Edexcel M1 — Question 3 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeUniform beam on two supports
DifficultyStandard +0.3 This is a standard M1 moments equilibrium problem requiring taking moments about a point and resolving vertically. The setup is straightforward with clearly defined positions, and the equal forces condition simplifies the algebra. Part (c) requires basic conceptual understanding of tilting. Slightly easier than average due to the helpful constraint that both pivot forces are equal.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

\includegraphics{figure_1} Figure 1 shows a uniform plank \(AB\) of length 8 m and mass 30 kg. It is supported in a horizontal position by two pivots, one situated at \(A\) and the other 2 m from \(B\). A man whose mass is 80 kg is standing on the plank 2 m from \(A\) when his dog steps onto the plank at \(B\). Given that the plank remains in equilibrium and that the magnitude of the forces exerted by each of the pivots on the plank are equal,
  1. calculate the magnitude of the force exerted on the plank by the pivot at \(A\), [5 marks]
  2. find the dog's mass. [3 marks]
If the dog was heavier and the plank was on the point of tilting,
  1. explain how the force exerted on the plank by each of the pivots would be changed. [2 marks]

AnswerMarks Guidance
(a) take moments about \(B\) (as mass of dog unknown) let reaction at \(A\) and \(B\) equal \(R\); \(8R + 2R = 30g(4) + 80g(6)\) \(10R = 600g\) so \(R = 60g\) (or 588 N)M1 M2 A1 A1
(b) resolve \(\uparrow\): \(R + R = 80g + 30g + Mg\) \(120g = 110g + Mg\) \(M = 10 \text{ kg}\)M1 M1 A1
(c) no reaction at \(A\); reaction at \(B\) greater (\(80g + 30g +\) weight of dog)B2 (10)
**(a)** take moments about $B$ (as mass of dog unknown) let reaction at $A$ and $B$ equal $R$; $8R + 2R = 30g(4) + 80g(6)$ $10R = 600g$ so $R = 60g$ (or 588 N) | M1 M2 A1 A1 |

**(b)** resolve $\uparrow$: $R + R = 80g + 30g + Mg$ $120g = 110g + Mg$ $M = 10 \text{ kg}$ | M1 M1 A1 |

**(c)** no reaction at $A$; reaction at $B$ greater ($80g + 30g +$ weight of dog) | B2 | (10)

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\includegraphics{figure_1}

Figure 1 shows a uniform plank $AB$ of length 8 m and mass 30 kg. It is supported in a horizontal position by two pivots, one situated at $A$ and the other 2 m from $B$. A man whose mass is 80 kg is standing on the plank 2 m from $A$ when his dog steps onto the plank at $B$.

Given that the plank remains in equilibrium and that the magnitude of the forces exerted by each of the pivots on the plank are equal,

\begin{enumerate}[label=(\alph*)]
\item calculate the magnitude of the force exerted on the plank by the pivot at $A$, [5 marks]
\item find the dog's mass. [3 marks]
\end{enumerate}

If the dog was heavier and the plank was on the point of tilting,

\begin{enumerate}[label=(\alph*)]\setcounter{enumi}{2}
\item explain how the force exerted on the plank by each of the pivots would be changed. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q3 [10]}}