Moderate -0.3 This is a standard M1 equilibrium problem requiring resolution of forces in two perpendicular directions to find two unknowns. While it involves trigonometry and simultaneous equations, it follows a routine method taught explicitly in mechanics courses with no novel problem-solving required, making it slightly easier than average.
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Three coplanar forces of magnitudes \(F \mathrm {~N} , 12 \mathrm {~N}\) and 15 N are in equilibrium acting at a point \(P\) in the directions shown in the diagram. Find \(\alpha\) and \(F\).
For using Lami's rule and \(\sin(180°-\alpha) = \sin\alpha\)
\(\alpha = 53.1\)
A1
\([F \div \sin143.1 = 15 \div \sin90]\)
M1
For using Lami's rule and value of \(\alpha\) to find \(F\)
\(F = 9\) N
A1
[4]
SR (max 2/4) for candidates who have sin and cos interchanged. Allow B1 for \(\alpha = 36.9\) and B1 for \(F = 9\) following correct work relative to the cos/sin interchange error.
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[12 = 15\sin\alpha]$ | M1 | For resolving forces in the direction of the force of magnitude 12 N |
| $\alpha = 53.1$ | A1 | |
| $[F = 15\cos\alpha]$ | M1 | For resolving forces in the direction of the force of magnitude $F$ N |
| $F = 9$ N | A1 | [4] |
**Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F\sin\alpha = 12\cos\alpha$ and $F\cos\alpha + 12\sin\alpha = 15 \Rightarrow \sin\alpha \div \cos\alpha = 12\cos\alpha \div 15 - 12\sin\alpha$ | M1 | For resolving forces in the $x$ and $y$ directions and eliminating $F$ from the resultant equations |
| $15\sin\alpha - 12\sin^2\alpha = 12\cos^2\alpha \Rightarrow 15\sin\alpha = 12 \Rightarrow \alpha = 53.1$ | A1 | |
| | M1 | For substituting into $F\sin\alpha = 12\cos\alpha$ or $F\cos\alpha + 12\sin\alpha = 15$ |
| $F = 9$ N | A1 | [4] |
**Alternative 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\sin\alpha = 12/15]$ | M1 | For using correct triangle of forces to find $\alpha$ |
| $\alpha = 53.1$ | A1 | |
| $[F^2 = 15^2 - 12^2]$ | M1 | For using correct triangle of forces to find $F$ |
| $F = 9$ N | A1 | [4] |
**Alternative 3:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[12 \div \sin(180-\alpha) = 15 \div \sin90 \Rightarrow 12 = 15\sin\alpha]$ | M1 | For using Lami's rule and $\sin(180°-\alpha) = \sin\alpha$ |
| $\alpha = 53.1$ | A1 | |
| $[F \div \sin143.1 = 15 \div \sin90]$ | M1 | For using Lami's rule and value of $\alpha$ to find $F$ |
| $F = 9$ N | A1 | [4] |
SR (max 2/4) for candidates who have sin and cos interchanged. Allow B1 for $\alpha = 36.9$ and B1 for $F = 9$ following correct work relative to the cos/sin interchange error.
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\includegraphics[max width=\textwidth, alt={}, center]{fa0e0e0d-b0a6-44e0-8b4f-4923e235c6c6-2_465_478_479_836}
Three coplanar forces of magnitudes $F \mathrm {~N} , 12 \mathrm {~N}$ and 15 N are in equilibrium acting at a point $P$ in the directions shown in the diagram. Find $\alpha$ and $F$.
\hfill \mbox{\textit{CAIE M1 2012 Q2 [4]}}