| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod against wall and ground |
| Difficulty | Standard +0.3 This is a standard M2 statics problem requiring resolution of forces in two directions and taking moments about one point. While it involves friction at both ends and the rod being on the point of slipping, the setup is routine: students apply F=μR at both contacts, resolve horizontally and vertically for three equations, then take moments to find the center of mass position. The calculations are straightforward with the given coefficients, making this slightly easier than average for M2. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(R = F\) | B1 | Resolve horizontally |
| \(S + Q = mg\) | B1 | Resolve vertically (requires \(Q\) acting upwards) |
| \(Q = \frac{2}{3}R\), \(F = \frac{1}{4}S\) | B1 | Use both coefficients of friction |
| \(Q = \frac{2}{3}R = \frac{2}{3}\times\frac{1}{4}S\), \(S + \frac{1}{6}S = mg\), \(S = \frac{6}{7}mg\) | M1 | Solve to find \(S\) in terms of \(m\) & \(g\). (Can be scored if \(Q\) is acting downwards) |
| A1 (5) | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| M(\(A\)): \(mg\cdot x\cos60 = Q\times2l\cos60 + R\times2l\sin60\) | M1 | Moments equation – must include all terms. Condone sign errors and sin/cos confusion |
| M(\(B\)): \(mg(2l-x)\cos60 = F\times2l\sin60 + S\times2l\cos60\) | — | — |
| M(c of m): \(Sx\cos60 = Fx\sin60 + R(2l-x)\sin60 + Q(2l-x)\cos60\) | A2 | Correct unsimplified equation (for their \(S\)). -1 each error |
| \(mgx\cos60 = \frac{1}{6}\times\frac{6}{7}mg\times2l\cos60 + \frac{1}{4}\times\frac{6}{7}mg\times2l\sin60\) | DM1 | Form an equation in \(x\). Depends on the preceding M |
| \(\frac{1}{2}x = \frac{1}{7}\times2l\times\frac{1}{2} + \frac{3}{14}\times l\sqrt{3}\) | — | — |
| \(AG = x = 1.028…l \quad x = 1.03l\) | A1 (5) | 1.03\(l\) or better \(\dfrac{l(2+3\sqrt{3})}{7}\) |
# Question 3:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $R = F$ | B1 | Resolve horizontally |
| $S + Q = mg$ | B1 | Resolve vertically (requires $Q$ acting upwards) |
| $Q = \frac{2}{3}R$, $F = \frac{1}{4}S$ | B1 | Use both coefficients of friction |
| $Q = \frac{2}{3}R = \frac{2}{3}\times\frac{1}{4}S$, $S + \frac{1}{6}S = mg$, $S = \frac{6}{7}mg$ | M1 | Solve to find $S$ in terms of $m$ & $g$. (Can be scored if $Q$ is acting downwards) |
| | A1 (5) | — |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| M($A$): $mg\cdot x\cos60 = Q\times2l\cos60 + R\times2l\sin60$ | M1 | Moments equation – must include all terms. Condone sign errors and sin/cos confusion |
| M($B$): $mg(2l-x)\cos60 = F\times2l\sin60 + S\times2l\cos60$ | — | — |
| M(c of m): $Sx\cos60 = Fx\sin60 + R(2l-x)\sin60 + Q(2l-x)\cos60$ | A2 | Correct unsimplified equation (for their $S$). -1 each error |
| $mgx\cos60 = \frac{1}{6}\times\frac{6}{7}mg\times2l\cos60 + \frac{1}{4}\times\frac{6}{7}mg\times2l\sin60$ | DM1 | Form an equation in $x$. Depends on the preceding M |
| $\frac{1}{2}x = \frac{1}{7}\times2l\times\frac{1}{2} + \frac{3}{14}\times l\sqrt{3}$ | — | — |
| $AG = x = 1.028…l \quad x = 1.03l$ | A1 (5) | 1.03$l$ or better $\dfrac{l(2+3\sqrt{3})}{7}$ |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{82cadc37-4cb0-455e-9531-e09ec0c19533-05_617_604_226_678}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A non-uniform rod, $A B$, of mass $m$ and length 2l, rests in equilibrium with one end $A$ on a rough horizontal floor and the other end $B$ against a rough vertical wall. The rod is in a vertical plane perpendicular to the wall and makes an angle of $60 ^ { \circ }$ with the floor as shown in Figure 1. The coefficient of friction between the rod and the floor is $\frac { 1 } { 4 }$ and the coefficient of friction between the rod and the wall is $\frac { 2 } { 3 }$. The rod is on the point of slipping at both ends.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the vertical component of the force exerted on the rod by the floor.
The centre of mass of the rod is at $G$.
\item Find the distance $A G$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q3 [10]}}