OCR MEI M2 2016 June — Question 3 18 marks

Exam BoardOCR MEI
ModuleM2 (Mechanics 2)
Year2016
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeEquilibrium with applied force
DifficultyStandard +0.3 This is a standard M2 centre of mass question requiring systematic application of moments about a point to find the centre of mass of a composite body, followed by equilibrium calculations. The multi-part structure and need to handle composite bodies with reinforcements adds some length, but the techniques are routine for M2 students with no novel problem-solving required.
Spec3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

  1. Use an energy method to find the magnitude of the frictional force acting on the block. Calculate the coefficient of friction between the block and the plane.
  2. Calculate the power of the tension in the string when the block has a speed of \(7 \mathrm {~ms} ^ { - 1 }\). Fig. 3.1 shows a thin planar uniform rigid rectangular sheet of metal, OPQR, of width 1.65 m and height 1.2 m . The mass of the sheet is \(M \mathrm {~kg}\). The sides OP and PQ have thin rigid uniform reinforcements attached with masses \(0.6 M \mathrm {~kg}\) and \(0.4 M \mathrm {~kg}\), respectively. Fig. 3.1 also shows coordinate axes with origin at O . The sheet with its reinforcements is to be used as an inn sign.
  3. Calculate the coordinates of the centre of mass of the inn sign. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-4_421_492_210_1334} \captionsetup{labelformat=empty} \caption{Fig. 3.1}
    \end{figure} The inn sign has a weight of 300 N . It hangs in equilibrium with QR horizontal when vertical forces \(Y _ { \mathrm { Q } } \mathrm { N }\) and \(Y _ { \mathrm { R } } \mathrm { N }\) act at Q and R respectively.
  4. Calculate the value of \(Y _ { \mathrm { Q } }\) and show that \(Y _ { \mathrm { R } } = 120\). The inn sign is hung from a framework, ABCD , by means of two light vertical inextensible wires attached to the sign at Q and R and the framework at B and C , as shown in Fig. 3.2. QR and BC are horizontal. The framework is made from light rigid rods \(\mathrm { AB } , \mathrm { BC } , \mathrm { CA }\) and CD freely pin-jointed together at \(\mathrm { A } , \mathrm { B }\) and C and to a vertical wall at A and D . Fig. 3.3 shows the dimensions of the framework in metres as well as the external forces \(X _ { \mathrm { A } } \mathrm { N } , Y _ { \mathrm { A } } \mathrm { N }\) acting at A and \(X _ { \mathrm { D } } \mathrm { N } , Y _ { \mathrm { D } } \mathrm { N }\) acting at D . You are given that \(\sin \alpha = \frac { 5 } { 13 } , \cos \alpha = \frac { 12 } { 13 } , \sin \beta = \frac { 4 } { 5 }\) and \(\cos \beta = \frac { 3 } { 5 }\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-4_543_526_1420_253} \captionsetup{labelformat=empty} \caption{Fig. 3.2}
    \end{figure} \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-4_629_793_1343_964} \captionsetup{labelformat=empty} \caption{Fig. 3.3}
    \end{figure}
  5. Mark on the diagram in your Printed Answer Book all the forces acting on the pin-joints at \(\mathrm { A } , \mathrm { B } , \mathrm { C }\) and D , including those internal to the rods, when the inn sign is hanging from the framework.
  6. Show that \(X _ { \mathrm { D } } = 261\).
  7. Calculate the forces internal to the rods \(\mathrm { AB } , \mathrm { BC }\) and CD , stating whether each rod is in tension or thrust (compression). Calculate also the values of \(Y _ { \mathrm { D } }\) and \(Y _ { \mathrm { A } }\). [Your working in this part should correspond to your diagram in part (iii).]

