| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Block on rough horizontal surface – equilibrium (finding friction, normal reaction, or coefficient of friction) |
| Difficulty | Moderate -0.3 This is a standard two-part equilibrium problem with friction requiring resolution of forces and application of F=μR. The angle is given in a convenient form (3-4-5 triangle), making the trigonometry straightforward. While it involves multiple steps (resolving vertically, finding R, then applying friction condition horizontally), these are routine mechanics techniques with no novel insight required. Slightly easier than average due to the structured parts and standard approach. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(N + X\cos\theta = mg\) | M1 | For resolving forces vertically (3 terms needed) |
| \(N + X\left(\frac{24}{25}\right) = 320 \times 10\) | A1 | |
| \(N = 3200 - \frac{24}{25}X\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = X\sin\theta\) | M1 | For resolving forces horizontally |
| \(\frac{7}{25}X = \frac{3}{8}\left(3200 - \frac{24}{25}X\right)\) | M1 | For using \(F = \mu N\) to obtain an equation in \(X\) only |
| \(X = 1875\) | A1 |
## Question 4:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N + X\cos\theta = mg$ | M1 | For resolving forces vertically (3 terms needed) |
| $N + X\left(\frac{24}{25}\right) = 320 \times 10$ | A1 | |
| $N = 3200 - \frac{24}{25}X$ | A1 | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = X\sin\theta$ | M1 | For resolving forces horizontally |
| $\frac{7}{25}X = \frac{3}{8}\left(3200 - \frac{24}{25}X\right)$ | M1 | For using $F = \mu N$ to obtain an equation in $X$ only |
| $X = 1875$ | A1 | |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{2026cad4-8494-4139-ad21-d8a17ac2b955-3_276_570_264_790}
A stone slab of mass 320 kg rests in equilibrium on rough horizontal ground. A force of magnitude $X \mathrm {~N}$ acts upwards on the slab at an angle of $\theta$ to the vertical, where $\tan \theta = \frac { 7 } { 24 }$ (see diagram).\\
(i) Find, in terms of $X$, the normal component of the force exerted on the slab by the ground.\\
(ii) Given that the coefficient of friction between the slab and the ground is $\frac { 3 } { 8 }$, find the value of $X$ for which the slab is about to slip.
\hfill \mbox{\textit{CAIE M1 2005 Q4 [6]}}