Moderate -0.3 This is a straightforward moments equilibrium problem requiring students to take moments about one point and apply the principle that the system is balanced. The setup is clear, the method is standard (moments about the suspension point), and the algebra is routine. It's slightly easier than average because it's a 'show that' question with a given answer, requiring only systematic application of a well-practiced technique with no problem-solving insight needed.
16 A straight uniform rod, \(A B\), has length 6 m and mass 0.2 kg
A particle of weight \(w\) newtons is fixed at \(A\).
A second particle of weight \(3 w\) newtons is fixed at \(B\).
The rod is suspended by a string from a point \(x\) metres from \(B\).
The rod rests in equilibrium with \(A B\) horizontal and the string hanging vertically as shown in the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-24_410_1148_767_445}
Show that
$$x = \frac { 3 w + 0.3 g } { 2 w + 0.1 g }$$
\includegraphics[max width=\textwidth, alt={}, center]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-25_2488_1716_219_153}
Obtains given answer showing at least the step \(x(4w+0.2g)=6w+0.6g\); OE; must have fully corrected any error involving weight and mass throughout; AG
## Question 16:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Take moments about point of suspension: $3wx = (3-x)0.2g + (6-x)w$ | M1 (AO3.3) | Forms moments equation involving $w$ and $x$ with one term correctly expressed; or moments about any other point involving $w$, $x$, $T$ |
| $4wx + 0.2gx = 6w + 0.6g \Rightarrow 2(2w+0.1g)x = 6w + 0.6g$ | M1 (AO1.1a) | Forms dimensionally correct moments equation with two terms correctly expressed |
| $x = \frac{3w + 0.3g}{2w + 0.1g}$ | A1 (AO1.1b) | Fully correct equation involving $w$, $x$ and $g$ only; if moments taken about $A$ or $B$ then $T = 4w + 0.2g$ must have been substituted |
| $x(4w + 0.2g) = 6w + 0.6g$ or $x = \frac{6w+0.6g}{4w+0.2g}$ | R1 (AO2.1) | Obtains given answer showing at least the step $x(4w+0.2g)=6w+0.6g$; OE; must have fully corrected any error involving weight and mass throughout; AG |
16 A straight uniform rod, $A B$, has length 6 m and mass 0.2 kg
A particle of weight $w$ newtons is fixed at $A$.\\
A second particle of weight $3 w$ newtons is fixed at $B$.\\
The rod is suspended by a string from a point $x$ metres from $B$.\\
The rod rests in equilibrium with $A B$ horizontal and the string hanging vertically as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-24_410_1148_767_445}
Show that
$$x = \frac { 3 w + 0.3 g } { 2 w + 0.1 g }$$
\includegraphics[max width=\textwidth, alt={}, center]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-25_2488_1716_219_153}
\hfill \mbox{\textit{AQA Paper 2 2021 Q16 [4]}}