CAIE FP2 2014 June — Question 4

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeCoplanar forces in equilibrium
DifficultyStandard +0.3 This is a standard mechanics problem involving moments about a hinge with a uniform rod. Students need to find the normal reaction and deduce tension equals weight, then verify the angle relationship - straightforward application of equilibrium conditions with minimal algebraic complexity. Slightly easier than average as it's a guided multi-part question with clear structure.
Spec3.03a Force: vector nature and diagrams3.04a Calculate moments: about a point

4 \includegraphics[max width=\textwidth, alt={}, center]{f8961f84-c080-4407-a178-45b76f200111-3_561_606_260_767} A uniform rod \(A B\) has mass \(m\) and length \(2 d\). The rod rests in equilibrium on a smooth peg \(C\), with the end \(A\) resting on a rough horizontal plane. The distance \(A C\) is \(2 a\) and the angle between \(A B\) and the horizontal is \(\alpha\), where \(\cos \alpha = \frac { 3 } { 5 }\). A particle of mass \(\frac { 1 } { 2 } m\) is attached to the rod at \(B\) (see diagram). Find the normal reaction at \(A\) and deduce that \(d < \frac { 25 } { 6 } a\). The coefficient of friction between the rod and the plane is \(\mu\). Show that \(\mu \geqslant \frac { 8 d } { 25 a - 6 d }\).

Question 4:
EITHER method:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(R_A + R_C \cos\alpha = \frac{3mg}{2}\)B1 Resolve vertically
\(R_C \cdot 2a = mgd\cos\alpha + \frac{1}{2}mg \cdot 2d\cos\alpha\)M1 A1 Take moments about \(A\)
\(R_C = \frac{mgd}{a}\cos\alpha = \frac{3mgd}{5a}\) Eliminate \(R_C\) to find \(R_A\)
\(R_A = \frac{3mg}{2} - \frac{9mgd}{25a}\) or \(\frac{3mg(25a-6d)}{50a}\)M1 A1 A.E.F.
OR method:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(R_A\sin\alpha + F_A\cos\alpha = mg\sin\alpha + \frac{1}{2}mg\sin\alpha\)B1 Resolve along \(AB\)
\((R_A\cos\alpha)2a - (F_A\sin\alpha)2a = (mg\cos\alpha)(2a-d) - (\frac{1}{2}mg\cos\alpha)(2d-2a)\)M1 A1 Take moments about \(C\)
\(R_A = \frac{3mg(25a-6d)}{50a}\)M1 A1 Eliminate \(F_A\), A.E.F.
Answer/WorkingMarks Guidance
\(25a - 6d > 0\), \(d < \frac{25a}{6}\)B1 Find limit on \(d\) from \(R_A > 0\)
\(F_A = R_C\sin\alpha = \frac{3mgd}{5a} \cdot \frac{4}{5} = \frac{12mgd}{25a}\)M1 A1 Find \(F_A\) by horizontal resolution
\(\mu \geq \frac{8d}{(25a-6d)}\)M1 A1 Find inequality for \(\mu\) from \(F_A \leq \mu R_A\)
## Question 4:

**EITHER method:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $R_A + R_C \cos\alpha = \frac{3mg}{2}$ | B1 | Resolve vertically |
| $R_C \cdot 2a = mgd\cos\alpha + \frac{1}{2}mg \cdot 2d\cos\alpha$ | M1 A1 | Take moments about $A$ |
| $R_C = \frac{mgd}{a}\cos\alpha = \frac{3mgd}{5a}$ | | Eliminate $R_C$ to find $R_A$ |
| $R_A = \frac{3mg}{2} - \frac{9mgd}{25a}$ or $\frac{3mg(25a-6d)}{50a}$ | M1 A1 | A.E.F. |

**OR method:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $R_A\sin\alpha + F_A\cos\alpha = mg\sin\alpha + \frac{1}{2}mg\sin\alpha$ | B1 | Resolve along $AB$ |
| $(R_A\cos\alpha)2a - (F_A\sin\alpha)2a = (mg\cos\alpha)(2a-d) - (\frac{1}{2}mg\cos\alpha)(2d-2a)$ | M1 A1 | Take moments about $C$ |
| $R_A = \frac{3mg(25a-6d)}{50a}$ | M1 A1 | Eliminate $F_A$, A.E.F. |

| Answer/Working | Marks | Guidance |
|---|---|---|
| $25a - 6d > 0$, $d < \frac{25a}{6}$ | B1 | Find limit on $d$ from $R_A > 0$ | 
| $F_A = R_C\sin\alpha = \frac{3mgd}{5a} \cdot \frac{4}{5} = \frac{12mgd}{25a}$ | M1 A1 | Find $F_A$ by horizontal resolution |
| $\mu \geq \frac{8d}{(25a-6d)}$ | M1 A1 | Find inequality for $\mu$ from $F_A \leq \mu R_A$ |

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\includegraphics[max width=\textwidth, alt={}, center]{f8961f84-c080-4407-a178-45b76f200111-3_561_606_260_767}

A uniform rod $A B$ has mass $m$ and length $2 d$. The rod rests in equilibrium on a smooth peg $C$, with the end $A$ resting on a rough horizontal plane. The distance $A C$ is $2 a$ and the angle between $A B$ and the horizontal is $\alpha$, where $\cos \alpha = \frac { 3 } { 5 }$. A particle of mass $\frac { 1 } { 2 } m$ is attached to the rod at $B$ (see diagram). Find the normal reaction at $A$ and deduce that $d < \frac { 25 } { 6 } a$.

The coefficient of friction between the rod and the plane is $\mu$. Show that $\mu \geqslant \frac { 8 d } { 25 a - 6 d }$.

\hfill \mbox{\textit{CAIE FP2 2014 Q4}}