| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cylinder or hemisphere from solid |
| Difficulty | Standard +0.3 This is a straightforward centre of mass problem requiring standard techniques: decomposing the solid into components (main cylinder minus two removed cylinders), applying the formula for composite bodies, and using equilibrium conditions. The calculations are routine for Further Maths students, though the multi-part structure and need for careful coordinate work place it slightly above average difficulty for A-level. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | Area of S π ( 5 0 2 3 0 2 1 0 2 ) 1 5 0 0 = − − = |
| Answer | Marks | Guidance |
|---|---|---|
| 0 4 0 0 y | M1 | 1.1 |
| ⟹ 𝑥̅ = 12 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (b) | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Angle is 77.47119…° | M1 | |
| A1 FT | 3.1b | |
| 1.1 | tan𝛼 = 12÷their | 𝑦̅ |
| Answer | Marks |
|---|---|
| Accept 102.5°, −77.5° FT their | 𝑦̅ |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | F 5 0 = W 1 2 | M1 |
| F = 0 .2 4 W | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (d) | Let the angle between R and the horizontal be . |
| Answer | Marks |
|---|---|
| A B | Resolving horizontally. Must attempt to resolve |
| Answer | Marks |
|---|---|
| A B | Resolving vertically. Must include W and attempt |
| Answer | Marks |
|---|---|
| B | Moments about A. Both moments present. Must |
| Answer | Marks | Guidance |
|---|---|---|
| M1 M1 | 3.3 | Any two of the above. See first lines and notes |
| OR | 18𝑊+60𝐹cos𝜃 = 60𝑅 sin𝜃 | |
| A | M2 | Moments about B. Must attempt to ‘resolve’ R |
| Answer | Marks | Guidance |
|---|---|---|
| A | and F (or their distances) | |
| M1 | 1.1 | Correct values for c o s and s in used in at least |
| Answer | Marks | Guidance |
|---|---|---|
| A B | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.555W 37 | A1 FT | 2.2a |
Question 5:
5 | (a) | Area of S π ( 5 0 2 3 0 2 1 0 2 ) 1 5 0 0 = − − = | B1 | 1.1 | Or ratio of masses (e.g. 25:9:1:15) used
throughout
0 0 2 0 x −
2 5 0 0 1 5 0 0 1 0 0 9 0 0 = + +
0 4 0 0 y | M1 | 1.1 | For (𝜋×302)(20) or (𝜋×102)(40) seen
⟹ 𝑥̅ = 12 | A1 | 1.1 | AG Needs convincing working
8
𝑦̅ = −
3 | A1 | 1.1 | Allow −2.67
[4]
(b) | 8
arctan(12÷ )
3
Angle is 77.47119…° | M1
A1 FT | 3.1b
1.1 | tan𝛼 = 12÷their |𝑦̅| (allow reciprocal)
Just arctan(𝑥̅/𝑦̅) is not sufficient
Accept 102.5°, −77.5° FT their |𝑦̅|
[2]
(c) | F 5 0 = W 1 2 | M1 | 1.1 | Taking moments about O – both moments present
F = 0 .2 4 W | A1 | 1.1 | λ = 6/25 (or 0.24)
[2]
(d) | Let the angle between R and the horizontal be .
A
𝑅 cos𝜃+𝐹sin𝜃 = 𝑅 cos𝜃
A B
0.6𝑅 +0.8(0.24𝑊) = 0.6𝑅
A B | Resolving horizontally. Must attempt to resolve
all three forces. F may be their 0.24W
𝑅 sin𝜃+𝑅 sin𝜃 = 𝑊+𝐹cos𝜃
A B
0.8𝑅 +0.8𝑅 = 𝑊+0.6(0.24𝑊)
A B | Resolving vertically. Must include W and attempt
to resolve the other three forces
60𝑅 sin𝜃 = 42𝑊
B
0.8×60𝑅 = 42𝑊
B | Moments about A. Both moments present. Must
attempt to ‘resolve’ R (or its distance)
B
M1 M1 | 3.3 | Any two of the above. See first lines and notes
OR | 18𝑊+60𝐹cos𝜃 = 60𝑅 sin𝜃
A | M2 | Moments about B. Must attempt to ‘resolve’ R
A
18𝑊+0.6×60(0.24𝑊) = 0.8×60𝑅
A | and F (or their distances)
M1 | 1.1 | Correct values for c o s and s in used in at least
one equation for which M1 has been earned.
See second lines above. Allow cos(53.1°) etc
Note Only s in needed in the third equation
⇒ 𝑅 = 0.555𝑊 ( 𝑅 = 0.875𝑊 )
A B | A1 | 1.1 | cao
= 0.24W =16 (=0.432432 )
0.555W 37 | A1 FT | 2.2a | FT their F and R given as multiples of W,
A
dependent on M3
[5]5 Fig. 5.1 shows the uniform cross-section of a solid S which is formed from a cylinder by boring two cylindrical tunnels the entire way through the cylinder. The radius of S is 50 cm , and the two tunnels have radii 10 cm and 30 cm .
The material making up $S$ has uniform density.\\
Coordinates refer to the axes shown in Fig. 5.1 and the units are centimetres.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
\includegraphics[alt={},max width=\textwidth]{a87d62b8-406d-44cd-9ffa-384005329566-6_684_666_708_278}
\end{center}
\end{figure}
The centre of mass of $S$ is ( $\mathrm { x } , \mathrm { y }$ ).
\begin{enumerate}[label=(\alph*)]
\item Show that $\bar { x } = 12$ and find the value of $\bar { y }$.
Solid $S$ is placed onto two rails, $A$ and $B$, whose point of contacts with $S$ are at ( $- 30 , - 40$ ) and $( 30 , - 40 )$ as shown in Fig. 5.2. Two points, $\mathrm { P } ( 0,50 )$ and $\mathrm { Q } ( 0 , - 50 )$, are marked on Fig. 5.2.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 5.2}
\includegraphics[alt={},max width=\textwidth]{a87d62b8-406d-44cd-9ffa-384005329566-6_654_640_1875_251}
\end{center}
\end{figure}
At first, you should assume that the contact between S and the two rails is smooth.
\item Determine the angle PQ makes with the vertical, after S settles into equilibrium.
For the remainder of the question, you should assume that the contact between S and A is rough, that the contact between $S$ and $B$ is smooth, and that $S$ does not move when placed on the rails. Fig. 5.3 shows only the forces exerted on S by the rails. The normal contact forces exerted by A and B on S have magnitude $R _ { \mathrm { A } } \mathrm { N }$ and $R _ { \mathrm { B } } \mathrm { N }$ respectively. The frictional force exerted by A on S has magnitude $F$ N.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 5.3}
\includegraphics[alt={},max width=\textwidth]{a87d62b8-406d-44cd-9ffa-384005329566-7_652_641_593_248}
\end{center}
\end{figure}
The weight of S is $W \mathrm {~N}$.
\item By taking moments about the origin, express $F$ in the form $\lambda W$, where $\lambda$ is a constant to be determined.
\item Given that S is in limiting equilibrium, find the coefficient of friction between A and S .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2023 Q5 [13]}}