Edexcel M1 2010 January — Question 4 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2010
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeBeam suspended by vertical ropes
DifficultyModerate -0.3 This is a standard M1 moments question requiring taking moments about a point, resolving forces vertically, and solving simultaneous equations. The multi-part structure guides students through the solution methodically. While it requires careful bookkeeping of distances and forces, it involves only routine application of equilibrium conditions with no novel insight needed—slightly easier than average for A-level.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

\includegraphics{figure_2} A pole \(AB\) has length 3 m and weight \(W\) newtons. The pole is held in a horizontal position in equilibrium by two vertical ropes attached to the pole at the points \(A\) and \(C\) where \(AC = 1.8\) m, as shown in Figure 2. A load of weight 20 N is attached to the rod at \(B\). The pole is modelled as a uniform rod, the ropes as light inextensible strings and the load as a particle.
  1. Show that the tension in the rope attached to the pole at \(C\) is \(\left(\frac{5}{6}W + \frac{100}{3}\right)\) N. [4]
  2. Find, in terms of \(W\), the tension in the rope attached to the pole at \(A\). [3]
Given that the tension in the rope attached to the pole at \(C\) is eight times the tension in the rope attached to the pole at \(A\),
  1. find the value of \(W\). [3]

Part (a):
M(A): \(W \times 1.5 + 20 \times 3 = Y \times 1.8\)
AnswerMarks Guidance
\(y = \frac{5}{6}W + \frac{100}{3}\) ✱M1 A2 (1,0) A1 (4)
Part (b):
\(\uparrow\): \(X + Y = W + 20\) or equivalent
AnswerMarks Guidance
\(X = \frac{1}{6}W - \frac{40}{3}\)M1 A1 A1 (3)
Part (c):
\(\frac{5}{6}W + \frac{100}{3} = 8\left(\frac{1}{6}W - \frac{40}{3}\right)\)
AnswerMarks Guidance
\(W = 280\)M1 A1 ft A1 (3) [10]
Alternative to (b):
M(C): \(X \times 1.8 + 20 \times 1.2 = W \times 0.3\)
AnswerMarks
\(X = \frac{1}{6}W - \frac{40}{3}\)M1 A1 A1
**Part (a):**
M(A): $W \times 1.5 + 20 \times 3 = Y \times 1.8$
$y = \frac{5}{6}W + \frac{100}{3}$ ✱ | M1 A2 (1,0) A1 | (4) | cso

**Part (b):**
$\uparrow$: $X + Y = W + 20$ or equivalent
$X = \frac{1}{6}W - \frac{40}{3}$ | M1 A1 A1 | (3)

**Part (c):**
$\frac{5}{6}W + \frac{100}{3} = 8\left(\frac{1}{6}W - \frac{40}{3}\right)$
$W = 280$ | M1 A1 ft A1 | (3) [10]

**Alternative to (b):**
M(C): $X \times 1.8 + 20 \times 1.2 = W \times 0.3$
$X = \frac{1}{6}W - \frac{40}{3}$ | M1 A1 A1 |
\includegraphics{figure_2}

A pole $AB$ has length 3 m and weight $W$ newtons. The pole is held in a horizontal position in equilibrium by two vertical ropes attached to the pole at the points $A$ and $C$ where $AC = 1.8$ m, as shown in Figure 2. A load of weight 20 N is attached to the rod at $B$. The pole is modelled as a uniform rod, the ropes as light inextensible strings and the load as a particle.

\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the rope attached to the pole at $C$ is $\left(\frac{5}{6}W + \frac{100}{3}\right)$ N. [4]
\item Find, in terms of $W$, the tension in the rope attached to the pole at $A$. [3]
\end{enumerate}

Given that the tension in the rope attached to the pole at $C$ is eight times the tension in the rope attached to the pole at $A$,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the value of $W$. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2010 Q4 [10]}}