Question 3:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
\(2M\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix}=M\begin{pmatrix}0.825\\0.6\end{pmatrix}+0.6M\begin{pmatrix}0.825\\0\end{pmatrix}+0.4M\begin{pmatrix}1.65\\0.6\end{pmatrix}\)M1 Complete method
A1At least 2 RHS vector terms or 3 component terms correct
\(\bar{x}=0.99\)A1
\(\bar{y}=0.42\)A1
[4]
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
c.w moments about R: \(300\times0.99-1.65\times Y_Q=0\)M1 Or moments about Q to find \(Y_R\)
\(Y_Q=180\)A1 Must be established not using given \(Y_R\)
\(Y_Q+Y_R=300\) so \(Y_R=120\)E1 AG
[3]
Part (iii)
AnswerMarks Guidance
AnswerMarks Guidance
Mark in 120 N and 180 N and all internal forcesB1 Accept labelled internal forces marked as T or C
[1]
Part (iv)
AnswerMarks Guidance
AnswerMarks Guidance
c.w moments about A: \(120\times0.75+180\times2.4-2\times X_D=0\)M1 Appropriate moments considered
Or \(300\times(0.99+0.75)-2\times X_D=0\), so \(X_D=261\)E1 Convincingly shown
[2]
Question (v):
AnswerMarks Guidance
AnswerMarks Guidance
At B: \(\uparrow T_{AB}\sin\alpha - 180 = 0\)M1 Equilibrium at a pin-joint to find one of required forces (all relevant forces). This solution takes all internal forces +ve when T(ensions)
\(T_{AB} = 180 \times \frac{13}{5} = 468\) so force in AB is 468 N (T)A1 Do not need T/C here
\(\leftarrow T_{BC} + T_{AB}\cos\alpha = 0\)M1 2nd equilibrium at the same or another pin-joint to find another required force (all relevant forces)
\(T_{BC} = -468 \times \frac{12}{13} = -432\) so force in BC is 432 N (C)F1 FT their values. Do not need T/C here
At D: \(\rightarrow T_{DC}\cos\beta + X_D = 0\)M1 3rd equilibrium at a pin-joint (complete method to find third required force)
\(T_{DC} = -261 \times \frac{5}{3} = -435\) so force in DC is 435 N (C)F1 FT their values and all T/C correct
\(\uparrow Y_D + T_{DC}\sin\beta = 0\) so \(Y_D = 435 \times \frac{4}{5} = 348\) so 348 NB1 cao for the first of \(Y_D\) or \(Y_A\) found
\(Y_D - Y_A = 300\) so \(Y_A = 48\) so 48 NB1 ft for the second of \(Y_D\) or \(Y_A\): difference = 300
[8]
# Question 3:

## Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2M\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix}=M\begin{pmatrix}0.825\\0.6\end{pmatrix}+0.6M\begin{pmatrix}0.825\\0\end{pmatrix}+0.4M\begin{pmatrix}1.65\\0.6\end{pmatrix}$ | M1 | Complete method |
| | A1 | At least 2 RHS vector terms or 3 component terms correct |
| $\bar{x}=0.99$ | A1 | |
| $\bar{y}=0.42$ | A1 | |
| **[4]** | | |

## Part (ii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| c.w moments about R: $300\times0.99-1.65\times Y_Q=0$ | M1 | Or moments about Q to find $Y_R$ |
| $Y_Q=180$ | A1 | Must be established not using given $Y_R$ |
| $Y_Q+Y_R=300$ so $Y_R=120$ | E1 | AG |
| **[3]** | | |

## Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Mark in 120 N and 180 N and all internal forces | B1 | Accept labelled internal forces marked as T or C |
| **[1]** | | |

## Part (iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| c.w moments about A: $120\times0.75+180\times2.4-2\times X_D=0$ | M1 | Appropriate moments considered |
| Or $300\times(0.99+0.75)-2\times X_D=0$, so $X_D=261$ | E1 | Convincingly shown |
| **[2]** | | |

