Edexcel M1 — Question 3 7 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeResultant force on lamina
DifficultyStandard +0.3 This is a standard M1 moments question requiring resolution of forces and calculation of moments about two points. The square geometry simplifies calculations, and part (b) involves straightforward algebraic manipulation with the diagonal force. Slightly easier than average due to the symmetric setup and routine application of moment principles.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e0de1908-cf67-460f-9473-b2dfded95b33-2_387_460_1626_726} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} Figure 2 shows 4 points \(A , B , C\) and \(D\) arranged such that they form the corners of a square of side 2 m . Forces of \(5 \mathrm {~N} , 3 \mathrm {~N} , 2 \mathrm {~N}\) and 4 N act in the directions \(\overrightarrow { A B } , \overrightarrow { B C } , \overrightarrow { D C }\) and \(\overrightarrow { D A }\) respectively.
  1. Calculate the magnitude and sense of the resultant moment about \(A\). An additional force of magnitude \(X\) Newtons is added in the direction \(\overrightarrow { C A }\). The resultant moment of all the forces about \(D\) is now zero.
  2. Find, in the form \(k \sqrt { } 2\), the value of \(X\).

AnswerMarks Guidance
(a) moments about \(A\) (anticlockwise +ve) \(= 3(2) - 2(2) = 2 \text{ Nm}\) (anticlockwise)M2, A1
(b) dist. of \(X\) from \(D\) is \(2\sqrt{2}\) (by Pythagoras) moments about \(D\): \(X(2\sqrt{2}) = 5(2) + 3(2)\) \(X = \frac{8}{\sqrt{2}} = 4\sqrt{2}\)M1, M1, M1 A1 (7)
| (a) moments about $A$ (anticlockwise +ve) $= 3(2) - 2(2) = 2 \text{ Nm}$ (anticlockwise) | M2, A1 | |
| (b) dist. of $X$ from $D$ is $2\sqrt{2}$ (by Pythagoras) moments about $D$: $X(2\sqrt{2}) = 5(2) + 3(2)$ $X = \frac{8}{\sqrt{2}} = 4\sqrt{2}$ | M1, M1, M1 A1 | (7) |
3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{e0de1908-cf67-460f-9473-b2dfded95b33-2_387_460_1626_726}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

Figure 2 shows 4 points $A , B , C$ and $D$ arranged such that they form the corners of a square of side 2 m . Forces of $5 \mathrm {~N} , 3 \mathrm {~N} , 2 \mathrm {~N}$ and 4 N act in the directions $\overrightarrow { A B } , \overrightarrow { B C } , \overrightarrow { D C }$ and $\overrightarrow { D A }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Calculate the magnitude and sense of the resultant moment about $A$.

An additional force of magnitude $X$ Newtons is added in the direction $\overrightarrow { C A }$. The resultant moment of all the forces about $D$ is now zero.
\item Find, in the form $k \sqrt { } 2$, the value of $X$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q3 [7]}}