| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Standard +0.3 This is a standard coplanar forces problem requiring resolution of forces in two perpendicular directions and basic trigonometry. Part (i) involves straightforward substitution and vector addition, while part (ii) requires setting up and solving two simultaneous equations from equilibrium conditions—routine techniques for M1 with no novel problem-solving insight required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | For resolving forces (either direction) | |
| \(X = 75 + 50\cos 60 (= 100)\), \(Y = 50\sin 60 (= 43.3)\) | A1 | For both equations, unevaluated |
| Resultant \(= \sqrt{100^2 + 43.3^2} = 109\) N | B1 | |
| Angle \(= \arctan\!\left(\dfrac{43.3}{100}\right) = 23.4°\) | B1 | Must state anticlockwise from the positive \(x\)-axis or show in a diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(50\cos\alpha - F\cos 50 = 0\) | B1 | Resolving forces horizontally |
| \(50\sin\alpha - 3F - F\sin 50 = 0\) | B1 | Resolving forces vertically |
| \(\tan\alpha = \dfrac{(3F + F\sin 50)}{(F\cos 50)}\) | M1 | For division to find \(\theta\) or for using Pythagoras to find \(F\) |
| \(\alpha = 80.3\) | A1 | |
| \(F = 13.1\) | A1 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | For resolving forces (either direction) |
| $X = 75 + 50\cos 60 (= 100)$, $Y = 50\sin 60 (= 43.3)$ | A1 | For both equations, unevaluated |
| Resultant $= \sqrt{100^2 + 43.3^2} = 109$ N | B1 | |
| Angle $= \arctan\!\left(\dfrac{43.3}{100}\right) = 23.4°$ | B1 | Must state anticlockwise from the positive $x$-axis or show in a diagram |
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## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $50\cos\alpha - F\cos 50 = 0$ | B1 | Resolving forces horizontally |
| $50\sin\alpha - 3F - F\sin 50 = 0$ | B1 | Resolving forces vertically |
| $\tan\alpha = \dfrac{(3F + F\sin 50)}{(F\cos 50)}$ | M1 | For division to find $\theta$ or for using Pythagoras to find $F$ |
| $\alpha = 80.3$ | A1 | |
| $F = 13.1$ | A1 | |
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\includegraphics[max width=\textwidth, alt={}, center]{db1b5f31-1a41-44dd-ae9a-0c67336997eb-08_529_606_260_767}
Coplanar forces, of magnitudes $F \mathrm {~N} , 3 F \mathrm {~N} , G \mathrm {~N}$ and 50 N , act at a point $P$, as shown in the diagram.\\
(i) Given that $F = 0 , G = 75$ and $\alpha = 60 ^ { \circ }$, find the magnitude and direction of the resultant force.\\
(ii) Given instead that $G = 0$ and the forces are in equilibrium, find the values of $F$ and $\alpha$.\\
\hfill \mbox{\textit{CAIE M1 2017 Q6 [9]}}