Moderate -0.3 This is a standard resultant force problem requiring resolution of forces into components using given sine values (from which cosines can be found via Pythagorean identity), followed by vector addition. It's slightly easier than average because the trigonometric values are given explicitly and the method is completely routine for M1 students, though it requires careful arithmetic across multiple steps.
2
\includegraphics[max width=\textwidth, alt={}, center]{41e63d05-d109-47dc-80a6-927953e3e607-03_659_655_258_744}
Coplanar forces of magnitudes \(34 \mathrm {~N} , 30 \mathrm {~N}\) and 26 N act at a point in the directions shown in the diagram.
Given that \(\sin \alpha = \frac { 5 } { 13 }\) and \(\sin \theta = \frac { 8 } { 17 }\), find the magnitude and direction of the resultant of the three forces.
\([\beta =] \tan^{-1}\left(\frac{Y}{X}\right)\) or \([\beta =] \tan^{-1}\left(\frac{X}{Y}\right)\)
M1
Attempt to solve for the direction of the resultant force
Question 2 (cont'd):
Answer
Marks
Guidance
Answer
Marks
Guidance
\(R = \sqrt{52} = 2\sqrt{13} = 7.21\) N and \(\beta = 56.3\) above 30N force or anticlockwise from 30N force
A1
Both correct with correct explanation of the direction. Must be a correct and clear explanation.
6
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolve either horizontally or vertically with correct number of terms | **M1** | Allow $\theta$ and $\alpha$ as in the question for this mark |
| $[X =] 30 - 34 \times \frac{8}{17} - 26 \times \frac{5}{13} [= 4]$ | **A1** | Allow $\pm X$; allow $[X=]30 - 34\sin 28 - 26\sin 23$ angle 2s.f. or better |
| $[Y =] 34 \times \frac{15}{17} - 26 \times \frac{12}{13} [= 6]$ | **A1** | Allow $\pm Y$; allow $[Y=]34\cos 28 - 26\cos 23$ angle 2s.f. or better |
| $[R =] \sqrt{X^2 + Y^2}$ | **M1** | Attempt to solve for the magnitude of the force |
| $[\beta =] \tan^{-1}\left(\frac{Y}{X}\right)$ or $[\beta =] \tan^{-1}\left(\frac{X}{Y}\right)$ | **M1** | Attempt to solve for the direction of the resultant force |
## Question 2 (cont'd):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = \sqrt{52} = 2\sqrt{13} = 7.21$ N and $\beta = 56.3$ above 30N force or anticlockwise from 30N force | A1 | Both correct with correct explanation of the direction. Must be a correct and clear explanation. |
| | **6** | |
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2\\
\includegraphics[max width=\textwidth, alt={}, center]{41e63d05-d109-47dc-80a6-927953e3e607-03_659_655_258_744}
Coplanar forces of magnitudes $34 \mathrm {~N} , 30 \mathrm {~N}$ and 26 N act at a point in the directions shown in the diagram.
Given that $\sin \alpha = \frac { 5 } { 13 }$ and $\sin \theta = \frac { 8 } { 17 }$, find the magnitude and direction of the resultant of the three forces.\\
\hfill \mbox{\textit{CAIE M1 2021 Q2 [6]}}