OCR H240/03 2018 June — Question 9 9 marks

Exam BoardOCR
ModuleH240/03 (Pure Mathematics and Mechanics)
Year2018
SessionJune
Marks9
PaperDownload PDF ↗
TopicMoments
TypeBeam on point of tilting
DifficultyStandard +0.3 This is a standard A-level mechanics question on moments and equilibrium. Parts (i)-(iii) involve straightforward applications of resolving forces vertically and taking moments about a point - routine techniques that follow directly from the setup. Part (iv) asks for standard modeling limitations. While it requires careful working through multiple parts, each step uses well-practiced methods without requiring novel insight or complex problem-solving.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

9 A uniform plank \(A B\) has weight 100 N and length 4 m . The plank rests horizontally in equilibrium on two smooth supports \(C\) and \(D\), where \(A C = x \mathrm {~m}\) and \(C D = 0.5 \mathrm {~m}\) (see diagram). \includegraphics[max width=\textwidth, alt={}, center]{d5ab20c8-afd5-473e-8238-96762bd3786d-6_181_1271_1101_395} The magnitude of the reaction of the support on the plank at \(C\) is 75 N . Modelling the plank as a rigid rod, find
  1. the magnitude of the reaction of the support on the plank at \(D\),
  2. the value of \(x\). A stone block, which is modelled as a particle, is now placed at the end of the plank at \(B\) and the plank is on the point of tilting about \(D\).
  3. Find the weight of the stone block.
  4. Explain the limitation of modelling
    1. the stone block as a particle,
    2. the plank as a rigid rod.

Question 9:
Part (i):
AnswerMarks
\(25\text{N}\)B1 [1]
Part (ii):
AnswerMarks Guidance
\(2(100) = 75x + (x+0.5)(25)\)A1ft Follow through their 25 only
\(x = 1.875\)A1 [3] eg moments about \(A\) – correct number of terms
Part (iii):
AnswerMarks Guidance
\((x + 0.5 - 2)(100) = W(4 - 0.5 - x)\)A1ft Follow through their \(x\) only
\(W = 23.1\text{N}\)A1 [3] Accept 23 or better; moments about \(D\) – correct number of terms
## Question 9:

### Part (i):
$25\text{N}$ | B1 [1] | |

### Part (ii):
$2(100) = 75x + (x+0.5)(25)$ | A1ft | Follow through their 25 only |

$x = 1.875$ | A1 [3] | eg moments about $A$ – correct number of terms |

### Part (iii):
$(x + 0.5 - 2)(100) = W(4 - 0.5 - x)$ | A1ft | Follow through their $x$ only |

$W = 23.1\text{N}$ | A1 [3] | Accept 23 or better; moments about $D$ – correct number of terms |

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9 A uniform plank $A B$ has weight 100 N and length 4 m . The plank rests horizontally in equilibrium on two smooth supports $C$ and $D$, where $A C = x \mathrm {~m}$ and $C D = 0.5 \mathrm {~m}$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{d5ab20c8-afd5-473e-8238-96762bd3786d-6_181_1271_1101_395}

The magnitude of the reaction of the support on the plank at $C$ is 75 N . Modelling the plank as a rigid rod, find\\
(i) the magnitude of the reaction of the support on the plank at $D$,\\
(ii) the value of $x$.

A stone block, which is modelled as a particle, is now placed at the end of the plank at $B$ and the plank is on the point of tilting about $D$.\\
(iii) Find the weight of the stone block.\\
(iv) Explain the limitation of modelling
\begin{enumerate}[label=(\alph*)]
\item the stone block as a particle,
\item the plank as a rigid rod.
\end{enumerate}

\hfill \mbox{\textit{OCR H240/03 2018 Q9 [9]}}