| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Moderate -0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions to find two unknowns. While it involves trigonometry and simultaneous equations, it's a routine textbook exercise with a straightforward method. Part (ii) is trivial once part (i) is solved. Slightly easier than average due to the mechanical nature of the solution process. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | For resolving forces in the \(i\) and \(j\) directions |
| \(F\cos\theta = 12\cos 30° (= 10.932)\) | A1 | |
| \(F\sin\theta = 10 - 12\sin 30° (= 4)\) | A1 | |
| M1 | For using \(F^2 = X^2 + Y^2\) or \(\tan\theta = Y/X\) | |
| \(F = 11.1\) or \(\theta = 21.1\) (accept 21.0) | A1 | |
| \(\theta = 21.1\) (accept 21.0) or \(F = 11.1\) | B1 | [6] |
| (ii) Magnitude is 12N | B1 | |
| Direction is 30° clockwise from \(+ve\) '\(x\)' axis | B1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| For triangle of forces with sides 12, F and 10 and at least one of the angles \((90° - \theta)\) or \(60° + (\theta + 30°)\) | B1 | |
| M1 | For use of cosine rule (with \(\theta\) absent) or use of sine rule (with F absent) and use of \(\sin(A \pm B) = \sin A\cos B \pm \sin B\cos A\) | |
| \(F^2 = 12^2 + 10^2 - 2 \times 12 \times 10\cos 60°\) or \((12\cos 30°)\sin\theta = (10 - 12\sin 30°)\cos\theta\) | A1 | |
| \(F = 11.1\) or \(\theta = 21.1\) (accept 21.0) | A1 | |
| M1 | For correct method for \(\theta\) or F | |
| \(\theta = 21.1\) (accept 21.0) or \(F = 11.1\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| For using Lami's theorem with 12 N and 10 N | M1 | |
| \(12/\sin(90° + \theta) = 10/\sin(150° - \theta)\) | A1 | |
| \(12/\cos\theta = 20 \div (\cos(\alpha + 3\frac{1}{2}\sin\theta)\) \(\to 12 \times 3^{\frac{1}{2}}\sin\theta = 8\cos\theta\) \(\to \tan\theta = 2 \div (3 \times 3^{-\frac{1}{2}})\) \(\to \theta = 21.1\) | A1 | |
| For using Lami's theorem with F N and (12 N or 10 N) | M1 | |
| \(F/\sin 120° = 12/\sin 111.1°\) (or \(10/\sin 128.9°\)) | A1 | |
| \(F = 11.1\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| For \(X = 11.1\cos 21.1°\) and \(Y = 11.1\sin 21.1° - 10\), \(R^2 = X^2 + Y^2\) and \(\tan\Phi = Y/X\) | M1 | |
| Magnitude 12 N and direction 30° clockwise from \(+ve\) x-axis | A1 | [2] |
**(i)** | M1 | For resolving forces in the $i$ and $j$ directions
$F\cos\theta = 12\cos 30° (= 10.932)$ | A1 |
$F\sin\theta = 10 - 12\sin 30° (= 4)$ | A1 |
| M1 | For using $F^2 = X^2 + Y^2$ or $\tan\theta = Y/X$
$F = 11.1$ or $\theta = 21.1$ (accept 21.0) | A1 |
$\theta = 21.1$ (accept 21.0) or $F = 11.1$ | B1 | [6] | SR for candidates who consistently have $\cos$ for $\sin$ and vice versa (max 4/6) M1 as above (resolving) A1 for $F\sin\theta = 12\sin 30°$ and $F\cos\theta = 10 - 12\cos 30°$ M1 as above $F^2 = .... \&$ tan $\theta = ....$ A1 for $F = 6.01$ and $\theta = 93.7$
**(ii) Magnitude is 12N** | B1 |
Direction is 30° clockwise from $+ve$ '$x$' axis | B1 | [2]
**alternative for 4(i)**
For triangle of forces with sides 12, F and 10 and at least one of the angles $(90° - \theta)$ or $60° + (\theta + 30°)$ | B1 |
| M1 | For use of cosine rule (with $\theta$ absent) or use of sine rule (with F absent) and use of $\sin(A \pm B) = \sin A\cos B \pm \sin B\cos A$
$F^2 = 12^2 + 10^2 - 2 \times 12 \times 10\cos 60°$ or $(12\cos 30°)\sin\theta = (10 - 12\sin 30°)\cos\theta$ | A1 |
$F = 11.1$ or $\theta = 21.1$ (accept 21.0) | A1 |
| M1 | For correct method for $\theta$ or F
$\theta = 21.1$ (accept 21.0) or $F = 11.1$ | A1 | [6]
**second alternative for 4(i)**
For using Lami's theorem with 12 N and 10 N | M1 |
$12/\sin(90° + \theta) = 10/\sin(150° - \theta)$ | A1 |
$12/\cos\theta = 20 \div (\cos(\alpha + 3\frac{1}{2}\sin\theta)$ $\to 12 \times 3^{\frac{1}{2}}\sin\theta = 8\cos\theta$ $\to \tan\theta = 2 \div (3 \times 3^{-\frac{1}{2}})$ $\to \theta = 21.1$ | A1 |
For using Lami's theorem with F N and (12 N or 10 N) | M1 |
$F/\sin 120° = 12/\sin 111.1°$ (or $10/\sin 128.9°$) | A1 |
$F = 11.1$ | A1 | [6]
**Alternative for 4(ii)**
For $X = 11.1\cos 21.1°$ and $Y = 11.1\sin 21.1° - 10$, $R^2 = X^2 + Y^2$ and $\tan\Phi = Y/X$ | M1 |
Magnitude 12 N and direction 30° clockwise from $+ve$ x-axis | A1 | [2]4\\
\begin{tikzpicture}[scale=1]
% Dashed horizontal line through P
\draw[dashed] (-3.5,0) -- (4.5,0);
% Point P
\fill (0,0) circle (1.5pt) node[below=2pt] {$P$};
% 12 N force: upward-left at 30 from horizontal
\draw[->, thick] (0,0) -- ({3*cos(150)},{3*sin(150)});
\node[above left] at ({3*cos(150)},{3*sin(150)}) {$12\;\mathrm{N}$};
% F N force: to the upper-right at angle theta above horizontal
\draw[->, thick] (0,0) -- ({3*cos(20)},{3*sin(20)});
\node[right] at ({3*cos(20)},{3*sin(20)}) {$F\;\mathrm{N}$};
% 10 N force: straight down
\draw[->, thick] (0,0) -- (0,-3);
\node[below] at (0,-3) {$10\;\mathrm{N}$};
% Angle arc for 30
\draw (0,0) +({-0.9},0) arc (180:150:0.9);
\node[above left] at ({-0.65},0.15) {$30°$};
% Angle arc for theta
\draw (0,0) +(0.9,0) arc (0:20:0.9);
\node[right] at (0.85,0.2) {$\theta°$};
\end{tikzpicture}
The three coplanar forces shown in the diagram act at a point $P$ and are in equilibrium.\\
(i) Find the values of $F$ and $\theta$.\\
(ii) State the magnitude and direction of the resultant force at $P$ when the force of magnitude 12 N is removed.
\hfill \mbox{\textit{CAIE M1 2011 Q4 [8]}}