| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Standard +0.3 This is a standard M2 moments question requiring taking moments about the hinge, resolving forces horizontally and vertically, and using a given angle condition. While it involves multiple parts and algebraic manipulation with parameters, the method is routine for M2 students: take moments about A, resolve in two directions, and apply the 45° condition. The 'show that' in part (a) provides a target to aim for, reducing difficulty. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Moments about \(A\): \(F\times\frac{5a}{4}=mga\cos\theta+2kmga\cos\theta\) | M1 | Requires all 3 terms. Condone trig & sign errors |
| A2 | -1 each error | |
| \(F=\frac{4mg\cos\theta}{5}(1+2k)=\frac{16}{25}mg(1+2k)\) | A1 | Substitute for \(\cos\theta\) and obtain GIVEN ANSWER |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H=F\sin\theta=\left(\frac{16}{25}mg(1+2k)\sin\theta\right)\) | M1 | Resolve horizontally |
| \(=\frac{48}{125}mg(1+2k)\) | A1 | |
| \(V=mg(1+k)-F\cos\theta\) | M1 | Resolve vertically. Need all three terms. Condone trig & sign errors |
| A1 | Correct unsimplified | |
| \(=mg(1+k)-\frac{64}{125}mg(1+2k)\left(=\frac{mg}{125}(61-3k)\right)\) | A1 | Correct unsimplified with \(k\) and \(\cos\theta\) substituted. Accept \(\pm\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{4}{5}H+\frac{3}{5}V=\frac{3}{5}mg+\frac{3}{5}kmg\) \((4H+3V=3mg(1+k))\) | M1 | Resolve parallel to the rod or perpendicular to the rod |
| \(\frac{4}{5}V+F=\frac{3}{5}H+\frac{4}{5}mg+\frac{4}{5}kmg\) \(\left(V=\frac{3}{4}H+mg(1+k)-\frac{4}{5}mg(1+2k)\right)\) | M1 | Obtain second equation in \(H\) and \(V\) and solve for \(H\) or \(V\) |
| A1 | Both equations correct unsimplified | |
| \(H=\frac{48}{125}mg(1+2k)\) | A1 | Or equivalent |
| \(V=mg(1+k)-\frac{64}{125}mg(1+2k)\left(=\frac{mg}{125}(61-3k)\right)\) | A1 | Or equivalent. Accept \(\pm\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R+F=mg(1+k)\times\frac{4}{5}\) \(\left(R=\frac{mg}{25}(4-12k)\right)\) | M1 | Component \(R\) perpendicular to the rod at \(A\) and \(L\) parallel to the rod. Attempt to find both |
| \(L=mg(1+k)\times\frac{3}{5}\) | ||
| \(H=\frac{4}{5}L-\frac{3}{5}R\), \(V=\frac{4}{5}R+\frac{3}{5}L\) | M1 A1 | Express \(V\) and \(H\) in terms of \(R\) and \(L\) |
| \(H=\frac{12mg}{25}(1+k)-\frac{3mg}{125}(4-12k)\left(=\frac{48mg}{125}(1+2k)\right)\) | A1 | Correct unsimplified |
| \(V=\frac{4}{5}\times\frac{mg}{25}(4-12k)+\frac{3}{5}\times\frac{3mg}{5}(1+k)\left(=\frac{mg}{125}(61-3k)\right)\) | A1 | Correct unsimplified. Accept \(\pm\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of \(H=V\) to form equation in \(k\) | M1 | |
| \(\frac{48}{125}mg(1+2k)=mg(1+k)-\frac{64}{125}mg(1+2k)\) | DM1 | Correct for their \(H\), \(V\) and solve for \(k\) |
| \(k=\frac{13}{99}\) | A1 | 0.13 or better \((0.1\dot{3})\) |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Moments about $A$: $F\times\frac{5a}{4}=mga\cos\theta+2kmga\cos\theta$ | M1 | Requires all 3 terms. Condone trig & sign errors |
| | A2 | -1 each error |
| $F=\frac{4mg\cos\theta}{5}(1+2k)=\frac{16}{25}mg(1+2k)$ | A1 | Substitute for $\cos\theta$ and obtain **GIVEN ANSWER** |
