CAIE M1 2011 November — Question 2 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2011
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeResultant of coplanar forces
DifficultyModerate -0.3 This is a standard resultant force problem requiring resolution of forces into components using basic trigonometry (finding sin α and cos α from tan α), summing horizontal and vertical components, then using Pythagoras and inverse tan to find magnitude and direction. It's slightly easier than average because it's a straightforward application of a well-practiced technique with no conceptual challenges, though the 3-4-5 triangle recognition and arithmetic with multiple forces requires care.
Spec3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors

2 \includegraphics[max width=\textwidth, alt={}, center]{155bc571-80e4-4c93-859f-bb150a109211-2_652_493_1457_826} Coplanar forces of magnitudes \(58 \mathrm {~N} , 31 \mathrm {~N}\) and 26 N act at a point in the directions shown in the diagram. Given that \(\tan \alpha = \frac { 5 } { 12 }\), find the magnitude and direction of the resultant of the three forces.
[0pt] [6]

AnswerMarks Guidance
\(X = 31 + 26\cos\alpha, Y = 58 - 26\sin\alpha\)A1
\(X = 55, Y = 48\)A1
dM1 For using \(R = (X^2 + Y^2)^{1/2}\) or \(\tan \theta = Y/X\)
Resultant is 73N or Direction is at 41.1° to i directionA1
Direction is at 41.1° to i direction or Resultant is 73NB1 6
Alternative solution for Q2:
AnswerMarks Guidance
\([\tan \theta_{12} = 58/31, R_{12} = \sqrt{31^2 + 58^2}]\)M1 For finding an angle and the hypotenuse of a right angled triangle whose other sides are 31 & 58
\(\theta_{12} = 61.9°\) and \(R_{12} = 65.76\)A1
\([\text{Incl. angle} = (180 - \theta_{12} - a)°, R^2 = 26^2 + R_{12}^2 - 2 \times 26R_{12}\cos(\text{incl. angle})]\)M1 For finding the included angle between sides \(R_{12}\) and 26 and using the cosine rule to find R
Incl. angle = 95.5°, Resultant is 73 NA1
\([\sin \beta = 26\sin 95.5/73; \theta = 61.9 - \beta]\)M1 For using the sine rule in the triangle to find the angle opposite 26 and subtracting this from \(\theta_{12}\)
Direction is at 41.1° to i directionA1
$X = 31 + 26\cos\alpha, Y = 58 - 26\sin\alpha$ | A1 | | For resolving in i and j directions
$X = 55, Y = 48$ | A1 | | May be implied
| dM1 | | For using $R = (X^2 + Y^2)^{1/2}$ or $\tan \theta = Y/X$
Resultant is 73N or Direction is at 41.1° to i direction | A1 |
Direction is at 41.1° to i direction or Resultant is 73N | B1 | 6

## Alternative solution for Q2:

$[\tan \theta_{12} = 58/31, R_{12} = \sqrt{31^2 + 58^2}]$ | M1 | For finding an angle and the hypotenuse of a right angled triangle whose other sides are 31 & 58
$\theta_{12} = 61.9°$ and $R_{12} = 65.76$ | A1 |
$[\text{Incl. angle} = (180 - \theta_{12} - a)°, R^2 = 26^2 + R_{12}^2 - 2 \times 26R_{12}\cos(\text{incl. angle})]$ | M1 | For finding the included angle between sides $R_{12}$ and 26 and using the cosine rule to find R
Incl. angle = 95.5°, Resultant is 73 N | A1 |
$[\sin \beta = 26\sin 95.5/73; \theta = 61.9 - \beta]$ | M1 | For using the sine rule in the triangle to find the angle opposite 26 and subtracting this from $\theta_{12}$
Direction is at 41.1° to i direction | A1 |
2\\
\includegraphics[max width=\textwidth, alt={}, center]{155bc571-80e4-4c93-859f-bb150a109211-2_652_493_1457_826}

Coplanar forces of magnitudes $58 \mathrm {~N} , 31 \mathrm {~N}$ and 26 N act at a point in the directions shown in the diagram. Given that $\tan \alpha = \frac { 5 } { 12 }$, find the magnitude and direction of the resultant of the three forces.\\[0pt]
[6]

\hfill \mbox{\textit{CAIE M1 2011 Q2 [6]}}