| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Standard +0.3 This is a standard two-part moments problem requiring systematic application of equilibrium conditions. Part (a) involves taking moments about a point to find x when reaction forces are equal. Part (b) requires finding when the beam lifts (reaction at A becomes zero) with a guided answer provided. Part (c) compares limiting friction to determine sliding vs lifting. While multi-step, it follows textbook procedures without requiring novel insight or complex geometric reasoning. |
| 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 force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | Each support must exert 12 ( 4 g + 1 0 g ) = 7 g N on the beam. |
| Answer | Marks |
|---|---|
| 7𝑔𝑥+7𝑔(𝑥+3.8) = 4𝑔×1.95+10𝑔×3 | M1 |
| A1 | 3.3 |
| 1.1 | Attempt to take moments about some point. All |
| Answer | Marks | Guidance |
|---|---|---|
| ⇒ 𝑥 = 0.8 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Moments about B, 5 .3 T s i n 6 0 = 1 0 g 2 .3 | |
| L | M1 | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| L | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (c) | Let the force at the supports be R and R N, and the friction |
| Answer | Marks | Guidance |
|---|---|---|
| B | B1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑆 𝑆 B | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑆 A B | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| A B | M1 | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| OR | 3.8𝑅 = 1.5×10𝑔+1.5×𝑇 sin60° | |
| B 𝑆 | M2 | Moments about A |
| Answer | Marks | Guidance |
|---|---|---|
| 49.1) | A1 | 2.2a |
Question 6:
6 | (a) | Each support must exert 12 ( 4 g + 1 0 g ) = 7 g N on the beam. | B1 | 1.1 | soi
Taking moments about the left-most point:
𝑅 𝑥+𝑅 (𝑥+3.8) = 4𝑔×1.95+10𝑔×3
A B
7𝑔𝑥+7𝑔(𝑥+3.8) = 4𝑔×1.95+10𝑔×3 | M1
A1 | 3.3
1.1 | Attempt to take moments about some point. All
moments present. Allow one error
Correct equation for x FT their 7g
Correct equation www implies B1
⇒ 𝑥 = 0.8 | A1 | 1.1 | cao
[4]
(b) | Moments about B, 5 .3 T s i n 6 0 = 1 0 g 2 .3
L | M1 | 3.1b | Attempt at moments. No force acting at A; allow
sin/cos switch.
If moments taken about A, 𝑅 = 0 must be soi
A
(e.g. by 𝑇 sin60°+𝑅 = 10𝑔 )
𝐿 B
Allow inequalities in (b) and (c)
T = 4 9 .1 0 7 4 5
L | A1 | 1.1 | AG Must be convincingly shown.
[2]
(c) | Let the force at the supports be R and R N, and the friction
A B
at B be F N.
F = 0 .4 R
B | B1 | 3.4 | Modelling friction. Must clearly be reaction at B
𝑇 cos60° = 𝐹 (⇒ 𝑇 = 0.8𝑅 )
𝑆 𝑆 B | M1 | 3.3 | Resolving horizontally
𝑇 sin60°+𝑅 +𝑅 = 10𝑔
𝑆 A B | M1 | 1.1 | Resolving vertically
Moments about LH end, 1.5R +5.3R =310g
A B | M1 | 3.1b | Attempt at moments. All moments present. Allow
one error
OR | 3.8𝑅 = 1.5×10𝑔+1.5×𝑇 sin60°
B 𝑆 | M2 | Moments about A
(⇒ 𝑅 = 53.2460… 𝑅 = 7.864…)
B A
T = 4 2 .5 9 6 8 so beam will slide first (since 42.6 <
S
49.1) | A1 | 2.2a | cao
[5]
PMT
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Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.6 A uniform beam of length 6 m and mass 10 kg rests horizontally on two supports A and B , which are 3.8 m apart. A particle $P$ of mass 4 kg is attached 1.95 m from one end of the beam (see Fig. 6.1).
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\includegraphics[alt={},max width=\textwidth]{a87d62b8-406d-44cd-9ffa-384005329566-8_257_1079_447_246}
\end{center}
\end{figure}
When A is $x \mathrm {~m}$ from the end of the beam, the supports exert forces of equal magnitude on the beam.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $x$.
P is now removed. The same beam is placed on the supports so that B is 0.7 m from the end of the beam. The supports remain 3.8 m apart (see Fig. 6.2).
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\includegraphics[alt={},max width=\textwidth]{a87d62b8-406d-44cd-9ffa-384005329566-8_296_1082_1162_246}
\end{center}
\end{figure}
The contact between A and the beam is smooth. The contact between B and the beam is rough, with coefficient of friction 0.4.
A small force of magnitude $T \mathrm {~N}$ is applied to one end of the beam. The force acts in the same vertical plane as the beam and the angle the force makes with the beam is $60 ^ { \circ }$.
As $T$ is increased, forces $\mathrm { T } _ { \mathrm { L } }$ and $\mathrm { T } _ { \mathrm { S } }$ are defined in the following way.
\begin{itemize}
\item $\quad \mathrm { T } _ { \mathrm { L } }$ is the value of $T$ at which the beam would start lifting, assuming that is not already sliding.
\item $\quad \mathrm { T } _ { \mathrm { S } }$ is the value of $T$ at which the beam would start sliding, assuming that it has not already lifted.
\item Show that $\mathrm { T } _ { \mathrm { L } } = 49.1$, correct to 3 significant figures.
\item Determine whether the beam will first slide or lift.
\end{itemize}
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2023 Q6 [11]}}