Challenging +1.2 This is a standard Further Maths mechanics problem requiring moments about a point, resolution of forces, and friction at limiting equilibrium. While it involves multiple steps and careful geometry with the 45° angle, the techniques are routine for FM students: taking moments about C to eliminate the reaction there, applying F=μR, and algebraic manipulation. The 'show that' part (ii) requires finding when x is physically valid (x≤2a), which adds modest problem-solving beyond pure calculation. Overall, moderately above average difficulty due to the multi-part nature and FM context, but follows standard equilibrium methods without requiring novel insight.
\includegraphics{figure_4}
A uniform rod \(AB\) of length \(2a\) and weight \(W\) rests against a smooth horizontal peg at a point \(C\) on the rod, where \(AC = x\). The lower end \(A\) of the rod rests on rough horizontal ground. The rod is in equilibrium inclined at an angle of \(45°\) to the horizontal (see diagram). The coefficient of friction between the rod and the ground is \(\mu\). The rod is about to slip at \(A\).
\begin{enumerate}[label=(\roman*)]
\item Find an expression for \(x\) in terms of \(a\) and \(\mu\).
[5]
\item Hence show that \(\mu \geqslant \frac{1}{3}\).
[2]
\item Given that \(x = \frac{5}{3}a\), find the value of \(\mu\) and the magnitude of the resultant force on the rod at \(A\).
[4]
\end{enumerate]
C: F cos 45° × x – R cos 45° × x + W cos 45° × (x – a) = 0
A A
G: F cos 45° × a – R cos 45° × a + R × (x – a) = 0
A A C
D: R cos 45° × a – R cos 45° × x cos 45°
A C
– R cos 45° × (x – a) cos 45° = 0
Answer
Marks
Guidance
C
B1
Take moments for rod about one chosen point
(F × may be replaced by µR and cos 45° by e.g. 1/√2)
A A
(A single resolution along the rod will then suffice since no R )
C
(G is mid-point of AB)
(D is on gound below G)
Horizontally: F – R cos 45° = 0
Answer
Marks
Guidance
A C
B1
Find two more indep. eqns, e.g. resolution of forces on rod
Vertically: R + R cos 45° – W = 0
Answer
Marks
Guidance
A C
B1
(a second moment eqn. may be used)
Along rod: F cos 45° + R cos 45° – W cos 45° = 0
Answer
Marks
A A
(B1
Perp. to rod: F cos 45° – R cos 45° – R + W cos 45° = 0
Answer
Marks
A A C
B1)
[R = W / (1+µ), F = µW / (1+µ), R = µW√ 2 / (1+µ)]
A A C
Answer
Marks
Guidance
x = a (1+µ) / 2µ or ½ a (1 + 1/µ)
M1A1
Combine to find x (using F = µR and cos 45° = 1/√2)
A A
5
Answer
Marks
Guidance
4(ii)
a (1+µ) / 2µ ⩽ 2a so µ ⩾ ⅓ AG
M1A1
2
Answer
Marks
Guidance
4(iii)
a (1+µ) / 2µ = 3a/2 so µ = ½
M1A1
F = W/3, R = 2W/3 [R = (√2) W/3]
A A C
N = √( F 2 + R 2) = (√5/3) W or 0.745 W
Answer
Marks
Guidance
A A A
M1A1
Find F , R and hence magnitude of resultant force N at A
A A A
4
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
--- 4(i) ---
4(i) | A: R × x – W cos 45° × a = 0
C
B: F cos 45° × 2a – R cos 45° × 2a – R × (2a – x)
A A C
+ W cos 45° × a = 0
C: F cos 45° × x – R cos 45° × x + W cos 45° × (x – a) = 0
A A
G: F cos 45° × a – R cos 45° × a + R × (x – a) = 0
A A C
D: R cos 45° × a – R cos 45° × x cos 45°
A C
– R cos 45° × (x – a) cos 45° = 0
C | B1 | Take moments for rod about one chosen point
(F × may be replaced by µR and cos 45° by e.g. 1/√2)
A A
(A single resolution along the rod will then suffice since no R )
C
(G is mid-point of AB)
(D is on gound below G)
Horizontally: F – R cos 45° = 0
A C | B1 | Find two more indep. eqns, e.g. resolution of forces on rod
Vertically: R + R cos 45° – W = 0
A C | B1 | (a second moment eqn. may be used)
Along rod: F cos 45° + R cos 45° – W cos 45° = 0
A A | (B1
Perp. to rod: F cos 45° – R cos 45° – R + W cos 45° = 0
A A C | B1)
[R = W / (1+µ), F = µW / (1+µ), R = µW√ 2 / (1+µ)]
A A C
x = a (1+µ) / 2µ or ½ a (1 + 1/µ) | M1A1 | Combine to find x (using F = µR and cos 45° = 1/√2)
A A
5
--- 4(ii) ---
4(ii) | a (1+µ) / 2µ ⩽ 2a so µ ⩾ ⅓ AG | M1A1 | Verify µ using x ⩽ 2a
2
--- 4(iii) ---
4(iii) | a (1+µ) / 2µ = 3a/2 so µ = ½ | M1A1 | Find µ when x = 3a/2 using result in (i)
F = W/3, R = 2W/3 [R = (√2) W/3]
A A C
N = √( F 2 + R 2) = (√5/3) W or 0.745 W
A A A | M1A1 | Find F , R and hence magnitude of resultant force N at A
A A A
4
Question | Answer | Marks | Guidance
\includegraphics{figure_4}
A uniform rod $AB$ of length $2a$ and weight $W$ rests against a smooth horizontal peg at a point $C$ on the rod, where $AC = x$. The lower end $A$ of the rod rests on rough horizontal ground. The rod is in equilibrium inclined at an angle of $45°$ to the horizontal (see diagram). The coefficient of friction between the rod and the ground is $\mu$. The rod is about to slip at $A$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $x$ in terms of $a$ and $\mu$.
[5]
\item Hence show that $\mu \geqslant \frac{1}{3}$.
[2]
\item Given that $x = \frac{5}{3}a$, find the value of $\mu$ and the magnitude of the resultant force on the rod at $A$.
[4]
\end{enumerate]
\hfill \mbox{\textit{CAIE FP2 2018 Q4 [11]}}