OCR MEI Further Mechanics Minor 2021 November — Question 2 7 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
Year2021
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeLadder against wall
DifficultyStandard +0.3 This is a standard statics problem requiring resolution of forces in two directions and taking moments about a point. While it involves three unknowns, the solution is methodical: resolve horizontally and vertically to find W and ΞΌ, then take moments to find ΞΈ. The given contact force magnitudes make this more straightforward than typical ladder problems where you must work with unknown reactions.
Spec3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force
2 The diagram shows a uniform beam AB that rests with its end A on rough horizontal ground and its end B against a smooth vertical wall. The beam makes an angle of \(\theta ^ { \circ }\) with the ground. \includegraphics[max width=\textwidth, alt={}, center]{b3e369f4-13f7-457b-9a43-04ed2e2a2bba-3_812_588_347_246} The weight of the beam is \(W N\). The beam is in limiting equilibrium and the coefficient of friction between the beam and the ground is \(\mu\). It is given that the magnitude of the contact force at A is 70 N and the magnitude of the contact force at B is 20 N . Determine, in any order,
  • the value of \(W\),
  • the value of \(\mu\),
  • the value of \(\theta\).
Question 2:
AnswerMarks
2Let the components of the force at A be (parallel to floor)
and (parallel to wall).
AnswerMarks Guidance
𝐹𝐹𝐴𝐴B1 1.1
𝑅𝑅𝐴𝐴
AnswerMarks Guidance
𝐹𝐹𝐴𝐴 =20M1 3.1b
normal contact force at A
2 2
AnswerMarks Guidance
𝑅𝑅𝐴𝐴 =οΏ½70 βˆ’20 = o√r4 50 0=30√5A1 1.1
or 0.30
βŸΉπ‘Šπ‘Š= 𝑅𝑅𝐴𝐴= 30√5 67
𝐹𝐹𝐴𝐴 20 2 2√5
AnswerMarks Guidance
πœ‡πœ‡ =π‘…π‘…π‘Žπ‘Ž=30√5=3√5= 15B1 3.4
equivalent exact form or 0.30 (2 sf
AnswerMarks
or better)0.29814…
e.g. Taking moments about A:
AnswerMarks Guidance
π‘Šπ‘Šπ‘Šπ‘Šcosπœƒπœƒ =𝐹𝐹𝐡𝐡2π‘Šπ‘ŠsinπœƒπœƒM1* 3.3
etc.) – correct number of terms.
Allow cos/sin errors but must
AnswerMarks
reflect ratio of distances.2√5
15
β‡’tanΞΈ= 3 5
AnswerMarks Guidance
4M1dep* 1.1
value for W and then obtain a value
for tan 𝐹𝐹𝐴𝐴 =2 0
AnswerMarks Guidance
A11.1 2 sf or better
[7]
Question 2:
2 | Let the components of the force at A be (parallel to floor)
and (parallel to wall).
𝐹𝐹𝐴𝐴 | B1 | 1.1
𝑅𝑅𝐴𝐴
𝐹𝐹𝐴𝐴 =20 | M1 | 3.1b | Using Pythagoras to find the
normal contact force at A
2 2
𝑅𝑅𝐴𝐴 =οΏ½70 βˆ’20 = o√r4 50 0=30√5 | A1 | 1.1 | Accept exact or to at least 2 sf | 67.082039…
or 0.30
βŸΉπ‘Šπ‘Š= 𝑅𝑅𝐴𝐴= 30√5 67
𝐹𝐹𝐴𝐴 20 2 2√5
πœ‡πœ‡ =π‘…π‘…π‘Žπ‘Ž=30√5=3√5= 15 | B1 | 3.4 | Using F =Β΅R - accept any
equivalent exact form or 0.30 (2 sf
or better) | 0.29814…
e.g. Taking moments about A:
π‘Šπ‘Šπ‘Šπ‘Šcosπœƒπœƒ =𝐹𝐹𝐡𝐡2π‘Šπ‘Šsinπœƒπœƒ | M1* | 3.3 | Taking moments about A (or B
etc.) – correct number of terms.
Allow cos/sin errors but must
reflect ratio of distances. | 2√5
15
β‡’tanΞΈ= 3 5
4 | M1dep* | 1.1 | Substituting and their
value for W and then obtain a value
for tan 𝐹𝐹𝐴𝐴 =2 0
A1 | 1.1 | 2 sf or better | 59.19301…
[7]
2 The diagram shows a uniform beam AB that rests with its end A on rough horizontal ground and its end B against a smooth vertical wall. The beam makes an angle of $\theta ^ { \circ }$ with the ground.\\
\includegraphics[max width=\textwidth, alt={}, center]{b3e369f4-13f7-457b-9a43-04ed2e2a2bba-3_812_588_347_246}

The weight of the beam is $W N$.

The beam is in limiting equilibrium and the coefficient of friction between the beam and the ground is $\mu$.

It is given that the magnitude of the contact force at A is 70 N and the magnitude of the contact force at B is 20 N .

Determine, in any order,

\begin{itemize}
  \item the value of $W$,
  \item the value of $\mu$,
  \item the value of $\theta$.
\end{itemize}

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2021 Q2 [7]}}
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