OCR PURE — Question 9 2 marks

Exam BoardOCR
ModulePURE
Marks2
PaperDownload PDF ↗
TopicVectors Introduction & 2D
TypeForces in equilibrium (find unknowns)
DifficultyModerate -0.8 This is a straightforward application of equilibrium conditions where forces sum to zero. Students simply need to add the two given force vectors and negate the result to find F. It requires only basic vector addition and understanding that equilibrium means ΣF = 0, making it easier than average with no problem-solving insight needed.
Spec1.10a Vectors in 2D: i,j notation and column vectors3.03m Equilibrium: sum of resolved forces = 0

9 Three forces \(\binom { 7 } { - 6 } \mathrm {~N} , \binom { 2 } { 5 } \mathrm {~N}\) and \(\mathbf { F N }\) act on a particle.
Given that the particle is in equilibrium under the action of these three forces, calculate \(\mathbf { F }\).

Question 9:
AnswerMarks Guidance
\(\mathbf{F} = \begin{pmatrix} -9 \\ 1 \end{pmatrix}\)B2 (B1 for one correct value) Allow \(-9\mathbf{i} + \mathbf{j}\); SC B1 for \((-9, 1)\) or \((-9 \quad 1)\)
## Question 9:

$\mathbf{F} = \begin{pmatrix} -9 \\ 1 \end{pmatrix}$ | **B2** (B1 for one correct value) | Allow $-9\mathbf{i} + \mathbf{j}$; SC B1 for $(-9, 1)$ or $(-9 \quad 1)$

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9 Three forces $\binom { 7 } { - 6 } \mathrm {~N} , \binom { 2 } { 5 } \mathrm {~N}$ and $\mathbf { F N }$ act on a particle.\\
Given that the particle is in equilibrium under the action of these three forces, calculate $\mathbf { F }$.

\hfill \mbox{\textit{OCR PURE  Q9 [2]}}