CAIE Further Paper 3 2024 November — Question 4 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod hinged to wall with rough contact at free end
DifficultyChallenging +1.8 This is a challenging statics problem requiring multiple equilibrium equations (forces and moments), geometric reasoning with the string constraint, and application of limiting friction. However, it follows a standard mechanics framework with clear setup, making it demanding but not exceptionally difficult for Further Maths students who have practiced similar problems.
Spec3.03v Motion on rough surface: including inclined planes6.04e Rigid body equilibrium: coplanar forces
\includegraphics{figure_4} The end \(A\) of a uniform rod \(AB\) of length \(6a\) and weight \(W\) is in contact with a rough vertical wall. One end of a light inextensible string of length \(3a\) is attached to the midpoint \(C\) of the rod. The other end of the string is attached to a point \(D\) on the wall, vertically above \(A\). The rod is in equilibrium when the angle between the rod and the wall is \(\theta\), where \(\tan \theta = \frac{3}{4}\). A particle of weight \(W\) is attached to the point \(E\) on the rod, where the distance \(AE\) is equal to \(ka\) (\(3 < k < 6\)) (see diagram). The rod and the string are in a vertical plane perpendicular to the wall. The coefficient of friction between the rod and the wall is \(\frac{1}{3}\). The rod is about to slip down the wall.
  1. Find the value of \(k\). [5]
  2. Find, in terms of \(W\), the magnitude of the frictional force between the rod and the wall. [2]
Question 4:
AnswerMarks
4(a)In this question, allow equivalent marks for resolutions in different directions and
moments about other points. Apply the guidance given in the main scheme.
AnswerMarks
Tcos+F =2WB1
→Tsin=RB1
Moments about A:
Tcos3asin+Tsin3acos=W3asin+Wkasin
OR
Moments about C:
AnswerMarks Guidance
R3acos=F3asin+W(ka−3a)sinM1 All relevant terms included, dimensionally correct,
forces must be resolved if appropriate.
Allow sin/cos mix, allow sign errors.
LHS: any equivalent expression, for example
3Tasin2, 3Tasin(180−2) .
All relevant terms included, dimensionally correct,
forces must be resolved if appropriate.
Allow sin/cos mix, allow sign errors.
T
[6aTcos=(3+k)aW and (3cos+sin)=2W give]  and giveand give
3
1
12cos=(3+k)(cos+ sin)
AnswerMarks Guidance
3M1 1
Use F = R and eliminate T and W to obtain an
3
expression in k and , dependent on a
dimensionally correct moments equation.
AnswerMarks
k=5A1
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
4(b)A complete method to find F in terms of W M1
For example, substitute into moments equation to
2
obtain T [T = 13 W, R=2W] .
3
2
F = W
AnswerMarks Guidance
3A1 Correct.
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
--- 4(a) ---
4(a) | In this question, allow equivalent marks for resolutions in different directions and
moments about other points. Apply the guidance given in the main scheme.
Tcos+F =2W | B1
→Tsin=R | B1
Moments about A:
Tcos3asin+Tsin3acos=W3asin+Wkasin
OR
Moments about C:
R3acos=F3asin+W(ka−3a)sin | M1 | All relevant terms included, dimensionally correct,
forces must be resolved if appropriate.
Allow sin/cos mix, allow sign errors.
LHS: any equivalent expression, for example
3Tasin2, 3Tasin(180−2) .
All relevant terms included, dimensionally correct,
forces must be resolved if appropriate.
Allow sin/cos mix, allow sign errors.
T
[6aTcos=(3+k)aW and (3cos+sin)=2W give]  and giveand give
3
1
12cos=(3+k)(cos+ sin)
3 | M1 | 1
Use F = R and eliminate T and W to obtain an
3
expression in k and , dependent on a
dimensionally correct moments equation.
k=5 | A1
5
Question | Answer | Marks | Guidance
--- 4(b) ---
4(b) | A complete method to find F in terms of W | M1 | Any complete method to find F.
For example, substitute into moments equation to
2
obtain T [T = 13 W, R=2W] .
3
2
F = W
3 | A1 | Correct.
2
Question | Answer | Marks | Guidance
\includegraphics{figure_4}

The end $A$ of a uniform rod $AB$ of length $6a$ and weight $W$ is in contact with a rough vertical wall. One end of a light inextensible string of length $3a$ is attached to the midpoint $C$ of the rod. The other end of the string is attached to a point $D$ on the wall, vertically above $A$. The rod is in equilibrium when the angle between the rod and the wall is $\theta$, where $\tan \theta = \frac{3}{4}$. A particle of weight $W$ is attached to the point $E$ on the rod, where the distance $AE$ is equal to $ka$ ($3 < k < 6$) (see diagram). The rod and the string are in a vertical plane perpendicular to the wall. The coefficient of friction between the rod and the wall is $\frac{1}{3}$. The rod is about to slip down the wall.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$. [5]
\item Find, in terms of $W$, the magnitude of the frictional force between the rod and the wall. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q4 [7]}}
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