| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Equilibrium on slope with horizontal force |
| Difficulty | Standard +0.3 This is a standard statics problem requiring resolution of forces and taking moments about a point. While it involves multiple steps (moments, resolving forces in two directions, and calculating coefficient of friction), the techniques are routine for M2 level and the question guides students through each part systematically. The geometry is straightforward with the rod perpendicular to the plane, making angle calculations simple. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (i) | M1 |
| Fx1.2sin60° = 15 × 0.6cos60° | A1 | |
| F = 4.33N AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) Fcos30° + Fr = 15cos60° | M1 | Resolving parallel to the plane |
| Fr = 3.75N | A1 | |
| OR | 15 × 0.6cos60° = 1.2Fr | M1 |
| Fr = 3.75N | A1 | |
| OR | Fcos30° × 0.6 = Fr x 0.6 | M1 |
| Fr = 3.75N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) R = 15cos30° + 4.33cos60° | M1 | |
| R = 15.2 | A1 | R = 15.155… Accept 15.1 |
| Answer | Marks |
|---|---|
| (= 3.75/15.2) = 0.247 | B1ft |
| [3] | From their F and R found but not R=W |
Question 4:
4 | (i) | M1 | Moments about A
Fx1.2sin60° = 15 × 0.6cos60° | A1
F = 4.33N AG | A1
[3]
(ii) Fcos30° + Fr = 15cos60° | M1 | Resolving parallel to the plane
Fr = 3.75N | A1
OR | 15 × 0.6cos60° = 1.2Fr | M1 | Moments about B
Fr = 3.75N | A1
OR | Fcos30° × 0.6 = Fr x 0.6 | M1 | Moments about centre of rod
Fr = 3.75N | A1
[2]
(iii) R = 15cos30° + 4.33cos60° | M1
R = 15.2 | A1 | R = 15.155… Accept 15.1
µ
(= 3.75/15.2) = 0.247 | B1ft
[3] | From their F and R found but not R=W\includegraphics{figure_4}
A uniform rod $AB$ has weight $15$ N and length $1.2$ m. The end $A$ of the rod is in contact with a rough plane inclined at $30°$ to the horizontal, and the rod is perpendicular to the plane. The rod is held in equilibrium in this position by means of a horizontal force applied at $B$, acting in the vertical plane containing the rod (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the magnitude of the force applied at $B$ is $4.33$ N, correct to $3$ significant figures. [3]
\item Find the magnitude of the frictional force exerted by the plane on the rod. [2]
\item Given that the rod is in limiting equilibrium, calculate the coefficient of friction between the rod and the plane. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q4 [8]}}