Edexcel FM2 2019 June — Question 4 12 marks

Exam BoardEdexcel
ModuleFM2 (Further Mechanics 2)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod hinged to wall with string support
DifficultyStandard +0.8 This is a Further Maths mechanics question requiring integration to find mass from variable density, then moments equilibrium with non-uniform rod. Part (a) is straightforward integration, but part (b) requires finding the center of mass of the non-uniform rod, then taking moments about point A with forces at angles. The multi-step nature and FM2 content places it moderately above average difficulty.
Spec3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
  1. A flagpole, \(A B\), is 4 m long. The flagpole is modelled as a non-uniform rod so that, at a distance \(x\) metres from \(A\), the mass per unit length of the flagpole, \(m \mathrm {~kg} \mathrm {~m} ^ { - 1 }\), is given by \(m = 18 - 3 x\).
    1. Show that the mass of the flagpole is 48 kg .
    \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{9b995178-a4be-4d5a-95f8-6c2978ff01b3-12_515_439_502_806} \captionsetup{labelformat=empty} \caption{Figure 3}
    \end{figure} The end \(A\) of the flagpole is fixed to a point on a vertical wall. A cable has one end attached to the midpoint of the flagpole and the other end attached to a point on the wall that is vertically above \(A\). The cable is perpendicular to the flagpole. The flagpole and the cable lie in the same vertical plane that is perpendicular to the wall. A small ball of mass 4 kg is attached to the flagpole at \(B\). The cable holds the flagpole and ball in equilibrium, with the flagpole at \(45 ^ { \circ }\) to the wall, as shown in Figure 3. The tension in the cable is \(T\) newtons.
    The cable is modelled as a light inextensible string and the ball is modelled as a particle.
  2. Using the model, find the value of \(T\).
  3. Give a reason why the answer to part (b) is not likely to be the true value of \(T\).
Question 4(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Total mass \(= \int_0^4(18-3x)\,dx\)M1 Use integration (usual rules) – do not need to see limits at this stage
\(= \left[18x - \frac{3x^2}{2}\right]_0^4\)A1 (M1 on open) Correct integration and correct limits seen
\(= 18\times4 - \frac{3\times16}{2} = 48\) (kg) *A1* Show sufficient working to justify given answer
Question 4(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Taking moments about the base: \(\int_0^4 x(18-3x)\,dx\)M1
\(= \left[9x^2 - x^3\right]_0^4 (= 80)\)A1
\(\Rightarrow 48d = 80\)M1
\(d = \frac{80}{48} = \frac{5}{3}\) (m)A1
Complete strategyM1
\(M(A): 2T = 4\cos45°\times4g + \frac{5}{3}\cos45°\times48g\)A1ft
\(\left(= \frac{96g}{\sqrt{2}}\right)\)A1ft
\(T = 333\) or \(330\)A1
Question 4(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Any appropriate comment e.g. the ball has been modelled as a point mass – its centre could be further from \(A\)B1
Question 4(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Use model to find moment of pole and ball about \(A\)M1 Usual rules for integration
Correct integrationA1
Use model to complete moments equation; 80 and 48 used correctlyM1
Any equivalent formA1
Complete strategy to find tension – e.g. locate c of m of pole and use momentsM1
Moments equation with at most one error; follow their c of m provided not at centreA1ft
Correct unsimplified moments equation for their c of m not at centreA1ft
Accept \(24\sqrt{2}g\), 333 or 330A1 ISW
(8)
Question 4(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Any one of: mass of cable ignored (unlikely negligible if holding pole this long); flagpole subject to cross winds; cable might be extensible; pole might not be rigidB1 NOT: wall might be rough/smooth. Ignore incorrect statements
(1) 
## Question 4(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Total mass $= \int_0^4(18-3x)\,dx$ | M1 | Use integration (usual rules) – do not need to see limits at this stage |
| $= \left[18x - \frac{3x^2}{2}\right]_0^4$ | A1 | (M1 on open) Correct integration and correct limits seen |
| $= 18\times4 - \frac{3\times16}{2} = 48$ (kg) * | A1* | Show sufficient working to justify given answer |

---

## Question 4(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Taking moments about the base: $\int_0^4 x(18-3x)\,dx$ | M1 | |
| $= \left[9x^2 - x^3\right]_0^4 (= 80)$ | A1 | |
| $\Rightarrow 48d = 80$ | M1 | |
| $d = \frac{80}{48} = \frac{5}{3}$ (m) | A1 | |
| Complete strategy | M1 | |
| $M(A): 2T = 4\cos45°\times4g + \frac{5}{3}\cos45°\times48g$ | A1ft | |
| $\left(= \frac{96g}{\sqrt{2}}\right)$ | A1ft | |
| $T = 333$ or $330$ | A1 | |

---

## Question 4(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Any appropriate comment e.g. the ball has been modelled as a point mass – its centre could be further from $A$ | B1 | |

## Question 4(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use model to find moment of pole and ball about $A$ | M1 | Usual rules for integration |
| Correct integration | A1 | |
| Use model to complete moments equation; 80 and 48 used correctly | M1 | |
| Any equivalent form | A1 | |
| Complete strategy to find tension – e.g. locate c of m of pole and use moments | M1 | |
| Moments equation with at most one error; follow their c of m provided not at centre | A1ft | |
| Correct unsimplified moments equation for their c of m not at centre | A1ft | |
| Accept $24\sqrt{2}g$, 333 or 330 | A1 | ISW |
| **(8)** | | |

## Question 4(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Any one of: mass of cable ignored (unlikely negligible if holding pole this long); flagpole subject to cross winds; cable might be extensible; pole might not be rigid | B1 | NOT: wall might be rough/smooth. Ignore incorrect statements |
| **(1)** | | |

---
\begin{enumerate}
  \item A flagpole, $A B$, is 4 m long. The flagpole is modelled as a non-uniform rod so that, at a distance $x$ metres from $A$, the mass per unit length of the flagpole, $m \mathrm {~kg} \mathrm {~m} ^ { - 1 }$, is given by $m = 18 - 3 x$.\\
(a) Show that the mass of the flagpole is 48 kg .
\end{enumerate}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{9b995178-a4be-4d5a-95f8-6c2978ff01b3-12_515_439_502_806}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

The end $A$ of the flagpole is fixed to a point on a vertical wall. A cable has one end attached to the midpoint of the flagpole and the other end attached to a point on the wall that is vertically above $A$. The cable is perpendicular to the flagpole. The flagpole and the cable lie in the same vertical plane that is perpendicular to the wall. A small ball of mass 4 kg is attached to the flagpole at $B$. The cable holds the flagpole and ball in equilibrium, with the flagpole at $45 ^ { \circ }$ to the wall, as shown in Figure 3.

The tension in the cable is $T$ newtons.\\
The cable is modelled as a light inextensible string and the ball is modelled as a particle.\\
(b) Using the model, find the value of $T$.\\
(c) Give a reason why the answer to part (b) is not likely to be the true value of $T$.

\hfill \mbox{\textit{Edexcel FM2 2019 Q4 [12]}}
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