:** For use of 680 N for weight of beam: (i) B1, M1, A0. In (ii) ft 680, so 3/3 possible.
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\includegraphics[max width=\textwidth, alt={}, center]{835616aa-0b2b-4e8c-bbbf-60b72dc5ea3e-3_737_700_264_721}
A uniform beam has length 2.4 m and weight 68 N . The beam is hinged at a fixed point of a vertical wall, and held in a horizontal position by a light rod of length 2.5 m . One end of the rod is attached to the beam at a point 0.7 m from the wall, and the other end of the rod is attached to the wall at a point vertically below the hinge. The beam carries a load of 750 N at its end (see diagram).\\
(i) Find the force in the rod.
The components of the force exerted by the hinge on the beam are $X \mathrm {~N}$ horizontally towards the wall and $Y \mathrm {~N}$ vertically downwards.\\
(ii) Find the values of $X$ and $Y$.
\hfill \mbox{\textit{CAIE M2 2004 Q4 [7]}}