| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Equilibrium with applied force |
| Difficulty | Standard +0.3 This is a standard centre of mass problem with composite shapes (subtraction method) followed by routine equilibrium and toppling conditions. Part (i) involves straightforward calculation using the formula for centre of mass of composite bodies. Parts (ii) and (iii) apply standard toppling criteria (reaction force at edge) with simple moment equations. All techniques are textbook exercises requiring no novel insight, though the multi-part nature and coordinate tracking add modest complexity above the most basic questions. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0.7 \times 0.4) \times 0.2 = (0.28 - 0.03)x + 0.03 \times 0.1\) | M1 | |
| \(x = 0.212\) | A1 | AG |
| \((0.7 \times 0.4) \times 0.35 = (0.28 - 0.03)y + 0.03 \times 0.2\) | M1 | |
| \(y = 0.368\) | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4F = 0.212 \times 70\) | M1 | Topples about A |
| \(F = 37.1\) | A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\theta = \tan^{-1}[(0.4 - 0.212)/0.368]\) | M1 | |
| \(\theta = 27.1\) | A1 (2) |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0.7 \times 0.4) \times 0.2 = (0.28 - 0.03)x + 0.03 \times 0.1$ | M1 | |
| $x = 0.212$ | A1 | **AG** |
| $(0.7 \times 0.4) \times 0.35 = (0.28 - 0.03)y + 0.03 \times 0.2$ | M1 | |
| $y = 0.368$ | A1 (4) | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4F = 0.212 \times 70$ | M1 | Topples about A |
| $F = 37.1$ | A1 (2) | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\theta = \tan^{-1}[(0.4 - 0.212)/0.368]$ | M1 | |
| $\theta = 27.1$ | A1 (2) | |
---4\\
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A uniform object is made by drilling a cylindrical hole through a rectangular block. The axis of the cylindrical hole is perpendicular to the cross-section $A B C D$ through the centre of mass of the object. $A B = C D = 0.7 \mathrm {~m} , B C = A D = 0.4 \mathrm {~m}$, and the centre of the hole is 0.1 m from $A B$ and 0.2 m from $A D$ (see diagram). The hole has a cross-section of area $0.03 \mathrm {~m} ^ { 2 }$.\\
(i) Show that the distance of the centre of mass of the object from $A B$ is 0.212 m , and calculate the distance of the centre of mass from $A D$.
The object has weight 70 N and is placed on a rough horizontal surface, with $A D$ in contact with the surface. A vertically upwards force of magnitude $F \mathrm {~N}$ acts on the object at $C$. The object is on the point of toppling.\\
(ii) Find the value of $F$.
The force acting at $C$ is removed, and the object is placed on a rough plane inclined at an angle $\theta ^ { \circ }$ to the horizontal. $A D$ lies along a line of greatest slope, with $A$ higher than $D$. The plane is sufficiently rough to prevent sliding, and the object does not topple.\\
(iii) Find the greatest possible value of $\theta$.
\hfill \mbox{\textit{CAIE M2 2016 Q4 [8]}}