---
## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H=F\sin\theta=\left(\frac{16}{25}mg(1+2k)\sin\theta\right)$ | M1 | Resolve horizontally |
| $=\frac{48}{125}mg(1+2k)$ | A1 | |
| $V=mg(1+k)-F\cos\theta$ | M1 | Resolve vertically. Need all three terms. Condone trig & sign errors |
| | A1 | Correct unsimplified |
| $=mg(1+k)-\frac{64}{125}mg(1+2k)\left(=\frac{mg}{125}(61-3k)\right)$ | A1 | Correct unsimplified with $k$ and $\cos\theta$ substituted. Accept $\pm$ |
**Alt 5(b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{4}{5}H+\frac{3}{5}V=\frac{3}{5}mg+\frac{3}{5}kmg$ $(4H+3V=3mg(1+k))$ | M1 | Resolve parallel to the rod or perpendicular to the rod |
| $\frac{4}{5}V+F=\frac{3}{5}H+\frac{4}{5}mg+\frac{4}{5}kmg$ $\left(V=\frac{3}{4}H+mg(1+k)-\frac{4}{5}mg(1+2k)\right)$ | M1 | Obtain second equation in $H$ and $V$ and solve for $H$ or $V$ |
| | A1 | Both equations correct unsimplified |
| $H=\frac{48}{125}mg(1+2k)$ | A1 | Or equivalent |
| $V=mg(1+k)-\frac{64}{125}mg(1+2k)\left(=\frac{mg}{125}(61-3k)\right)$ | A1 | Or equivalent. Accept $\pm$ |
**Alt2 5(b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R+F=mg(1+k)\times\frac{4}{5}$ $\left(R=\frac{mg}{25}(4-12k)\right)$ | M1 | Component $R$ perpendicular to the rod at $A$ and $L$ parallel to the rod. Attempt to find both |
| $L=mg(1+k)\times\frac{3}{5}$ | | |
| $H=\frac{4}{5}L-\frac{3}{5}R$, $V=\frac{4}{5}R+\frac{3}{5}L$ | M1 A1 | Express $V$ and $H$ in terms of $R$ and $L$ |
| $H=\frac{12mg}{25}(1+k)-\frac{3mg}{125}(4-12k)\left(=\frac{48mg}{125}(1+2k)\right)$ | A1 | Correct unsimplified |
| $V=\frac{4}{5}\times\frac{mg}{25}(4-12k)+\frac{3}{5}\times\frac{3mg}{5}(1+k)\left(=\frac{mg}{125}(61-3k)\right)$ | A1 | Correct unsimplified. Accept $\pm$ |
---
## Question 5(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $H=V$ to form equation in $k$ | M1 | |
| $\frac{48}{125}mg(1+2k)=mg(1+k)-\frac{64}{125}mg(1+2k)$ | DM1 | Correct for their $H$, $V$ and solve for $k$ |
| $k=\frac{13}{99}$ | A1 | 0.13 or better $(0.1\dot{3})$ |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{25c503ad-94c7-4137-83b5-c3e0aea62f0c-09_636_1143_251_468}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform rod $A B$, of mass $m$ and length $2 a$, is freely hinged to a fixed point $A$. A particle of mass $k m$ is fixed to the rod at $B$. The rod is held in equilibrium, at an angle $\theta$ to the horizontal, by a force of magnitude $F$ acting at the point $C$ on the rod, where $A C = \frac { 5 } { 4 } a$, as shown in Figure 2. The line of action of the force at $C$ is at right angles to $A B$ and in the vertical plane containing $A B$.
Given that $\tan \theta = \frac { 3 } { 4 }$
\begin{enumerate}[label=(\alph*)]
\item show that $F = \frac { 16 } { 25 } m g ( 1 + 2 k )$,
\item find, in terms of $m , g$ and $k$,
\begin{enumerate}[label=(\roman*)]
\item the horizontal component of the force exerted by the hinge on the rod at $A$,
\item the vertical component of the force exerted by the hinge on the rod at $A$.
Given also that the force acting on the rod at $A$ acts at $45 ^ { \circ }$ above the horizontal,
\end{enumerate}\item find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2015 Q5 [12]}}