| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Maximum or minimum mass |
| Difficulty | Standard +0.3 This is a standard M2 moments question involving tipping conditions and friction. Part (i) requires routine moment calculations about a pivot point, part (ii) is straightforward resolution of forces and moment calculation, and part (iii) combines equilibrium equations with given conditions. While multi-part, each step follows standard mechanics procedures without requiring novel insight. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(P \times 0.125 - 340 \times 0.5 = 0\) | M1 | Moments about C. All forces present. No extra forces. |
| \(P = 1360\) so 1360 N | A1, A1 | Distances correct; cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(P \times 2.125 - 340 \times (2 - 0.5) = 0\) | M1 | Moments about E. All forces present. No extra forces. |
| \(P = 240\) so 240 N | A1, A1 | Distances correct; cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q \sin \theta \times 2.125 + Q \cos \theta \times 0.9\) | M1, B1 | Moments expression. Accept \(s \leftrightarrow c\). Correct trig ratios or lengths |
| \(= \frac{2850}{13} + \frac{450}{13} = \frac{360}{13}\) so \(\frac{360}{13}\) N m | E1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| so \(Q = 221\) | M1, E1 | Moments eqn with all relevant forces; Shown |
| Answer | Marks | Guidance |
|---|---|---|
| so \(F = 85\) | M1, A1 | (blank) |
| Answer | Marks | Guidance |
|---|---|---|
| so \(R = 136\) | M1, A1 | (blank) |
| Answer | Marks | Guidance |
|---|---|---|
| so \(85 < 136\mu\) | M1, A1 | Accept \(\leq\) or =; FT their F and R |
| so \(\mu > \frac{5}{8}\) | E1 | (blank) |
## 3(i)
Let force be $P$
a.c. moments about C:
$P \times 0.125 - 340 \times 0.5 = 0$ | M1 | Moments about C. All forces present. No extra forces.
$P = 1360$ so 1360 N | A1, A1 | Distances correct; cao
## 3(i)(B)
Let force be $P$
c.w. moments about E:
$P \times 2.125 - 340 \times (2 - 0.5) = 0$ | M1 | Moments about E. All forces present. No extra forces.
$P = 240$ so 240 N | A1, A1 | Distances correct; cao
---
# Question 3 (continued)
## (ii)
$Q \sin \theta \times 2.125 + Q \cos \theta \times 0.9$ | M1, B1 | Moments expression. Accept $s \leftrightarrow c$. Correct trig ratios or lengths
$= \frac{2850}{13} + \frac{450}{13} = \frac{360}{13}$ so $\frac{360}{13}$ N m | E1 | Shown
## (iii)
We need $\frac{360}{13} = 340 \times 1.5$
so $Q = 221$ | M1, E1 | Moments eqn with all relevant forces; Shown
Let friction be $F$ and normal reaction $R$
Resolve $\to$: $221 \cos \theta - F = 0$
so $F = 85$ | M1, A1 | (blank)
Resolve $\uparrow$: $221 \sin \theta + R = 340$
so $R = 136$ | M1, A1 | (blank)
$F < \mu R$ as not on point of sliding
so $85 < 136\mu$ | M1, A1 | Accept $\leq$ or =; FT their F and R
so $\mu > \frac{5}{8}$ | E1 | (blank)
---3 A side view of a free-standing kitchen cupboard on a horizontal floor is shown in Fig. 3.1. The cupboard consists of: a base CE; vertical ends BC and DE; an overhanging horizontal top AD. The dimensions, in metres, of the cupboard are shown in the figure. The cupboard and contents have a weight of 340 N and centre of mass at G .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2aaae62-a5f3-47da-afa5-1dd4b37ea2d6-4_533_1356_477_392}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Calculate the magnitude of the vertical force required at A for the cupboard to be on the point of tipping in the cases where the force acts\\
(A) downwards,\\
(B) upwards.
A force of magnitude $Q \mathrm {~N}$ is now applied at A at an angle of $\theta$ to AB , as shown in Fig. 3.2, where $\cos \theta = \frac { 5 } { 13 } \left( \right.$ and $\left. \sin \theta = \frac { 12 } { 13 } \right)$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2aaae62-a5f3-47da-afa5-1dd4b37ea2d6-4_303_1134_1619_504}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
\item By considering the vertical and horizontal components of the force at A , show that the clockwise moment of this force about E is $\frac { 30 Q } { 13 } \mathrm { Nm }$.
With the force of magnitude $Q \mathrm {~N}$ acting as shown in Fig. 3.2, the cupboard is in equilibrium and is on the point of tipping but not on the point of sliding.
\item Show that $Q = 221$ and that the coefficient of friction between the cupboard base and the floor must be greater than $\frac { 5 } { 8 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2010 Q3 [18]}}