| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Coplanar forces in equilibrium |
| Difficulty | Standard +0.8 This is a substantial M2 framework question requiring resolution of forces at multiple joints, geometric reasoning with equilateral triangles, and systematic application of equilibrium conditions across 5 parts. While the methods are standard (resolving forces, method of joints), the multi-step nature, need for careful geometric analysis (60° angles), and the conceptual part (v) about invariance of forces elevate this above average difficulty. It's more challenging than routine statics but doesn't require exceptional 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 |
|---|---|---|
| Vertical: \(R - Y - 45 = 0\) | B1, B1 | Any form. Any form |
| Answer | Marks | Guidance |
|---|---|---|
| so \(135 - Y - 45 = 0\) and \(Y = 90\) | M1, E1, E1 | Clearly shown. Shown |
| Answer | Marks | Guidance |
|---|---|---|
| In analysis below all internal forces are taken as tensions | B1, B1 | Correct arrow pairs for all internal forces. Correct labels |
| Answer | Marks | Guidance |
|---|---|---|
| and the force in AB is \(60\sqrt{3}\) N (C) | B1, A1, F1, F1, F1, F1, B1 | At least 3 equations of resolution correct or follow through. At least 4 T/C correct |
| Answer | Marks | Guidance |
|---|---|---|
| The equilibria at C depend only on the framework geometry and the 45 N. These are not changed so forces in CB and CD are not changed | E1, E1 | Resolve in two directions at C and obtain same results as in (iv). M1A1 |
## Part (i)
Horizontal: $X - 50 = 0$
Vertical: $R - Y - 45 = 0$ | B1, B1 | Any form. Any form | 2 marks
## Part (ii)
a.c. moments about A
$1 \times R = 3 \times 45$
so $R = 135$
so $135 - Y - 45 = 0$ and $Y = 90$ | M1, E1, E1 | Clearly shown. Shown | 3 marks
## Part (iii)
In analysis below all internal forces are taken as tensions | B1, B1 | Correct arrow pairs for all internal forces. Correct labels | 2 marks
## Part (iv)
**At C**
$↑ T_{CD} \cos 30 - 45 = 0$ so $T_{CD} = 30\sqrt{3}$
and force in CD is $30\sqrt{3}$ N (T)
$← T_{BC} + T_{CD} \cos 60 = 0$ so $T_{BC} = -15\sqrt{3}$
and force in BC is $15\sqrt{3}$ N (C)
**At D**
$↓ T_{BD} \cos 30 + T_{CD} \cos 30 = 0$
so $T_{BD} = -30\sqrt{3}$
and force in BD is $30\sqrt{3}$ N (C)
$← T_{AD} + T_{BD} \cos 60 - T_{CD} \cos 60 - 50 = 0$
so $T_{AD} = 50 + 30\sqrt{3}$
and the force in AD is $50 + 30\sqrt{3}$ N (T)
**At A**
$↓ T_{AB} \cos 30 + 90 = 0$ so $T_{AB} = -60\sqrt{3}$
and the force in AB is $60\sqrt{3}$ N (C) | B1, A1, F1, F1, F1, F1, B1 | At least 3 equations of resolution correct or follow through. At least 4 T/C correct | 10 marks
## Part (v)
The equilibria at C depend only on the framework geometry and the 45 N. These are not changed so forces in CB and CD are not changed | E1, E1 | Resolve in two directions at C and obtain same results as in (iv). M1A1 | 2 marks
---\includegraphics{figure_3}
Fig. 3 shows a framework in equilibrium in a vertical plane. The framework is made from the equal, light, rigid rods AB, AD, BC, BD and CD so that ABD and BCD are equilateral triangles of side 2 m. AD and BC are horizontal.
The rods are freely pin-jointed to each other at A, B, C and D. The pin-joint at A is fixed to a wall and the pin-joint at B rests on a smooth horizontal support.
Fig. 3 also shows the external forces acting on the framework: there is a vertical load of 45 N at C and a horizontal force of 50 N applied at D; the normal reaction of the support on the framework at B is $R$ N; horizontal and vertical forces $X$ N and $Y$ N act at A.
\begin{enumerate}[label=(\roman*)]
\item Write down equations for the horizontal and vertical equilibrium of the framework. [2]
\item Show that $R = 135$ and $Y = 90$. [3]
\item On the diagram in your printed answer book, show the forces internal to the rods acting on the pin-joints. [2]
\item Calculate the forces internal to the five rods, stating whether each rod is in tension or compression (thrust). [You may leave your answers in surd form. Your working in this part should correspond to your diagram in part (iii).] [10]
\item Suppose that the force of magnitude 50 N applied at D is no longer horizontal, and the system remains in equilibrium in the same position.
By considering the equilibrium at C, show that the forces in rods CD and BC are not changed. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2011 Q3 [19]}}