Edexcel M1 — Question 2 8 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeResultant force on lamina
DifficultyModerate -0.3 This is a straightforward moments question requiring systematic application of moment = force × perpendicular distance. Part (a) involves simple arithmetic with all forces at the same distance. Part (b) requires setting up an equilibrium equation and solving for x. While it has multiple steps, it uses only basic M1 concepts with no geometric complexity or novel insight required, making it slightly easier than average.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

\includegraphics{figure_1} Figure 1 shows an aerial view of a revolving door consisting of 4 panels, each of width 1.2 m and set at 90° intervals, which are free to rotate about a fixed central column, \(O\). The revolving door is situated outside a lecture theatre and four students are trying to push the door. Two of the students are pushing panels \(OA\) and \(OD\) clockwise (as viewed from above) with horizontal forces of 70 N and 90 N respectively, whilst the other two are pushing panels \(OB\) and \(OC\) anti-clockwise with horizontal forces of 80 N and 60 N respectively.
  1. Calculate the total moment about \(O\) when the four students are pushing the panels at their outer edge, 1.2 m from \(O\). [3 marks]
The student at \(C\) moves her hand 0.2 m closer to \(O\) and the student at \(D\) moves his hand \(x\) m closer to \(O\). Given that the students all push in the same directions and with the same forces as in part (a), and that the door is in equilibrium,
  1. Find the value of \(x\). [5 marks]

(a) moments about O (clockwise +ve): \(90(1.2) + 70(1.2) - 80(1.2) - 60(1.2)\)
AnswerMarks Guidance
\(= 20(1.2) = 24\) Ns (clockwise)M1 M1 A1
(b) moments about O: \(90(d) + 70(1.2) - 80(1.2) - 60(1) = 0\)
AnswerMarks Guidance
\(90d = 72\) ∴ \(a = 0.8\) m ∴ \(x = 0.4\) mM2 A1 M1 A1
**(a)** moments about O (clockwise +ve): $90(1.2) + 70(1.2) - 80(1.2) - 60(1.2)$
$= 20(1.2) = 24$ Ns (clockwise) | M1 | M1 A1 |

**(b)** moments about O: $90(d) + 70(1.2) - 80(1.2) - 60(1) = 0$
$90d = 72$ ∴ $a = 0.8$ m ∴ $x = 0.4$ m | M2 A1 | M1 A1 | (8) |

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\includegraphics{figure_1}

Figure 1 shows an aerial view of a revolving door consisting of 4 panels, each of width 1.2 m and set at 90° intervals, which are free to rotate about a fixed central column, $O$.

The revolving door is situated outside a lecture theatre and four students are trying to push the door. Two of the students are pushing panels $OA$ and $OD$ clockwise (as viewed from above) with horizontal forces of 70 N and 90 N respectively, whilst the other two are pushing panels $OB$ and $OC$ anti-clockwise with horizontal forces of 80 N and 60 N respectively.

\begin{enumerate}[label=(\alph*)]
\item Calculate the total moment about $O$ when the four students are pushing the panels at their outer edge, 1.2 m from $O$. [3 marks]
\end{enumerate}

The student at $C$ moves her hand 0.2 m closer to $O$ and the student at $D$ moves his hand $x$ m closer to $O$. Given that the students all push in the same directions and with the same forces as in part (a), and that the door is in equilibrium,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $x$. [5 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q2 [8]}}