Moderate -0.3 This is a straightforward vector resolution problem requiring students to resolve three forces into components, sum them, and find the magnitude and direction of the resultant. While it involves multiple steps and trigonometry with given sin/cos values, it follows a standard mechanical procedure taught in M1 with no conceptual challenges or novel problem-solving required. The 7 marks reflect routine calculation work rather than difficulty.
\includegraphics{figure_4}
Coplanar forces of magnitudes \(250 \text{ N}\), \(160 \text{ N}\) and \(370 \text{ N}\) 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 N
A1ℹ
ℹ 264 N for inconsistent sin/cos mix
M1
For using \(\tan\theta = Y/X\)
Required angle is 36.9° (or 0.644 rads)
A1ℹ
ℹ 29.5° for inconsistent 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 N | A1ℹ | ℹ 264 N for inconsistent sin/cos mix
| M1 | For using $\tan\theta = Y/X$
Required angle is 36.9° (or 0.644 rads) | A1ℹ | ℹ 29.5° for inconsistent sin/cos mix
| | [7]
# 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 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]
\includegraphics{figure_4}
Coplanar forces of magnitudes $250 \text{ N}$, $160 \text{ N}$ and $370 \text{ N}$ 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]}}