Edexcel M1 2009 June — Question 7 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2009
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeBeam suspended by vertical ropes
DifficultyStandard +0.3 This is a standard M1 moments question requiring taking moments about two points and applying equilibrium conditions. The multi-part structure guides students through the problem systematically, and the techniques (moments about a point, resolving vertically, algebraic manipulation) are routine for this module. Slightly easier than average due to the scaffolded approach and straightforward algebra.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force3.04c Use moments: beams, ladders, static problems

\includegraphics{figure_2} A beam \(AB\) is supported by two vertical ropes, which are attached to the beam at points \(P\) and \(Q\), where \(AP = 0.3\) m and \(BQ = 0.3\) m. The beam is modelled as a uniform rod, of length 2 m and mass 20 kg. The ropes are modelled as light inextensible strings. A gymnast of mass 50 kg hangs on the beam between \(P\) and \(Q\). The gymnast is modelled as a particle attached to the beam at the point \(X\), where \(PX = x\) m, \(0 < x < 1.4\) as shown in Figure 2. The beam rests in equilibrium in a horizontal position.
  1. Show that the tension in the rope attached to the beam at \(P\) is \((588 - 350x)\) N. [3]
  2. Find, in terms of \(x\), the tension in the rope attached to the beam at \(Q\). [3]
  3. Hence find, justifying your answer carefully, the range of values of the tension which could occur in each rope. [3]
Given that the tension in the rope attached at \(Q\) is three times the tension in the rope attached at \(P\),
  1. find the value of \(x\). [3]

AnswerMarks
(a) \(M(Q)\), \(50g(1.4 - x) + 20g \times 0.7 = T_P \times 1.4\)M1 A1
\(T_P = 588 - 350x\)A1 Printed answer
(3)
(b) \(M(P)\), \(50gx + 20g \times 0.7 = T_Q \times 1.4\) or R(\(\uparrow\)), \(T_P + T_Q = 70g\)M1 A1
\(T_Q = 98 + 350x\)A1
(3)
(c) Since \(0 < x < 1.4\), \(98 < T_P < 588\) and \(98 < T_Q < 588\)M1 A1 A1
(3)
(d) \(98 + 350x = 3(588 - 350x)\)M1
\(x = 1.19\)DM1 A1
(3)
Total: [12]
**(a)** $M(Q)$, $50g(1.4 - x) + 20g \times 0.7 = T_P \times 1.4$ | M1 A1 |
$T_P = 588 - 350x$ | A1 Printed answer |
| **(3)** |

**(b)** $M(P)$, $50gx + 20g \times 0.7 = T_Q \times 1.4$ or R($\uparrow$), $T_P + T_Q = 70g$ | M1 A1 |
$T_Q = 98 + 350x$ | A1 |
| **(3)** |

**(c)** Since $0 < x < 1.4$, $98 < T_P < 588$ and $98 < T_Q < 588$ | M1 A1 A1 |
| **(3)** |

**(d)** $98 + 350x = 3(588 - 350x)$ | M1 |
$x = 1.19$ | DM1 A1 |
| **(3)** |
| **Total: [12]** |
\includegraphics{figure_2}

A beam $AB$ is supported by two vertical ropes, which are attached to the beam at points $P$ and $Q$, where $AP = 0.3$ m and $BQ = 0.3$ m. The beam is modelled as a uniform rod, of length 2 m and mass 20 kg. The ropes are modelled as light inextensible strings. A gymnast of mass 50 kg hangs on the beam between $P$ and $Q$. The gymnast is modelled as a particle attached to the beam at the point $X$, where $PX = x$ m, $0 < x < 1.4$ as shown in Figure 2. The beam rests in equilibrium in a horizontal position.

\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the rope attached to the beam at $P$ is $(588 - 350x)$ N. [3]
\item Find, in terms of $x$, the tension in the rope attached to the beam at $Q$. [3]
\item Hence find, justifying your answer carefully, the range of values of the tension which could occur in each rope. [3]
\end{enumerate}

Given that the tension in the rope attached at $Q$ is three times the tension in the rope attached at $P$,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item find the value of $x$. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2009 Q7 [12]}}