| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on peg or cylinder |
| Difficulty | Standard +0.3 This is a standard A-level mechanics question on moments and equilibrium. Part (i) is routine horizontal rod equilibrium requiring basic moment calculations. Part (ii)(A) involves resolving forces and taking moments with friction, leading to a straightforward proof. Part (ii)(B) requires similar techniques with different friction location. All parts use standard methods with clear setups, making this slightly easier than average despite multiple parts. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let vertical force from support be \(R\) N and tension in string \(T\) N. | ||
| moments about A: \(30 \times 0.5 \times 2.4 - R \times (2.4 - 0.6) = 0\) | M1 | Use of moments with all relevant moments attempted (FT from \(T\) if \(T\) found first) |
| \(R = 20\) so force from block is 20 N | A1 | |
| \(\uparrow \; R + T - 30 = 0\) | M1 | FT from \(R\) |
| \(T = 10\) so tension is 10 N | F1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\rightarrow \; R\sin\theta - F\cos\theta = 0\) | M1 | |
| A1 | Must be consideration of a force at A | |
| As on the point of slipping \(F = 0.6R\) | M1 | \(F\) and \(R\) must be identified, e.g. on a diagram |
| so \(R\sin\theta = 0.6R\cos\theta\) so \(\sin\theta = 0.6\cos\theta\) | M1 | |
| and \(\tan\theta = 0.6\) | E1 | Complete argument |
| [5] | ||
| OR \(F = mg\sin\theta - S\sin\theta\) | M1 | Resolve parallel and perpendicular to rod |
| \(R = mg\cos\theta - S\cos\theta\) | A1 | Both correct |
| As on the point of slipping \(F = 0.6R\) | M1 | \(F\) and \(R\) must be identified, e.g. on a diagram |
| \(\dfrac{F}{R} = \dfrac{(mg-S)\sin\theta}{(mg-S)\cos\theta} = \dfrac{\sin\theta}{\cos\theta}\) | M1 | Divide factored expressions with \(S\) included |
| \(\tan\theta = 0.6\) | E1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| AP is 1.5 m gives \(\sin\theta = 0.6\) or \(\cos\theta = 0.8\) | B1 | oe. or \(\theta = 36.9°\) |
| c.w. moments about A: \(1.5R' - 30 \times 1.2 \times \cos\theta = 0\) | M1 | Moments and all terms present. Accept \(\cos\theta\) or 0.8 |
| \(R' = 19.2\) so 19.2 N | A1 | cao |
| \(\uparrow \; S' + R'\cos\theta - 30 = 0\) | M1 | An equilibrium equation with all relevant forces, resolved appropriately, e.g. \(R' + S'\cos\theta = 30\cos\theta + F'\sin\theta\). Allow \(\sin \leftrightarrow \cos\) |
| \((S' = 14.64)\) | A1 | Correct equation involving only \(S'\). Numerical answer not required |
| \(\rightarrow \; R'\sin\theta - F' = 0\) | M1 | Second equilibrium equation with all relevant forces, resolved appropriately. e.g. \(F'\cos\theta + S'\sin\theta = 30\sin\theta\). Allow \(\sin \leftrightarrow \cos\) |
| \((F' = 11.52)\) | A1 | Correct equation involving only \(F'\). Numerical answer not required |
| \(\mu = \dfrac{11.52}{14.64}\) | M1 | Use of \(F' = \mu S'\) for a calculated \(F'\) and \(S'\) |
| \(= 0.78688\ldots\) so \(0.787\) (3 s.f.) | A1 | cao |
| [9] |
# Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let vertical force from support be $R$ N and tension in string $T$ N. | | |
| moments about A: $30 \times 0.5 \times 2.4 - R \times (2.4 - 0.6) = 0$ | M1 | Use of moments with all relevant moments attempted (FT from $T$ if $T$ found first) |
| $R = 20$ so force from block is 20 N | A1 | |
| $\uparrow \; R + T - 30 = 0$ | M1 | FT from $R$ |
| $T = 10$ so tension is 10 N | F1 | |
| **[4]** | | |
---
# Question 4(ii)(A):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\rightarrow \; R\sin\theta - F\cos\theta = 0$ | M1 | |
| | A1 | Must be consideration of a force at A |
