| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Standard +0.8 This is a statics problem requiring moments about two different points, resolution of forces in two directions, and friction inequality analysis. While the setup is standard Further Maths mechanics, it demands careful geometric reasoning to find the perpendicular distance from the peg to the rod's weight, systematic application of equilibrium conditions, and understanding that friction can act in either direction at the limiting case. The multi-step nature and need to consider friction bounds elevates this above routine exercises. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| 4(a) | 2×3gcos25=... |
| Answer | Marks |
|---|---|
| P | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Correct moment of the weight about A – must be part of |
| Answer | Marks |
|---|---|
| 4(b) | SEE APPENDIX FOR ALTERNATIVES |
| Answer | Marks |
|---|---|
| µ 0.402 | M1* |
| Answer | Marks |
|---|---|
| [5] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Attempt to resolve and equate either the vertical or |
Question 4:
--- 4(a) ---
4(a) | 2×3gcos25=...
1.5 1.5
×R or (1.5sin25)×R + ×R cos25
sin25 P P tan25 P
R =15.0(N)
P | B1
B1
B1
[3] | 3.3
3.1b
1.1 | Correct moment of the weight about A – must be part of
an equation (so must be equal to k×R where k ≠0) –
P
might see 53.29...=k×R
P
Correct moment of the normal contact force about A –
must be part of an equation – might see 3.549...×R =...
P
15.0144710… allow awrt 15.0 or 15 www
--- 4(b) ---
4(b) | SEE APPENDIX FOR ALTERNATIVES
F =R sin25
A P
R +R cos25=3g
A P
F A µR A
µ 0.402 | M1*
A1FT
A1FT
M1dep*
A1
[5] | 3.3
1.1
1.1
3.4
1.1 | Attempt to resolve and equate either the vertical or
horizontal forces (with correct number of terms) – allow
sign errors and sin/cos mix – but must be using the correct
angle – allowR(oe) for this mark only
Using their value of R from part (a) (for reference if
P
correct: F =6.34538967...) – must not be part of an
A
inequality. The follow through is only on their value of R
P
Using their value of R from part (a) (for reference: if
P
correct R =15.79226793...) – must not be part of an
A
inequality. The follow through is only on their value of R
P
Use of either F =µR or F µRwith their values of F
A
and R - allow this mark if earlier inequalities used when
A
resolving forces. Their values of F and R must have
A A
come from equations with the correct number of relevant
terms (so must have used components of R but not a
P
weight component) but allow sin/cos mix and sign errors
awrt 0.401 or awrt 0.402 or allow 0.40 www but not 0.4.
Any upper limit is A0. Condone µ=0.40replaced with
µ 0.40oe www\includegraphics{figure_4}
A uniform rod AB has mass 3 kg and length 4 m. The end A of the rod is in contact with rough horizontal ground.
The rod rests in equilibrium on a smooth horizontal peg 1.5 m above the ground, such that the rod is inclined at an angle of $25°$ to the ground (see diagram).
The rod is in a vertical plane perpendicular to the peg.
\begin{enumerate}[label=(\alph*)]
\item Determine the magnitude of the normal contact force between the peg and the rod. [3]
\item Determine the range of possible values of the coefficient of friction between the rod and the ground. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2024 Q4 [8]}}