CAIE M2 2012 November — Question 2 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2012
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod or block on rough surface in limiting equilibrium (no wall)
DifficultyStandard +0.8 This is a multi-step statics problem requiring resolution of forces in two directions, taking moments about a point, and applying limiting friction conditions. While the concepts are standard A-level mechanics (equilibrium, moments, friction), the combination of an inclined rod with a force at an angle to the rod (not horizontal/vertical) requires careful geometric reasoning and systematic application of multiple equilibrium conditions across 7 marks.
Spec3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_2} A uniform rod \(AB\) has weight \(6\) N and length \(0.8\) m. The rod rests in limiting equilibrium with \(B\) in contact with a rough horizontal surface and \(AB\) inclined at \(60°\) to the horizontal. Equilibrium is maintained by a force, in the vertical plane containing \(AB\), acting at \(A\) at an angle of \(45°\) to \(AB\) (see diagram). Calculate
  1. the magnitude of the force applied at \(A\), [3]
  2. the least possible value of the coefficient of friction at \(B\). [4]

(i)
AnswerMarks Guidance
Takes moments about \(B\)M1
\(6 \times 0.4\cos60 = 0.8 P\cos45\)A1
\(P = 2.12N\)A1 [3] \(P\) is the force at \(A\)
(ii)
AnswerMarks Guidance
\(F = P\sin75\) (F is friction force at B)B1 Must use correct angle (cos15)
\(R = 6 + P\cos75\) (R is normal reaction at B)B1 Must use correct angle (sin15)
\(\mu = (2.12\sin75)/(6 + 2.12\cos75)\)M1
\(\mu = 0.313\)A1 [4]
## (i)

Takes moments about $B$ | M1 |
$6 \times 0.4\cos60 = 0.8 P\cos45$ | A1 |
$P = 2.12N$ | A1 [3] | $P$ is the force at $A$

## (ii)

$F = P\sin75$ (F is friction force at B) | B1 | Must use correct angle (cos15)
$R = 6 + P\cos75$ (R is normal reaction at B) | B1 | Must use correct angle (sin15)
$\mu = (2.12\sin75)/(6 + 2.12\cos75)$ | M1 |
$\mu = 0.313$ | A1 [4] |
\includegraphics{figure_2}

A uniform rod $AB$ has weight $6$ N and length $0.8$ m. The rod rests in limiting equilibrium with $B$ in contact with a rough horizontal surface and $AB$ inclined at $60°$ to the horizontal. Equilibrium is maintained by a force, in the vertical plane containing $AB$, acting at $A$ at an angle of $45°$ to $AB$ (see diagram). Calculate

\begin{enumerate}[label=(\roman*)]
\item the magnitude of the force applied at $A$, [3]
\item the least possible value of the coefficient of friction at $B$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2012 Q2 [7]}}