| As on the point of slipping $F = 0.6R$ | M1 | $F$ and $R$ must be identified, e.g. on a diagram |
| so $R\sin\theta = 0.6R\cos\theta$ so $\sin\theta = 0.6\cos\theta$ | M1 | |
| and $\tan\theta = 0.6$ | E1 | Complete argument |
| **[5]** | | |
| **OR** $F = mg\sin\theta - S\sin\theta$ | M1 | Resolve parallel and perpendicular to rod |
| $R = mg\cos\theta - S\cos\theta$ | A1 | Both correct |
| As on the point of slipping $F = 0.6R$ | M1 | $F$ and $R$ must be identified, e.g. on a diagram |
| $\dfrac{F}{R} = \dfrac{(mg-S)\sin\theta}{(mg-S)\cos\theta} = \dfrac{\sin\theta}{\cos\theta}$ | M1 | Divide factored expressions with $S$ included |
| $\tan\theta = 0.6$ | E1 | |
| **[5]** | | |
---
# Question 4(ii)(B):
| Answer | Marks | Guidance |
|--------|-------|----------|
| AP is 1.5 m gives $\sin\theta = 0.6$ or $\cos\theta = 0.8$ | B1 | oe. or $\theta = 36.9°$ |
| c.w. moments about A: $1.5R' - 30 \times 1.2 \times \cos\theta = 0$ | M1 | Moments and all terms present. Accept $\cos\theta$ or 0.8 |
| $R' = 19.2$ so 19.2 N | A1 | cao |
| $\uparrow \; S' + R'\cos\theta - 30 = 0$ | M1 | An equilibrium equation with all relevant forces, resolved appropriately, e.g. $R' + S'\cos\theta = 30\cos\theta + F'\sin\theta$. Allow $\sin \leftrightarrow \cos$ |
| $(S' = 14.64)$ | A1 | Correct equation involving only $S'$. Numerical answer not required |
| $\rightarrow \; R'\sin\theta - F' = 0$ | M1 | Second equilibrium equation with all relevant forces, resolved appropriately. e.g. $F'\cos\theta + S'\sin\theta = 30\sin\theta$. Allow $\sin \leftrightarrow \cos$ |
| $(F' = 11.52)$ | A1 | Correct equation involving only $F'$. Numerical answer not required |
| $\mu = \dfrac{11.52}{14.64}$ | M1 | Use of $F' = \mu S'$ for a calculated $F'$ and $S'$ |
| $= 0.78688\ldots$ so $0.787$ (3 s.f.) | A1 | cao |
| **[9]** | | |4 A rigid thin uniform rod AB with length 2.4 m and weight 30 N is used in different situations.
\begin{enumerate}[label=(\roman*)]
\item In the first situation, the rod rests on a small support 0.6 m from B and is held horizontally in equilibrium by a vertical string attached to A, as shown in Fig. 4.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42b6ee17-f0ae-4687-8392-281ba724a607-5_196_707_456_680}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
Calculate the tension in the string and the force of the support on the rod.
\item In the second situation, the rod rests in equilibrium on the point of slipping with end A on a horizontal floor and the rod resting at P on a fixed block of height 0.9 m , as shown in Fig. 4.2. The rod is perpendicular to the edge of the block on which it rests and is inclined at $\theta$ to the horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42b6ee17-f0ae-4687-8392-281ba724a607-5_208_746_1101_657}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
(A) Suppose that the contact between the block and the rod is rough with coefficient of friction 0.6 and contact between the end A and the floor is smooth.
Show that $\tan \theta = 0.6$.\\
(B) Suppose instead that the contact between the block and the rod is smooth and the contact between the end A and the floor is rough. The rod is now in limiting equilibrium at a different angle $\theta$ such that the distance AP is 1.5 m .
Calculate the normal reaction of the block on the rod.
Calculate the coefficient of friction between the rod and the floor.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2013 Q4 [18]}}