| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod or block on rough surface in limiting equilibrium (no wall) |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring resolution of forces and taking moments about a point, with straightforward application of limiting equilibrium conditions. The question guides students by giving the tension answer in part (i), and part (ii) is a direct application of F = μR. While it involves multiple steps, all techniques are routine for M2 level with no novel insight required. |
| Spec | 3.03b Newton's first law: equilibrium6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moment of \(W\) about \(D\): \(W(0.4^2 + 0.6^2)^{1/2}\cos(20^\circ + \tan^{-1}\frac{2}{3})\) | ||
| or \(W(0.6\cos20^\circ - 0.4\sin20^\circ) = (0.427W)\) | B2 | |
| M1 | For taking moments about \(D\) | |
| \(0.8T = 350 \times 0.427\) | A1ft | |
| Tension is \(187\text{ N}\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = 350 + T\sin20^\circ\) | B1 | |
| \(F = T\cos20^\circ\) | B1 | |
| \(\mu = 176/414\) | M1 | For using \(\mu = F/R\) |
| Coefficient is \(0.424\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moment of \(W\) about \(E\): \(W\left[0.8/\sin20^\circ + (0.4^2+0.6^2)^{\frac{1}{2}}\cos(20^\circ+\tan^{-1}\frac{2}{3})\right]\) | B2 | |
| M1 | For taking moments about \(E\) | |
| \(2.34R = 2.766 \times 350\) | A1ft | |
| \(R = 350 + T\sin20^\circ\) | B1 | |
| Tension is \(187\text{ N}\) | A1 | |
| \(F = T\cos20^\circ\) | B1 | |
| \(\mu = 176/414\) | M1 | For using \(\mu = F/R\) |
| Coefficient is \(0.424\) | A1 | [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance of line of action of \(R\) from \(G\): \((0.4^2+0.6^2)^{\frac{1}{2}}\cos(20^\circ+\tan^{-1}\frac{2}{3})\) | ||
| Distance of line of action of \(F\) from \(G\): \((0.4^2+0.6^2)^{\frac{1}{2}}\sin(20^\circ+\tan^{-1}\frac{2}{3})\) | B2 | |
| M1 | For taking moments about \(G\), the centre of mass of the block | |
| \(0.4T + 0.581F = 0.427R\) | A1ft | |
| \(R = 350 + T\sin20^\circ\) | B1 | |
| \(F = T\cos20^\circ\) | B1 | |
| Tension is \(187\text{ N}\) | A1 | |
| \(\mu = 176/414\) | M1 | For using \(\mu = F/R\) |
| Coefficient is \(0.424\) | A1 | [9] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moment of $W$ about $D$: $W(0.4^2 + 0.6^2)^{1/2}\cos(20^\circ + \tan^{-1}\frac{2}{3})$ | | |
| or $W(0.6\cos20^\circ - 0.4\sin20^\circ) = (0.427W)$ | B2 | |
| | M1 | For taking moments about $D$ |
| $0.8T = 350 \times 0.427$ | A1ft | |
| Tension is $187\text{ N}$ | A1 | **[5]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 350 + T\sin20^\circ$ | B1 | |
| $F = T\cos20^\circ$ | B1 | |
| $\mu = 176/414$ | M1 | For using $\mu = F/R$ |
| Coefficient is $0.424$ | A1 | **[4]** |
---
### First Alternative - Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moment of $W$ about $E$: $W\left[0.8/\sin20^\circ + (0.4^2+0.6^2)^{\frac{1}{2}}\cos(20^\circ+\tan^{-1}\frac{2}{3})\right]$ | B2 | |
| | M1 | For taking moments about $E$ |
| $2.34R = 2.766 \times 350$ | A1ft | |
| $R = 350 + T\sin20^\circ$ | B1 | |
| Tension is $187\text{ N}$ | A1 | |
| $F = T\cos20^\circ$ | B1 | |
| $\mu = 176/414$ | M1 | For using $\mu = F/R$ |
| Coefficient is $0.424$ | A1 | **[9]** |
---
### Second Alternative - Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance of line of action of $R$ from $G$: $(0.4^2+0.6^2)^{\frac{1}{2}}\cos(20^\circ+\tan^{-1}\frac{2}{3})$ | | |
| Distance of line of action of $F$ from $G$: $(0.4^2+0.6^2)^{\frac{1}{2}}\sin(20^\circ+\tan^{-1}\frac{2}{3})$ | B2 | |
| | M1 | For taking moments about $G$, the centre of mass of the block |
| $0.4T + 0.581F = 0.427R$ | A1ft | |
| $R = 350 + T\sin20^\circ$ | B1 | |
| $F = T\cos20^\circ$ | B1 | |
| Tension is $187\text{ N}$ | A1 | |
| $\mu = 176/414$ | M1 | For using $\mu = F/R$ |
| Coefficient is $0.424$ | A1 | **[9]** |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{5109244c-3062-4f5f-9277-fc6b5b28f2d4-4_495_1405_264_370}\\
$A B C D$ is a central cross-section of a uniform rectangular block of mass 35 kg . The lengths of $A B$ and $B C$ are 1.2 m and 0.8 m respectively. The block is held in equilibrium by a rope, one end of which is attached to the point $E$ of a rough horizontal floor. The other end of the rope is attached to the block at $A$. The rope is in the same vertical plane as $A B C D$, and $E A B$ is a straight line making an angle of $20 ^ { \circ }$ with the horizontal (see diagram).\\
(i) Show that the tension in the rope is 187 N , correct to the nearest whole number.\\
(ii) The block is on the point of slipping. Find the coefficient of friction between the block and the floor.
\hfill \mbox{\textit{CAIE M2 2008 Q5 [9]}}