| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod or block on rough surface in limiting equilibrium (no wall) |
| Difficulty | Standard +0.8 This is a challenging M3 statics problem requiring moments about multiple points, resolution of forces at a joint, and friction analysis across a two-rod system. The multi-step nature (finding equations, proving a specific result, then analyzing limiting friction) combined with the geometric complexity and the need to determine which rod slips first makes this significantly harder than average A-level questions, though it follows standard M3 techniques. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(3W + 8F = 6R\) oe | M1, A1 | Allow if sign errors, sin\(θ\) / cos\(θ\) present; dim correct. Allow sign errors |
| A1 3 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q = \frac{77}{50}W\) | M1, A1, B1, M1, A1 | Allow if sign errors, sin\(θ\) / cos\(θ\) present. Allow sign errors. \((16W + 12G = 16Q\) oe) |
| Answer | Marks | Guidance |
|---|---|---|
| A1 6 | If < 4 marks, sc B1 for \(R + Q =\) \(W + 2W\) | Or 1.54\(W\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(2W × 2l\frac{4}{5} + W\left(4l\frac{1}{5} + l\frac{3}{5}\right) - R\left(4l\frac{1}{5} + 2l\frac{3}{5}\right) = F\left(4l\frac{3}{5} - 2l\frac{1}{5}\right)\) | (M1), (A1), (A1) | Allow if sign errors, sin\(θ\) / cos\(θ\) present. Need 4 terms. Allow sign errors. Correct |
| Answer | Marks | Guidance |
|---|---|---|
| Solve with eqn from (i) | (M1) | Final A1A1 as main scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to find \(F\) (\(= G\)) = \(\frac{18}{25}W\) or \(F(= G) = \frac{36}{73}R\) | M1 | Or 0.72 W. Ft their normal forces if at least M1 in (i) and (ii). Can be implied |
| Answer | Marks | Guidance |
|---|---|---|
| Coeff of friction = \(\frac{18}{25}W ÷ \frac{73}{50}W\) | B1, M1 | Or 0.72W/1.46W |
| \(\frac{36}{73}\) or 0.493(15) | A1 4 | CAO |
## (i)
Moments about $A$ for $AB$
$W1\frac{3}{5} + F2l\frac{4}{5} = R2l\frac{3}{5}$
$3W + 8F = 6R$ oe | M1, A1 | Allow if sign errors, sin$θ$ / cos$θ$ present; dim correct. Allow sign errors | Allow sin/cos confusion.
A1 3 | CAO
## (ii)
Moments about $A$ for $AC$
$2W2l\frac{4}{5} + 64l\frac{3}{5} = 04l\frac{4}{5}$
$F = G$
Solve with eqn from (i) and $R + Q = 3W$
$R = \frac{73}{50}W$
$Q = \frac{77}{50}W$ | M1, A1, B1, M1, A1 | Allow if sign errors, sin$θ$ / cos$θ$ present. Allow sign errors. $(16W + 12G = 16Q$ oe) | $Q$ is normal reaction at $C$. Allow sin/cos confusion. $G$ is friction force at $C$
dep moments equation. Accept 1.46$W$
A1 6 | If < 4 marks, sc B1 for $R + Q =$ $W + 2W$ | Or 1.54$W$
Or moments about $C$ for whole system
$2W × 2l\frac{4}{5} + W\left(4l\frac{1}{5} + l\frac{3}{5}\right) - R\left(4l\frac{1}{5} + 2l\frac{3}{5}\right) = F\left(4l\frac{3}{5} - 2l\frac{1}{5}\right)$ | (M1), (A1), (A1) | Allow if sign errors, sin$θ$ / cos$θ$ present. Need 4 terms. Allow sign errors. Correct | dep moments equation
$(35W - 22R = 4F)$
Solve with eqn from (i) | (M1) | Final A1A1 as main scheme | dep moments equation
## (iii)
Attempt to find $F$ ($= G$) = $\frac{18}{25}W$ or $F(= G) = \frac{36}{73}R$ | M1 | Or 0.72 W. Ft their normal forces if at least M1 in (i) and (ii). Can be implied | Or use (i) with F=$μR$ and $R = \frac{73}{50}W$ M1A1. $\frac{584μW = 438W - 150W$. Solve A1; B1 as scheme
$AB$
Coeff of friction = $\frac{18}{25}W ÷ \frac{73}{50}W$ | B1, M1 | Or 0.72W/1.46W | Or use (i) with F=$μR$ and $R = \frac{73}{50}W$ M1A1
$\frac{36}{73}$ or 0.493(15) | A1 4 | CAO | May see $\frac{18}{25}W - \frac{77}{50}W$ at $C\left(\frac{36}{77}\right)$
---\includegraphics{figure_6}
Two uniform rods $AB$ and $AC$ are freely jointed at $A$. Rod $AB$ is of length $2l$ and weight $W$; rod $AC$ is of length $4l$ and weight $2W$. The rods rest in equilibrium in a vertical plane on two rough horizontal steps, so that $AB$ makes an angle of $\theta$ with the horizontal, where $\sin \theta = \frac{3}{5}$, and $AC$ makes an angle of $\varphi$ with the horizontal, where $\sin \varphi = \frac{1}{5}$ (see diagram). The force of the step acting on $AB$ at $B$ has vertical component $R$ and horizontal component $F$.
\begin{enumerate}[label=(\roman*)]
\item By taking moments about $A$ for the rod $AB$, find an equation relating $W$, $R$ and $F$. [3]
\item Show that $R = \frac{75}{68}W$, and find the vertical component of the force acting on $AC$ at $C$. [6]
\item The coefficient of friction at $B$ is equal to that at $C$. Given that one of the rods is on the point of slipping, explain which rod this must be, and find the coefficient of friction. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M3 2016 Q6 [13]}}