| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Moderate -0.3 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions. Part (a) involves basic trigonometry with a 3-4-5 triangle (recognizing sin α = 3/5 = 0.6), and part (b) requires setting up and solving two simultaneous equations from ΣFx = 0 and ΣFy = 0. While it involves multiple steps and three unknowns, the techniques are standard and the geometry is simple, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | (i) |
| Answer | Marks | Guidance |
|---|---|---|
| ∠OAB = 𝛼𝛼 𝑠𝑠𝑠𝑠𝑠𝑠(∠OAB) = 5 = 0.6 | B1 | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| [1] | ∠OAB = 𝛼𝛼 | |
| (ii) | cosα=0.8 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Moments about O: 3⋅10sinα=4T | |
| 2 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| Answer | Marks |
|---|---|
| 1 2 1 2 1 | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Attempt at resolving: one equation |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 1.1 | Correct method for finding T or θ |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | A1 | 1.1 |
Question 2:
2 | (a) | (i) | and
3
∠OAB = 𝛼𝛼 𝑠𝑠𝑠𝑠𝑠𝑠(∠OAB) = 5 = 0.6 | B1 | 3.1a | AG
Need not justify why .
Allow finding angles 53.1°, 36.9°
[1] | ∠OAB = 𝛼𝛼
(ii) | cosα=0.8 | B1 | 1.1
[1]
(b) | Moments about O: 3⋅10sinα=4T
2 | M1 | 3.3 | Taking moments about some point.
⇒T =4.5
2 | A1 | 1.1
T cosθ=10sinα or T cosθ=6 or 3T cosθ=4T
1 1 1 2
T sinθ=10cosα+T or T sinθ=8+T or 4T sinθ=10(5)
1 2 1 2 1 | M1
A1
A1 | 3.3
1.1
1.1 | Attempt at resolving: one equation
sufficient (all terms present, but
sin↔cos)
condone
Or Taking moments about a
different point
Correct equation involving T cosθ
1
Correct equation involving T sinθ
1
M1 | 1.1 | Correct method for finding T or θ
1
(from values of T cosθ and T sinθ)
1 1
Dependent on previous M1M1
tanθ=12.5⇒θ≈64.4°
6
T = 12.52 +62 ≈13.9
1 | A1 | 1.1 | cao Both correct
[7]2 The vertices of a triangular lamina, which is in the $x - y$ plane, are at the origin O and the points $\mathrm { A } ( 4,0 )$ and $\mathrm { B } ( 0,3 )$. Forces, of magnitude $T _ { 1 } \mathrm {~N} , T _ { 2 } \mathrm {~N}$ and 10 N , whose lines of action are in the $x - y$ plane, are applied to the lamina at $\mathrm { O } , \mathrm { A }$ and B respectively, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{5c1cfe41-d7a2-4f69-ae79-67d9f023c246-2_814_922_1135_246}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\sin \alpha = 0.6$.
\item Write down the value of $\cos \alpha$.
The lamina is in equilibrium.
\end{enumerate}\item Determine the values of $T _ { 1 } , T _ { 2 }$ and $\theta$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2021 Q2 [9]}}