| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard statics problem with a ladder against a wall, requiring basic geometry (3-4-5 triangle), taking moments about a point, and solving simultaneous equations. The steps are routine for Further Mechanics students: find geometry, apply equilibrium conditions, and solve for unknowns. Part (a) is straightforward trigonometry, part (b) is a standard moment calculation with given answer to verify, and part (c) requires simple substitution. No novel insight required, just methodical application of standard techniques. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | c o s 45 = |
| BC=22cos=3.2 | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | T s i n ( 3 .2 ) 6 g s i n ( 2 .5 ) 7 5 g s i n ( 5 x ) R s i n ( 5 ) + + − = | M1 |
| Answer | Marks |
|---|---|
| A1ft | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Attempt at taking moments: four |
| Answer | Marks | Guidance |
|---|---|---|
| 2 5 R + 3 6 7 5 x − 1 6 T = 1 9 1 1 0 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | 6 g 7 5 g R T c o s 2 + = + | M1 |
| M1 | 3.3 | |
| 3.1a | Attempt to “resolve” all forces |
| Answer | Marks | Guidance |
|---|---|---|
| ⇒ 16×720+19110 = 25 (793.8−0.28T)+3675x | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| x=4.3061 | A1 | 2.2b |
Question 4:
4 | (a) | c o s 45 = | B1 | 1.1a | soi
BC=22cos=3.2 | B1 | 1.1
[2]
(b) | T s i n ( 3 .2 ) 6 g s i n ( 2 .5 ) 7 5 g s i n ( 5 x ) R s i n ( 5 ) + + − = | M1
A1ft
A1ft | 3.3
1.1
1.1 | Attempt at taking moments: four
terms present, condone sincos
confusion.
Two or three terms correct.
Trigonometric factors must be seen
to start with (even if visibly
cancelled immediately). Ft their BC
Fully correct (unsimplified)
equation ft their BC
3 .2 T + 1 5 g + 3 7 5 g − 7 5 g x = 5 R
3 .2 T + 3 8 2 2 = 5 R + 7 3 5 x
2 5 R + 3 6 7 5 x − 1 6 T = 1 9 1 1 0 | A1 | 1.1 | AG
[4]
(c) | 6 g 7 5 g R T c o s 2 + = + | M1
M1 | 3.3
3.1a | Attempt to “resolve” all forces
vertically.
Using c o s 2 .
R=793.8−0.28T
⇒ 16×720+19110 = 25 (793.8−0.28T)+3675x | M1 | 1.1 | Substitution of their R and using
T = 7 2 0
3 0 6 3 0 = 1 4 8 0 5 + 3 6 7 5 x
x=4.3061 | A1 | 2.2b
[4]4 A uniform beam AB of mass 6 kg and length 5 m rests with its end A on smooth horizontal ground and its end B against a smooth vertical wall. The vertical distance between the ground and B is 4 m , and the angle between the beam and the downward vertical is $\theta$. To prevent the beam from sliding, one end of a light taut rope of length 2 m is attached to the beam at C and the other end of the rope is attached to a point on the wall 2 m above the ground, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{9b624694-edb6-4000-838f-3557e078952d-5_558_556_500_251}
\begin{enumerate}[label=(\alph*)]
\item By considering the value of $\cos \theta$, determine the distance BC .
An object of mass 75 kg is placed on the beam at a point which is $x \mathrm {~m}$ from A .
It is given that the tension in the rope is $T \mathrm {~N}$ and the magnitude of the normal contact force between the ground and the beam is $R \mathrm {~N}$.
\item By taking moments about B for the beam, show that $25 \mathrm { R } + 3675 \mathrm { x } - 16 \mathrm {~T} = 19110$.
\item Given that the rope can withstand a maximum tension of 720 N , determine the largest possible value of $x$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2022 Q4 [10]}}