Edexcel M2 2024 October — Question 6

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2024
SessionOctober
PaperDownload PDF ↗
TopicMoments
TypeRod on smooth peg or cylinder
DifficultyStandard +0.3 This is a standard M2 moments problem requiring taking moments about a point, resolving forces, and applying friction laws. The geometry is straightforward (given tan θ), and the method is routine: moments about C to find k, then resolve horizontally/vertically and use F=μR. Slightly easier than average due to clear setup and standard technique.
Spec3.04b Equilibrium: zero resultant moment and force
6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-18_419_1307_315_379} \captionsetup{labelformat=empty} \caption{Figure 5}
\end{figure} A uniform beam \(A B\), of weight \(5 W\) and length \(12 a\), rests with end \(A\) on rough horizontal ground.
A package of weight \(W\) is attached to the beam at \(B\).
The beam rests in equilibrium on a smooth horizontal peg at \(C\), with \(A C = 9 a\), as shown in Figure 5.
The beam is inclined at an angle \(\theta\) to the ground, where \(\tan \theta = \frac { 5 } { 12 }\) The beam is modelled as a rod that lies in a vertical plane perpendicular to the peg. The package is modelled as a particle. The normal reaction between the beam and the peg at \(C\) has magnitude \(k W\) Using the model,
  1. show that \(k = \frac { 56 } { 13 }\) The coefficient of friction between \(A\) and the ground is \(\mu\) Given that the beam is resting in limiting equilibrium,
  2. find the value of \(\mu\)
Question 6:
AnswerMarks
6(a)kW
3a B
R W
9a C
5W
A θ
F
AnswerMarks
Moments about AM1
5 W 6 a c o s W 1 2 a c o s 9 a k W    +  =
 1 2 1 2 
3 0  + 1 2  = 9 k
AnswerMarks
1 3 1 3A1
A1
56
k = *
AnswerMarks
13A1*
(4)
6(a)
AnswerMarks
M1Complete method eg moments about A. All required terms present
and no extra. Dimensionally correct, product of perpendicular force
and distance (a and W present throughout). Condone sign errors and
sin/cos confusion. Condone R (or similar) instead of kW.
c
AnswerMarks
A1Unsimplified equation with at most one error
A1Correct unsimplified equation
A1*Obtain given answer from correct working. Trig replaced and R (or
c
similar) replaced in terms of k. At least one line of working required
between the equation and the given answer. Do not accept
embedded value for k.
AnswerMarks Guidance
6(b)First relevant equation M1
Correct unsimplified equation
Relevant equations:
 3 4 2 
• Vert: R k W c o s 6 W  + = R = W
1 6 9
 280 
• Horiz: F =kWsin F = W 
 169 
• // to AB: F c o s R s i n 6 W s i n    + =
• Perp to AB: k W R c o s 6 W c o s F s i n    + = +
• M(G): Rcos6a+W cos6a= Fsin6a+kW3a
• M(C): W cos3a+Rcos9a=5W cos3a+Fsin9a
AnswerMarks
• M(B): kW3a+Rcos12a=5W cos6a+Fsin12aA1
Second relevant equationM1
Correct unsimplified equationA1
Use of F R  = to form an equation in onlyDM1
2 8 0 1 4 0  
0 .8 1 8 7 ....  = = =
AnswerMarks
3 4 2 1 7 1A1
(6)
(10)
Notes
Note that an extra g in a resolution or moments equation is an
accuracy error and not a method error.
6(b)
AnswerMarks
M1First relevant equation. All required terms present and no extra.
Dimensionally consistent. Condone sign errors and sin/cos
confusion.
AnswerMarks
A1Correct unsimplified equation. Trig does not need to be replaced.
Condone R (or similar) instead of kW.
c
AnswerMarks
M1Second relevant equation. To be relevant it must be possible to use
with the first equation to find (at least one equation must be a
horizontal, vertical, parallel or perpendicular resolution). All
required terms present and no extra. Dimensionally consistent.
Condone sign errors and sin/cos confusion.
AnswerMarks
A1Correct unsimplified equation. Trig does not need to be replaced.
