| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam on point of tilting |
| Difficulty | Standard +0.3 This is a standard moments problem requiring calculation of center of mass for composite bodies and taking moments about a pivot. The two-part structure is straightforward: part (i) uses moments equilibrium with given dimensions, part (ii) requires finding the combined center of mass and using geometry. All techniques are routine for M2 level with clear setup and no novel insight required, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(50 \times 9.8 \times 2 = R \times 3.75 + 80 \times 9.8 \times 0.25\) | M1 | moments about D. |
| A1 | SR/no g/ R = 21.3 (M1A1A0) | |
| \(R = 209 \text{ N}\) | A1 | 3 |
| (ii) \(130\bar{x} = 50 \times 2 + 80 \times 4.25\) | M1 | moments about BC or FE..... |
| A1 | \(130\bar{x} = 80 \times 0.25 + 50 \times 2.5\) | |
| \(\bar{x} = 3.385\) | A1 | \(\bar{x} = 1.115\) |
| \(130\bar{y} = 50 \times 0.125 + 80 \times 0.25\) | M1 | moments about EC |
| A1 | ||
| \(\bar{y} = 0.202\) | A1 | |
| \(\tan\theta = 0.615/0.202\) | M1 | |
| \(\theta = 71.8°\) to the horizontal | A1 | 8 |
(i) $50 \times 9.8 \times 2 = R \times 3.75 + 80 \times 9.8 \times 0.25$ | M1 | moments about D.
| A1 | SR/no g/ R = 21.3 (M1A1A0)
$R = 209 \text{ N}$ | A1 | 3 |
(ii) $130\bar{x} = 50 \times 2 + 80 \times 4.25$ | M1 | moments about BC or FE.....
| A1 | $130\bar{x} = 80 \times 0.25 + 50 \times 2.5$
$\bar{x} = 3.385$ | A1 | $\bar{x} = 1.115$
$130\bar{y} = 50 \times 0.125 + 80 \times 0.25$ | M1 | moments about EC
| A1 |
$\bar{y} = 0.202$ | A1 |
$\tan\theta = 0.615/0.202$ | M1 |
$\theta = 71.8°$ to the horizontal | A1 | 8 | $71.6°$ to $72.0°$ | 117\\
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\includegraphics[max width=\textwidth, alt={}, center]{35477eb8-59e0-4de6-889c-1f5841f65eec-4_332_1427_322_360}
A barrier is modelled as a uniform rectangular plank of wood, $A B C D$, rigidly joined to a uniform square metal plate, $D E F G$. The plank of wood has mass 50 kg and dimensions 4.0 m by 0.25 m . The metal plate has mass 80 kg and side 0.5 m . The plank and plate are joined in such a way that $C D E$ is a straight line (see diagram). The barrier is smoothly pivoted at the point $D$. In the closed position, the barrier rests on a thin post at $H$. The distance $C H$ is 0.25 m .\\
(i) Calculate the contact force at $H$ when the barrier is in the closed position.
In the open position, the centre of mass of the barrier is vertically above $D$.\\
(ii) Calculate the angle between $A B$ and the horizontal when the barrier is in the open position.
\hfill \mbox{\textit{OCR M2 2005 Q7 [11]}}