Edexcel M1 2004 June — Question 4 11 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2004
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeBeam on point of tilting
DifficultyModerate -0.3 This is a standard M1 moments question requiring taking moments about two points and resolving vertically. The setup is straightforward with clear geometry, and part (c) uses the standard 'on the point of tilting' condition (reaction at B = 0). While it requires careful bookkeeping across multiple parts, it involves only routine application of equilibrium principles with no novel problem-solving insight.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

\includegraphics{figure_2} A plank \(AE\), of length \(6\) m and mass \(10\) kg, rests in a horizontal position on supports at \(B\) and \(D\), where \(AB = 1\) m and \(DE = 2\) m. A child of mass \(20\) kg stands at \(C\), the mid-point of \(BD\), as shown in Fig. 2. The child is modelled as a particle and the plank as a uniform rod. The child and the plank are in equilibrium. Calculate
  1. the magnitude of the force exerted by the support on the plank at \(B\), [4]
  2. the magnitude of the force exerted by the support on the plank at \(D\). [3]
The child now stands at a point \(F\) on the plank. The plank is in equilibrium and on the point of tilting about \(D\).
  1. Calculate the distance \(DF\). [4]

Question 4:
AnswerMarks
4(a) M(D): 20g x 1.5 + 10g x 1 = R x 3 M1 A1
B
⇒ R = 40g/3 ≈ 131 or 130 N M1 A1
B
(4)
[NB For moments about another point, allow M1 A1 for moments equation dimensionally
correct and with correct number of terms; second M1 is for complete method to find R .]
B
(b) R(↑): R + 40g/3 = 20g + 10g M1 A1√
D
⇒ R = 50g/3 ≈ 163 or 160 N A1
D
(3)
or M(B): 20g x 1.5 + 10g x 2 = R x 3 M1 A1
D
⇒ R = 50g/3 ≈ 163 or 160 N A1
D
(3)
[NB For moments about another point, allow M1 for a complete method to find R , A1 for a correct
D
equation for R .]
D
(c) R = 0 M1
B
M(D): 20g x x = 10g x 1 M1 A1
x = DF = 0.5 m A1
(4)
For weight/mass confusion, A0 A0 in (a) but allow f.t. in (b) (ans 50/3 = 16.7)
General rule of deducting max. 1 per question for > 3 s.f
(c) 2nd M1: must have correct no. of non=zero terms, and equation in x only
If use value(s) of R’s from (a) or (b): M0.
Question 4:
4 | (a) M(D): 20g x 1.5 + 10g x 1 = R x 3 M1 A1
B
↓
⇒ R = 40g/3 ≈ 131 or 130 N M1 A1
B
(4)
[NB For moments about another point, allow M1 A1 for moments equation dimensionally
correct and with correct number of terms; second M1 is for complete method to find R .]
B
(b) R(↑): R + 40g/3 = 20g + 10g M1 A1√
D
⇒ R = 50g/3 ≈ 163 or 160 N A1
D
(3)
or M(B): 20g x 1.5 + 10g x 2 = R x 3 M1 A1
D
⇒ R = 50g/3 ≈ 163 or 160 N A1
D
(3)
[NB For moments about another point, allow M1 for a complete method to find R , A1 for a correct
D
equation for R .]
D
(c) R = 0 M1
B
M(D): 20g x x = 10g x 1 M1 A1
x = DF = 0.5 m A1
(4)
For weight/mass confusion, A0 A0 in (a) but allow f.t. in (b) (ans 50/3 = 16.7)
General rule of deducting max. 1 per question for > 3 s.f
(c) 2nd M1: must have correct no. of non=zero terms, and equation in x only
If use value(s) of R’s from (a) or (b): M0.
\includegraphics{figure_2}

A plank $AE$, of length $6$ m and mass $10$ kg, rests in a horizontal position on supports at $B$ and $D$, where $AB = 1$ m and $DE = 2$ m. A child of mass $20$ kg stands at $C$, the mid-point of $BD$, as shown in Fig. 2. The child is modelled as a particle and the plank as a uniform rod. The child and the plank are in equilibrium. Calculate

\begin{enumerate}[label=(\alph*)]
\item the magnitude of the force exerted by the support on the plank at $B$, [4]
\item the magnitude of the force exerted by the support on the plank at $D$. [3]
\end{enumerate}

The child now stands at a point $F$ on the plank. The plank is in equilibrium and on the point of tilting about $D$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Calculate the distance $DF$. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2004 Q4 [11]}}