Standard +0.3 This is a standard mechanics resultant force problem requiring resolution of forces into components and Pythagoras/trigonometry. While it involves multiple forces and given trigonometric values, it follows a routine algorithmic approach (resolve horizontally, resolve vertically, find magnitude and direction) that is well-practiced at this level. Slightly above average difficulty due to three forces and the need for careful component calculation, but no novel problem-solving required.
\includegraphics{figure_4}
Coplanar forces of magnitudes 250 N, 160 N and 370 N act at a point \(O\) in the directions shown in the diagram, where the angle \(\alpha\) is such that \(\sin \alpha = 0.28\) and \(\cos \alpha = 0.96\). Calculate the magnitude of the resultant of the three forces. Calculate also the angle that the resultant makes with the \(x\)-direction. [7]
For resolving forces in the \(x\)-direction or in the \(y\)-direction
\(X = 160 + 250\cos \alpha\)
A1
\(Y = 370 - 250\sin \alpha\)
A1
M1
For using \(R^2 = X^2 + Y^2\)
Magnitude is \(500 \text{ N}\)
A1ft
ft 264 N for consistent sin/cos mix
M1
For using \(\tan\theta = \frac{Y}{X}\)
Required angle is \(36.9°\) (or \(0.644 \text{ rads}\))
A1ft
ft 29.5° for consistent sin/cos mix
[7]
Question 4 Alternative
Answer
Marks
Guidance
M1
For finding the resultant in magnitude and direction of two forces and obtaining a triangle enabling the calculation of the resultant of the three forces
| M1 | For resolving forces in the $x$-direction or in the $y$-direction
$X = 160 + 250\cos \alpha$ | A1 |
$Y = 370 - 250\sin \alpha$ | A1 |
| | M1 | For using $R^2 = X^2 + Y^2$
Magnitude is $500 \text{ N}$ | A1ft | ft 264 N for consistent sin/cos mix
| | M1 | For using $\tan\theta = \frac{Y}{X}$
Required angle is $36.9°$ (or $0.644 \text{ rads}$) | A1ft | ft 29.5° for consistent sin/cos mix
| | | **[7]**
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## Question 4 Alternative
| M1 | For finding the resultant in magnitude and direction of two forces and obtaining a triangle enabling the calculation of the resultant of the three forces
Triangle has sides 403, 250 and R | A1 | or equivalent for different choice of two forces*
Triangle has angle opposite R equal to $97.1°$ | A1 | As *
$[R^2 = 403^2 + 250^2 - 2 \times 403 \times 250\cos97.1°]$ | M1 | For using cosine rule to find R
Magnitude is $500 \text{ N}$ | A1 |
$[\sin(66.6° - z) + 250 = \sin97.1° ÷ R]$ | M1 | For using sine rule to find $z$
Required angle is $36.9°$ | A1 |
| | | **[7]**
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\includegraphics{figure_4}
Coplanar forces of magnitudes 250 N, 160 N and 370 N act at a point $O$ in the directions shown in the diagram, where the angle $\alpha$ is such that $\sin \alpha = 0.28$ and $\cos \alpha = 0.96$. Calculate the magnitude of the resultant of the three forces. Calculate also the angle that the resultant makes with the $x$-direction. [7]
\hfill \mbox{\textit{CAIE M1 2010 Q4 [7]}}