CAIE M1 2010 November — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2010
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeEquilibrium of particle under coplanar forces
DifficultyModerate -0.8 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions and solving two simultaneous equations. The setup is clear with given angles, and the method is standard textbook procedure with no conceptual challenges beyond basic trigonometry and Newton's first law.
Spec3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0
3 \includegraphics[max width=\textwidth, alt={}, center]{881993e1-71ea-4801-bfc8-40c17a1387a9-2_597_616_888_762} A particle \(P\) is in equilibrium on a smooth horizontal table under the action of four horizontal forces of magnitudes \(6 \mathrm {~N} , 5 \mathrm {~N} , F \mathrm {~N}\) and \(F \mathrm {~N}\) acting in the directions shown. Find the values of \(\alpha\) and \(F\).
AnswerMarks Guidance
M1For resolving forces in i and j directions (3 terms in at least one of the equations)
\(6\cos\alpha° + 5\cos(90° - \alpha°) = F\) and \(6\sin\alpha° - 5\sin(90° - \alpha°) = F\)A1
\([6\cos\alpha° + 5\sin\alpha° = 6\sin\alpha° - 5\cos\alpha°\) \(\Rightarrow 11\cos\alpha° = \sin\alpha°]\)DM1 For attempting to solve for \(\alpha°\). Dependent on 1st M1
\(\alpha = 84.8\)A1
\([F = 6\cos84.8° + 5\sin84.8°; F = 6\sin84.8° - 5\cos84.8°]\)DM1 For substituting to find F; dependent on the 1st M1
\(F = 5.52\)A1 [6]
Question 3 (First alternative scheme)
AnswerMarks Guidance
\([2F^2 = 25 + 36]\)M1 For using '(resultant of forces of magnitude F)² = (resultant of forces of magnitudes 5 and 6)²'
\(F = 5.52\)A1
M1For using 'resultant of forces of magnitudes 5 and 6 makes angle 45° with x-axis'
M1For using relevant trigonometry
\(\tan(\alpha° - 45°) = 5/6\) or \(\tan(135° - \alpha°) = 6/5\) or \(\cos(\alpha° - 45°) = 6/\sqrt{61}\) or \(\sin(135° - \alpha°) = 6/\sqrt{61}\) or \(\sin(\alpha° - 45°) = 5/\sqrt{61}\) or \(\cos(135° - \alpha°) = 5/\sqrt{61}\)A1
\(\alpha = 84.8\)A1
Question 3 (Second alternative scheme)
AnswerMarks Guidance
\([6\cos\alpha° + 5\cos(90° - \alpha°) = 6\sin\alpha° - 5\sin(90° - \alpha°)]\)M1 For using \(Rx = Ry\)
\([11\cos\alpha° - \sin\alpha° = 0]\)M1 For attempting to solve for \(\alpha°\)
\(\alpha = 84.8\)A1
For \(F = 6\cos\alpha° + 5\cos(90° - \alpha°)\) or \(F = 6\sin\alpha° - 5\sin(90° - \alpha°)\)B1
M1For substituting for \(\alpha\)
\(F = 5.52\)A1 
| M1 | For resolving forces in i and j directions (3 terms in at least one of the equations)

$6\cos\alpha° + 5\cos(90° - \alpha°) = F$ and $6\sin\alpha° - 5\sin(90° - \alpha°) = F$ | A1 |

$[6\cos\alpha° + 5\sin\alpha° = 6\sin\alpha° - 5\cos\alpha°$ $\Rightarrow 11\cos\alpha° = \sin\alpha°]$ | DM1 | For attempting to solve for $\alpha°$. Dependent on 1st M1

$\alpha = 84.8$ | A1 |

$[F = 6\cos84.8° + 5\sin84.8°; F = 6\sin84.8° - 5\cos84.8°]$ | DM1 | For substituting to find F; dependent on the 1st M1

$F = 5.52$ | A1 | [6]

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# Question 3 (First alternative scheme)

$[2F^2 = 25 + 36]$ | M1 | For using '(resultant of forces of magnitude F)² = (resultant of forces of magnitudes 5 and 6)²'

$F = 5.52$ | A1 |
| M1 | For using 'resultant of forces of magnitudes 5 and 6 makes angle 45° with x-axis'

| M1 | For using relevant trigonometry

$\tan(\alpha° - 45°) = 5/6$ or $\tan(135° - \alpha°) = 6/5$ or $\cos(\alpha° - 45°) = 6/\sqrt{61}$ or $\sin(135° - \alpha°) = 6/\sqrt{61}$ or $\sin(\alpha° - 45°) = 5/\sqrt{61}$ or $\cos(135° - \alpha°) = 5/\sqrt{61}$ | A1 |

$\alpha = 84.8$ | A1 |

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# Question 3 (Second alternative scheme)

$[6\cos\alpha° + 5\cos(90° - \alpha°) = 6\sin\alpha° - 5\sin(90° - \alpha°)]$ | M1 | For using $Rx = Ry$

$[11\cos\alpha° - \sin\alpha° = 0]$ | M1 | For attempting to solve for $\alpha°$

$\alpha = 84.8$ | A1 |

For $F = 6\cos\alpha° + 5\cos(90° - \alpha°)$ or $F = 6\sin\alpha° - 5\sin(90° - \alpha°)$ | B1 |

| M1 | For substituting for $\alpha$

$F = 5.52$ | A1 |

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\includegraphics[max width=\textwidth, alt={}, center]{881993e1-71ea-4801-bfc8-40c17a1387a9-2_597_616_888_762}

A particle $P$ is in equilibrium on a smooth horizontal table under the action of four horizontal forces of magnitudes $6 \mathrm {~N} , 5 \mathrm {~N} , F \mathrm {~N}$ and $F \mathrm {~N}$ acting in the directions shown. Find the values of $\alpha$ and $F$.

\hfill \mbox{\textit{CAIE M1 2010 Q3 [6]}}
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