| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam on point of tilting |
| Difficulty | Moderate -0.3 This is a standard M1 moments question with routine application of equilibrium principles. Part (a) is explanation only, part (b) uses tilting condition (one reaction = 0) with straightforward moment calculation, and part (c) applies moment equilibrium about a point to find reaction ratio. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| (a) uniform – same density throughout; rod – bench probably fairly rigid, doesn't bend very much | B1 B1 | |
| (b) [diagram showing rod on two supports with \(Mg\) at center and \(55g\) at right support] | ||
| bench on pt. of tilting so \(R = 0\); moments about \(S\): \(55g(0.3) - Mg(1.1) = 0\) | B1 M2 | |
| \(1.1M = 16.5 \therefore M = 15 \text{ kg}\) | A1 | |
| (c) [diagram showing rod on two supports with \(15g\) and \(33g\) marked] | ||
| resolve \(\uparrow\): \(R + S = 33g + 15g = 48g\) | M1 M1 | |
| moments about \(S\): \(33g(0.7) + 15g(1.1) - R(2.2) = 0\) | M1 | |
| \(2.2R = 23.1g + 16.5g \therefore R = 18g\) | M1 A1 | |
| \(S = 30g\). \(S : R = 30g : 18g = 5 : 3\) | M1 A1 | (12) |
**(a)** uniform – same density throughout; rod – bench probably fairly rigid, doesn't bend very much | B1 B1 |
**(b)** [diagram showing rod on two supports with $Mg$ at center and $55g$ at right support] |
bench on pt. of tilting so $R = 0$; moments about $S$: $55g(0.3) - Mg(1.1) = 0$ | B1 M2 |
$1.1M = 16.5 \therefore M = 15 \text{ kg}$ | A1 |
**(c)** [diagram showing rod on two supports with $15g$ and $33g$ marked] |
resolve $\uparrow$: $R + S = 33g + 15g = 48g$ | M1 M1 |
moments about $S$: $33g(0.7) + 15g(1.1) - R(2.2) = 0$ | M1 |
$2.2R = 23.1g + 16.5g \therefore R = 18g$ | M1 A1 |
$S = 30g$. $S : R = 30g : 18g = 5 : 3$ | M1 A1 | (12)6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{10b4d660-3980-4204-b18d-5240dea61a45-4_250_1036_1251_422}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Figure 2 shows a bench of length 3 m being used in a gymnasium.\\
The bench rests horizontally on two identical supports which are 2.2 m apart and equidistant from the middle of the bench.
\begin{enumerate}[label=(\alph*)]
\item Explain why it is reasonable to model the bench as a uniform rod.
When a gymnast of mass 55 kg stands on the bench 0.1 m from one end, the bench is on the point of tilting.
\item Find the mass of the bench.
The first gymnast dismounts and a second gymnast of mass 33 kg steps onto the bench at a distance of 0.4 m from its centre.
\item Show that the magnitudes of the reaction forces on the two supports are in the ratio $5 : 3$.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [12]}}