## Question (v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| At B: $\uparrow T_{AB}\sin\alpha - 180 = 0$ | M1 | Equilibrium at a pin-joint to find one of required forces (all relevant forces). This solution takes all internal forces +ve when T(ensions) |
| $T_{AB} = 180 \times \frac{13}{5} = 468$ so force in AB is 468 N (T) | A1 | Do not need T/C here |
| $\leftarrow T_{BC} + T_{AB}\cos\alpha = 0$ | M1 | 2nd equilibrium at the same or another pin-joint to find another required force (all relevant forces) |
| $T_{BC} = -468 \times \frac{12}{13} = -432$ so force in BC is 432 N (C) | F1 | FT their values. Do not need T/C here |
| At D: $\rightarrow T_{DC}\cos\beta + X_D = 0$ | M1 | 3rd equilibrium at a pin-joint (complete method to find third required force) |
| $T_{DC} = -261 \times \frac{5}{3} = -435$ so force in DC is 435 N (C) | F1 | FT their values and all T/C correct |
| $\uparrow Y_D + T_{DC}\sin\beta = 0$ so $Y_D = 435 \times \frac{4}{5} = 348$ so 348 N | B1 | cao for the first of $Y_D$ or $Y_A$ found |
| $Y_D - Y_A = 300$ so $Y_A = 48$ so 48 N | B1 | ft for the second of $Y_D$ or $Y_A$: difference = 300 |
| **[8]** | | |

---
(i) Use an energy method to find the magnitude of the frictional force acting on the block.

Calculate the coefficient of friction between the block and the plane.\\
(ii) Calculate the power of the tension in the string when the block has a speed of $7 \mathrm {~ms} ^ { - 1 }$.

Fig. 3.1 shows a thin planar uniform rigid rectangular sheet of metal, OPQR, of width 1.65 m and height 1.2 m . The mass of the sheet is $M \mathrm {~kg}$. The sides OP and PQ have thin rigid uniform reinforcements attached with masses $0.6 M \mathrm {~kg}$ and $0.4 M \mathrm {~kg}$, respectively. Fig. 3.1 also shows coordinate axes with origin at O .

The sheet with its reinforcements is to be used as an inn sign.\\
(i) Calculate the coordinates of the centre of mass of the inn sign.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-4_421_492_210_1334}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}

The inn sign has a weight of 300 N . It hangs in equilibrium with QR horizontal when vertical forces $Y _ { \mathrm { Q } } \mathrm { N }$ and $Y _ { \mathrm { R } } \mathrm { N }$ act at Q and R respectively.\\
(ii) Calculate the value of $Y _ { \mathrm { Q } }$ and show that $Y _ { \mathrm { R } } = 120$.

The inn sign is hung from a framework, ABCD , by means of two light vertical inextensible wires attached to the sign at Q and R and the framework at B and C , as shown in Fig. 3.2. QR and BC are horizontal. The framework is made from light rigid rods $\mathrm { AB } , \mathrm { BC } , \mathrm { CA }$ and CD freely pin-jointed together at $\mathrm { A } , \mathrm { B }$ and C and to a vertical wall at A and D . Fig. 3.3 shows the dimensions of the framework in metres as well as the external forces $X _ { \mathrm { A } } \mathrm { N } , Y _ { \mathrm { A } } \mathrm { N }$ acting at A and $X _ { \mathrm { D } } \mathrm { N } , Y _ { \mathrm { D } } \mathrm { N }$ acting at D .

You are given that $\sin \alpha = \frac { 5 } { 13 } , \cos \alpha = \frac { 12 } { 13 } , \sin \beta = \frac { 4 } { 5 }$ and $\cos \beta = \frac { 3 } { 5 }$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-4_543_526_1420_253}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-4_629_793_1343_964}
\captionsetup{labelformat=empty}
\caption{Fig. 3.3}
\end{center}
\end{figure}

(iii) Mark on the diagram in your Printed Answer Book all the forces acting on the pin-joints at $\mathrm { A } , \mathrm { B } , \mathrm { C }$ and D , including those internal to the rods, when the inn sign is hanging from the framework.\\
(iv) Show that $X _ { \mathrm { D } } = 261$.\\
(v) Calculate the forces internal to the rods $\mathrm { AB } , \mathrm { BC }$ and CD , stating whether each rod is in tension or thrust (compression). Calculate also the values of $Y _ { \mathrm { D } }$ and $Y _ { \mathrm { A } }$. [Your working in this part should correspond to your diagram in part (iii).]

\hfill \mbox{\textit{OCR MEI M2 2016 Q3 [18]}}