Condone R (or similar) instead of kW.
c
AnswerMarks
DM1Depending on the two previous M marks. Use of F R  = to reach
=…
AnswerMarks
A12 8 0  1 4 0 
= = 0 . 8 1 8 7 . . . . Accept 0.82 or better
3 4 2 1 7 1
AnswerMarks Guidance
QuestionScheme Marks
2u 3u
P Q
5m 2m
x y
Question 6:
--- 6(a) ---
6(a) | kW
3a B
R W
9a C
5W
A θ
F
Moments about A | M1
5 W 6 a c o s W 1 2 a c o s 9 a k W    +  =
 1 2 1 2 
3 0  + 1 2  = 9 k
1 3 1 3 | A1
A1
56
k = *
13 | A1*
(4)
6(a)
M1 | Complete method eg moments about A. All required terms present
and no extra. Dimensionally correct, product of perpendicular force
and distance (a and W present throughout). Condone sign errors and
sin/cos confusion. Condone R (or similar) instead of kW.
c
A1 | Unsimplified equation with at most one error
A1 | Correct unsimplified equation
A1* | Obtain given answer from correct working. Trig replaced and R (or
c
similar) replaced in terms of k. At least one line of working required
between the equation and the given answer. Do not accept
embedded value for k.
--- 6(b) ---
6(b) | First relevant equation | M1
Correct unsimplified equation
Relevant equations:
 3 4 2 
• Vert: R k W c o s 6 W  + = R = W
1 6 9
 280 
• Horiz: F =kWsin F = W 
 169 
• // to AB: F c o s R s i n 6 W s i n    + =
• Perp to AB: k W R c o s 6 W c o s F s i n    + = +
• M(G): Rcos6a+W cos6a= Fsin6a+kW3a
• M(C): W cos3a+Rcos9a=5W cos3a+Fsin9a
• M(B): kW3a+Rcos12a=5W cos6a+Fsin12a | A1
Second relevant equation | M1
Correct unsimplified equation | A1
Use of F R  = to form an equation in only | DM1
2 8 0 1 4 0  
0 .8 1 8 7 ....  = = =
3 4 2 1 7 1 | A1
(6)
(10)
Notes
Note that an extra g in a resolution or moments equation is an
accuracy error and not a method error.
6(b)
M1 | First relevant equation. All required terms present and no extra.
Dimensionally consistent. Condone sign errors and sin/cos
confusion.
A1 | Correct unsimplified equation. Trig does not need to be replaced.
Condone R (or similar) instead of kW.
c
M1 | Second relevant equation. To be relevant it must be possible to use
with the first equation to find (at least one equation must be a
horizontal, vertical, parallel or perpendicular resolution). All
required terms present and no extra. Dimensionally consistent.
Condone sign errors and sin/cos confusion.
A1 | Correct unsimplified equation. Trig does not need to be replaced.
Condone R (or similar) instead of kW.
c
DM1 | Depending on the two previous M marks. Use of F R  = to reach
=…
A1 | 2 8 0  1 4 0 
= = 0 . 8 1 8 7 . . . . Accept 0.82 or better
3 4 2 1 7 1
Question | Scheme | Marks
2u 3u
P Q
5m 2m
x y
6.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-18_419_1307_315_379}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}

A uniform beam $A B$, of weight $5 W$ and length $12 a$, rests with end $A$ on rough horizontal ground.\\
A package of weight $W$ is attached to the beam at $B$.\\
The beam rests in equilibrium on a smooth horizontal peg at $C$, with $A C = 9 a$, as shown in Figure 5.\\
The beam is inclined at an angle $\theta$ to the ground, where $\tan \theta = \frac { 5 } { 12 }$\\
The beam is modelled as a rod that lies in a vertical plane perpendicular to the peg. The package is modelled as a particle.

The normal reaction between the beam and the peg at $C$ has magnitude $k W$\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item show that $k = \frac { 56 } { 13 }$

The coefficient of friction between $A$ and the ground is $\mu$\\
Given that the beam is resting in limiting equilibrium,
\item find the value of $\mu$
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2024 Q6}}
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