CAIE M2 2019 June — Question 6 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeLamina hinged at point with string support
DifficultyStandard +0.3 This is a standard mechanics question on moments with a triangular lamina. Part (i) requires recall of the centroid formula (1/3 from each side), part (ii) is a straightforward equilibrium problem using tan θ, and part (iii) involves taking moments about B with a given force. All steps are routine applications of standard techniques with no novel problem-solving required, making it slightly easier than average.
Spec3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry

6 \(A B C\) is a uniform lamina in the form of a triangle with \(A B = 0.3 \mathrm {~m} , B C = 0.6 \mathrm {~m}\) and a right angle at \(B\) (see diagram).
  1. State the distances of the centre of mass of the lamina from \(A B\) and from \(B C\). Distance from \(A B\) Distance from \(B C\) \(\_\_\_\_\) The lamina is freely suspended at \(B\) and hangs in equilibrium.
  2. Find the angle between \(A B\) and the horizontal.
    A force of magnitude 12 N is applied along the edge \(A C\) of the lamina in the direction from \(A\) towards \(C\). The lamina, still suspended at \(B\), is now in equilibrium with \(A B\) vertical.
  3. Calculate the weight of the lamina.

Question 6(i):
AnswerMarks Guidance
AnswerMarks Guidance
From \(AB = 0.2\)B1
From \(BC = 0.1\)B1
Question 6(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\tan\theta = \frac{0.1}{0.2}\)M1 \(\theta\) is the angle between \(AB\) and the horizontal
\(\theta = 26.6°\)A1
Question 6(iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(12\cos 26.6 \times 0.3 = W \times 0.2\)M1A1 Take moments about \(B\). (\(W\) is the weight of the lamina)
\(W = 16.1 \text{ N}\)A1
## Question 6(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| From $AB = 0.2$ | B1 | |
| From $BC = 0.1$ | B1 | |

---

## Question 6(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan\theta = \frac{0.1}{0.2}$ | M1 | $\theta$ is the angle between $AB$ and the horizontal |
| $\theta = 26.6°$ | A1 | |

---

## Question 6(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $12\cos 26.6 \times 0.3 = W \times 0.2$ | M1A1 | Take moments about $B$. ($W$ is the weight of the lamina) |
| $W = 16.1 \text{ N}$ | A1 | |

---
6\\
$A B C$ is a uniform lamina in the form of a triangle with $A B = 0.3 \mathrm {~m} , B C = 0.6 \mathrm {~m}$ and a right angle at $B$ (see diagram).\\
(i) State the distances of the centre of mass of the lamina from $A B$ and from $B C$.

Distance from $A B$\\

Distance from $B C$ $\_\_\_\_$\\

The lamina is freely suspended at $B$ and hangs in equilibrium.\\
(ii) Find the angle between $A B$ and the horizontal.\\

A force of magnitude 12 N is applied along the edge $A C$ of the lamina in the direction from $A$ towards $C$. The lamina, still suspended at $B$, is now in equilibrium with $A B$ vertical.\\
(iii) Calculate the weight of the lamina.\\

\hfill \mbox{\textit{CAIE M2 2019 Q6 [7]}}