Edexcel M2 — Question 5 10 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks10
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, and finding resultant reaction. The setup is straightforward with clearly defined geometry (20° angle, horizontal force), and follows a routine method: moments equation for F, then resolution for reaction components. Slightly easier than average due to simple geometry and standard technique application.
Spec3.04b Equilibrium: zero resultant moment and force

5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{086ace58-0aa9-4f36-95c3-5698d14f511e-3_417_851_778_614} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} A uniform rod \(A B\) of length \(2 a\) and mass 8 kg is smoothly hinged to a vertical wall at \(A\). The rod is held in equilibrium inclined at an angle of \(20 ^ { \circ }\) to the horizontal by a force of magnitude \(F\) newtons acting horizontally at \(B\) which is below the level of \(A\) as shown in Figure 2.
  1. Find, correct to 3 significant figures, the value of \(F\).
  2. Show that the magnitude of the reaction at the hinge is 133 N , correct to 3 significant figures, and find to the nearest degree the acute angle which the reaction makes with the vertical.

AnswerMarks Guidance
(a) mom. about A: \(8g(\cos 20°) - F(2\sin 20°) = 0\)M1 A1
\(F = \frac{4g}{\tan 20°} = 108\) N (3sf)M1 A1
(b) resolve \(\rightarrow\): \(F + X = 0 \therefore X = 108\)A1
resolve \(\uparrow\): \(Y - 8g = 0 \therefore Y = 78.4\) NA1
mag. of reaction at hinge = \(\sqrt{(108)^2 + (78.4)^2} = 133\) N (3sf)M1 A1
req'd angle = \(\tan^{-1}\frac{108}{78.4} = 54°\) (nearest degree) to the verticalM1 A1 (10)
**(a)** mom. about A: $8g(\cos 20°) - F(2\sin 20°) = 0$ | M1 A1 |
$F = \frac{4g}{\tan 20°} = 108$ N (3sf) | M1 A1 |

**(b)** resolve $\rightarrow$: $F + X = 0 \therefore X = 108$ | A1 |
resolve $\uparrow$: $Y - 8g = 0 \therefore Y = 78.4$ N | A1 |
mag. of reaction at hinge = $\sqrt{(108)^2 + (78.4)^2} = 133$ N (3sf) | M1 A1 |
req'd angle = $\tan^{-1}\frac{108}{78.4} = 54°$ (nearest degree) to the vertical | M1 A1 | (10)
5.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{086ace58-0aa9-4f36-95c3-5698d14f511e-3_417_851_778_614}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

A uniform rod $A B$ of length $2 a$ and mass 8 kg is smoothly hinged to a vertical wall at $A$.

The rod is held in equilibrium inclined at an angle of $20 ^ { \circ }$ to the horizontal by a force of magnitude $F$ newtons acting horizontally at $B$ which is below the level of $A$ as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find, correct to 3 significant figures, the value of $F$.
\item Show that the magnitude of the reaction at the hinge is 133 N , correct to 3 significant figures, and find to the nearest degree the acute angle which the reaction makes with the vertical.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2  Q5 [10]}}