Moderate -0.3 This is a straightforward application of the sine rule to a force triangle. Students need to recognize the geometry (angles in the triangle are 40°, 80°, and 60°), then apply sine rule: Q/sin(40°) = 12/sin(60°). This requires basic trigonometry and understanding of vector addition, making it slightly easier than average for M1.
3
\includegraphics[max width=\textwidth, alt={}, center]{a9f3480e-7a8a-497d-a26a-b2aba9b05512-2_462_721_1672_712}
Two forces have magnitudes \(P \mathrm {~N}\) and \(Q \mathrm {~N}\). The resultant of the two forces has magnitude 12 N and acts in a direction \(40 ^ { \circ }\) clockwise from the force of magnitude \(P \mathrm {~N}\) and \(80 ^ { \circ }\) anticlockwise from the force of magnitude \(Q \mathrm {~N}\) (see diagram). Find the value of \(Q\).
For triangle of forces with sides P, Q and 12, and values of any 2 angles shown or implied (Note P, Q and –R are in equil.)
Angles opposite Q and 12 are \(40°\) and \(60°\) respectively
M1
\(Q/\sin 40° = 12/\sin 60°\)
A1
For using the sine rule
\(Q = 8.91\)
M1
4
(Third alternative)
Answer
Marks
Guidance
Angle between P and Q is \(120°\) and between P and –R is \(140°\)
M1
For force diagram showing P, Q and –R, and values of any 2 angles at 'O' shown.
\([Q/\sin 140° = 12/\sin 120°]\)
A1
\(Q = 8.91\)
M1
4
$Q - P\cos 60° = 12\cos 80°$ and $P\sin 60° = 12\sin 80°$ | M1 | For resolving forces parallel to or perpendicular to: the force of magnitude QN, or the resultant
$Q\cos 80° + P\cos 40° = 12$ and $P\sin 40° = Q\sin 80°$ | A1 |
$[Q - 12\sin 80°\cos 60°/\sin 60° = 12\cos 80°]$ | M1 | For eliminating P
$Q\cos 80° + Q\sin 80°\cos 40°/\sin 40° = 12]$ | A1 | 4 |
$Q = 8.91$ | | |
**(First alternative)**
$Q\cos 30° = 12\cos 50°$ | M2 | For resolving forces perp. to P
$Q = 8.91$ | A1 | 4 |
**(Second alternative)**
| | | For triangle of forces with sides P, Q and 12, and values of any 2 angles shown or implied (Note P, Q and –R are in equil.)
Angles opposite Q and 12 are $40°$ and $60°$ respectively | M1 |
$Q/\sin 40° = 12/\sin 60°$ | A1 | For using the sine rule
$Q = 8.91$ | M1 | 4 |
**(Third alternative)**
Angle between P and Q is $120°$ and between P and –R is $140°$ | M1 | For force diagram showing P, Q and –R, and values of any 2 angles at 'O' shown.
$[Q/\sin 140° = 12/\sin 120°]$ | A1 |
$Q = 8.91$ | M1 | 4 | For using Lami's theorem
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\includegraphics[max width=\textwidth, alt={}, center]{a9f3480e-7a8a-497d-a26a-b2aba9b05512-2_462_721_1672_712}
Two forces have magnitudes $P \mathrm {~N}$ and $Q \mathrm {~N}$. The resultant of the two forces has magnitude 12 N and acts in a direction $40 ^ { \circ }$ clockwise from the force of magnitude $P \mathrm {~N}$ and $80 ^ { \circ }$ anticlockwise from the force of magnitude $Q \mathrm {~N}$ (see diagram). Find the value of $Q$.
\hfill \mbox{\textit{CAIE M1 2009 Q3 [4]}}