Edexcel M2 — Question 4 12 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod hinged to wall with string support
DifficultyStandard +0.3 This is a standard M2 moments problem requiring taking moments about the hinge, resolving forces horizontally and vertically, and finding resultant force magnitude/direction. The setup is straightforward with clear geometry (60° angle, midpoint attachment), and part (a) guides students by giving the answer to show. While it requires multiple steps and careful handling of distances and angles, it follows a well-practiced procedure with no novel insight needed, making it slightly easier than average.
Spec3.04b Equilibrium: zero resultant moment and force

4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f5449ec3-ead0-464f-9d03-f225cd21bca6-3_390_725_191_575} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} Figure 1 shows a uniform rod \(A B\) of mass 2 kg and length \(2 a\). The end \(A\) is attached by a smooth hinge to a fixed point on a vertical wall so that the rod can rotate freely in a vertical plane. A mass of 6 kg is placed at \(B\) and the rod is held in a horizontal position by a light string joining the midpoint of the rod to a point \(C\) on the wall, vertically above \(A\). The string is inclined at an angle of \(60 ^ { \circ }\) to the wall.
  1. Show that the tension in the string is \(28 g\).
  2. Find the magnitude and direction of the force exerted by the hinge on the rod, giving your answers correct to 3 significant figures.

AnswerMarks Guidance
(a)Mom. about \(A\): \(2ga + 6g(2a) - Ta\cos 60° = 0\); \(14ga = \frac{1}{2}Ta\) \(\therefore T = 28g\) M1 A1, M1 A1
(b)Resolve \(\uparrow\): \(Y + T\cos 60° - 8g = 0\) \(\therefore Y = -6g\) M1 A1
Resolve \(\rightarrow\): \(X - T\sin 60° = 0\) \(\therefore X = 14\sqrt{3}g\)M1 A1
Mag. of force at hinge: \(\sqrt{(14\sqrt{3}g)^2 + (-6g)^2} = 245\text{ N}\) (3sf)M1 A1
Req'd angle \(= \tan^{-1}\frac{-6g}{14\sqrt{3}g} = 13.9°\) (3sf) below horizontal (away from wall)M1 A1 (12 marks total)
(a) | Mom. about $A$: $2ga + 6g(2a) - Ta\cos 60° = 0$; $14ga = \frac{1}{2}Ta$ $\therefore T = 28g$ | M1 A1, M1 A1 |

(b) | Resolve $\uparrow$: $Y + T\cos 60° - 8g = 0$ $\therefore Y = -6g$ | M1 A1 |
| Resolve $\rightarrow$: $X - T\sin 60° = 0$ $\therefore X = 14\sqrt{3}g$ | M1 A1 |
| Mag. of force at hinge: $\sqrt{(14\sqrt{3}g)^2 + (-6g)^2} = 245\text{ N}$ (3sf) | M1 A1 |
| Req'd angle $= \tan^{-1}\frac{-6g}{14\sqrt{3}g} = 13.9°$ (3sf) below horizontal (away from wall) | M1 A1 | (12 marks total)

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4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f5449ec3-ead0-464f-9d03-f225cd21bca6-3_390_725_191_575}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

Figure 1 shows a uniform rod $A B$ of mass 2 kg and length $2 a$. The end $A$ is attached by a smooth hinge to a fixed point on a vertical wall so that the rod can rotate freely in a vertical plane. A mass of 6 kg is placed at $B$ and the rod is held in a horizontal position by a light string joining the midpoint of the rod to a point $C$ on the wall, vertically above $A$. The string is inclined at an angle of $60 ^ { \circ }$ to the wall.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the string is $28 g$.
\item Find the magnitude and direction of the force exerted by the hinge on the rod, giving your answers correct to 3 significant figures.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2  Q4 [12]}}