| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod with string perpendicular |
| Difficulty | Standard +0.3 This is a standard moments equilibrium problem requiring taking moments about the hinge, resolving forces, and finding a resultant angle. The perpendicular rope simplifies calculations significantly. While it involves multiple steps (moments, horizontal/vertical resolution, arctangent), these are routine techniques for M2 level with no conceptual surprises or geometric complexity. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (i) T x 0.8 = 70x1sin α + 220x2sin α | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| sin = 1.5/1.7 | A1 | cos α = 0.8/1.7 α = 61.9° |
| T = 562.5 N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) H = 562.5cos α = 265 N | B1 | H = 264.70 N |
| Answer | Marks | Guidance |
|---|---|---|
| V = 562.5sin – 70 – 220 | M1 | V = 206.3 N |
| Answer | Marks | Guidance |
|---|---|---|
| tan = 265/206.3 | M1 | |
| α = 52.1° (with vertical) | A1 | Or 37.9 (with horizontal) |
| OR | α | |
| X = (70+220)cos = 136.6 | B1 | Resolving along the rod AB |
| Answer | Marks | Guidance |
|---|---|---|
| Y = 562.5 – (70+220)sin = 306.7 | M1 | Resolving perpendicular to AB |
| Answer | Marks |
|---|---|
| tan = 306.7/136.6 | M1 |
| θ = 65.99° or 66.0° (with beam) | A1 |
Question 4:
4 | (i) T x 0.8 = 70x1sin α + 220x2sin α | M1 | Moments about A (3 terms)
α
sin = 1.5/1.7 | A1 | cos α = 0.8/1.7 α = 61.9°
T = 562.5 N | A1
[3]
(ii) H = 562.5cos α = 265 N | B1 | H = 264.70 N
α
V = 562.5sin – 70 – 220 | M1 | V = 206.3 N
α
tan = 265/206.3 | M1
α = 52.1° (with vertical) | A1 | Or 37.9 (with horizontal)
OR | α
X = (70+220)cos = 136.6 | B1 | Resolving along the rod AB
α
Y = 562.5 – (70+220)sin = 306.7 | M1 | Resolving perpendicular to AB
θ
tan = 306.7/136.6 | M1
θ = 65.99° or 66.0° (with beam) | A1
[4]\includegraphics{figure_4}
A uniform beam $AB$ has length $2 \text{ m}$ and weight $70 \text{ N}$. The beam is hinged at $A$ to a fixed point on a vertical wall, and is held in equilibrium by a light inextensible rope. One end of the rope is attached to the wall at a point $1.7 \text{ m}$ vertically above the hinge. The other end of the rope is attached to the beam at a point $0.8 \text{ m}$ from $A$. The rope is at right angles to $AB$. The beam carries a load of weight $220 \text{ N}$ at $B$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Find the tension in the rope. [3]
\item Find the direction of the force exerted on the beam at $A$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q4 [7]}}