Edexcel M1 — Question 3 9 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeForces in equilibrium (find unknowns)
DifficultyModerate -0.3 This is a standard M1 equilibrium problem requiring conversion of bearings to component form and applying the equilibrium condition (sum of forces = 0). While it involves multiple steps (converting bearings, resolving forces, solving simultaneous equations, finding magnitude/direction), these are all routine procedures taught explicitly in M1 with no novel insight required. Slightly easier than average due to the straightforward nature of the calculations.
Spec3.03a Force: vector nature and diagrams3.03n Equilibrium in 2D: particle under forces3.03p Resultant forces: using vectors

3. A particle is in equilibrium under the action of three forces \(\mathbf { P } , \mathbf { Q }\) and \(\mathbf { R }\) acting in the same horizontal plane. \(P\) has magnitude 9 N and acts on a bearing of \(030 ^ { \circ } . Q\) has magnitude 12 N . and acts on a bearing of \(225 ^ { \circ }\).
  1. Find the values of \(a\) and \(b\) such that \(\mathbf { R } = ( a \mathbf { i } + b \mathbf { j } ) \mathrm { N }\), where \(\mathbf { i }\) and \(\mathbf { j }\) are unit vectors in the directions due East and due North respectively.
  2. Calculate the magnitude and direction of \(\mathbf { R }\)

AnswerMarks Guidance
(a) Total force to north: \(9\cos 30° - 12\cos 45° = -0.691 \text{ N}\)M1 A1
Total force to east: \(9\sin 30° - 12\sin 45° = -3.985 \text{ N}\)M1 A1
So \(\mathbf{R} = -3.99\mathbf{i} + 0.69\mathbf{j}\); \(a = 3.99\), \(b = 0.69\)A1
(b) \(\mathbf{R} = \sqrt{(3.985)^2 + (0.691)^2} = 4.04 \text{ N}\), at \(\tan^{-1} 5.77 = 080.1°\)
(a) Total force to north: $9\cos 30° - 12\cos 45° = -0.691 \text{ N}$ | M1 A1 |

Total force to east: $9\sin 30° - 12\sin 45° = -3.985 \text{ N}$ | M1 A1 |

So $\mathbf{R} = -3.99\mathbf{i} + 0.69\mathbf{j}$; $a = 3.99$, $b = 0.69$ | A1 |

(b) $|\mathbf{R}| = \sqrt{(3.985)^2 + (0.691)^2} = 4.04 \text{ N}$, at $\tan^{-1} 5.77 = 080.1°$ | M1 A1 M1 A1 | 9 marks
3. A particle is in equilibrium under the action of three forces $\mathbf { P } , \mathbf { Q }$ and $\mathbf { R }$ acting in the same horizontal plane. $P$ has magnitude 9 N and acts on a bearing of $030 ^ { \circ } . Q$ has magnitude 12 N . and acts on a bearing of $225 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $a$ and $b$ such that $\mathbf { R } = ( a \mathbf { i } + b \mathbf { j } ) \mathrm { N }$, where $\mathbf { i }$ and $\mathbf { j }$ are unit vectors in the directions due East and due North respectively.
\item Calculate the magnitude and direction of $\mathbf { R }$
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q3 [9]}}