OCR M2 2012 January — Question 3 8 marks

Exam BoardOCR
ModuleM2 (Mechanics 2)
Year2012
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod on smooth peg or cylinder
DifficultyStandard +0.3 This is a standard M2 statics problem requiring resolution of forces and taking moments about a point. The setup is straightforward with given values, requiring routine application of equilibrium conditions (ΣF=0, Σmoments=0) and friction inequality μR≥F. While it involves multiple steps and careful geometry, it follows a well-practiced template with no novel insight required, making it slightly easier than average.
Spec3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces6.04e Rigid body equilibrium: coplanar forces

3 \includegraphics[max width=\textwidth, alt={}, center]{5addd79d-d502-455c-936f-27005483164e-3_483_787_260_641} A uniform rod \(A B\) of mass 10 kg and length 2.4 m rests with \(A\) on rough horizontal ground. The rod makes an angle of \(60 ^ { \circ }\) with the horizontal and is supported by a fixed smooth peg \(P\). The distance \(A P\) is 1.6 m (see diagram).
  1. Calculate the magnitude of the force exerted by the peg on the rod.
  2. Find the least value of the coefficient of friction between the rod and the ground needed to maintain equilibrium.

Question 3(i):
AnswerMarks Guidance
AnswerMarks Guidance
M1Moments about \(A\)
\(P \times 1.6 = 10g\cos60 \times 1.2\)A1
\(P = 36.75\) NA1 Allow 36.8
Question 3(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(R + 36.75\sin30 = 10g\)M1 Attempt at resolving vertically or taking moments
A1 FTMay be implied. \(R = 79.6(25)\)
\(F = 36.75\cos30\)B1 FT Expect 31.8. Or second correct equation involving \(F\) or \(R\) or both
\(\mu = 31.8/79.6\)M1 For use of \((\text{their})F = \mu(\text{their})R\); \(R\) not \(= 10g\) or their \(P\) from (i)
\(\mu = 0.4(00)\)A1 AWRT; www. Allow inequality
## Question 3(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Moments about $A$ |
| $P \times 1.6 = 10g\cos60 \times 1.2$ | A1 | |
| $P = 36.75$ N | A1 | Allow 36.8 |

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## Question 3(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $R + 36.75\sin30 = 10g$ | M1 | Attempt at resolving vertically or taking moments |
| | A1 FT | May be implied. $R = 79.6(25)$ |
| $F = 36.75\cos30$ | B1 FT | Expect 31.8. Or second correct equation involving $F$ or $R$ or both |
| $\mu = 31.8/79.6$ | M1 | For use of $(\text{their})F = \mu(\text{their})R$; $R$ not $= 10g$ or their $P$ from (i) |
| $\mu = 0.4(00)$ | A1 | AWRT; www. Allow inequality |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{5addd79d-d502-455c-936f-27005483164e-3_483_787_260_641}

A uniform rod $A B$ of mass 10 kg and length 2.4 m rests with $A$ on rough horizontal ground. The rod makes an angle of $60 ^ { \circ }$ with the horizontal and is supported by a fixed smooth peg $P$. The distance $A P$ is 1.6 m (see diagram).\\
(i) Calculate the magnitude of the force exerted by the peg on the rod.\\
(ii) Find the least value of the coefficient of friction between the rod and the ground needed to maintain equilibrium.

\hfill \mbox{\textit{OCR M2 2012 Q3 [8]}}