| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Coplanar forces in equilibrium |
| Difficulty | Moderate -0.8 This is a straightforward coplanar forces problem requiring resolution of forces into components and basic vector addition. Part (i) is routine verification, part (ii) involves standard calculations with given trig values, and part (iii) requires simple reasoning about force reversal. All techniques are standard M1 material with no novel problem-solving required. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([X = 25 \times 0.96 - 30 \times 0.8 = 0]\) | M1 | For resolving forces in the \(x\) direction |
| Component in \(x\)-direction is zero | A1 | 2 marks total; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([Y = 25 \times 0.28 - 20 + 30 \times 0.6 = 5]\) | M1 | For resolving forces in the \(y\) direction |
| Resultant has magnitude 5 N and acts in the positive \(y\) direction | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Replacement has magnitude 30 N and acts in the \(-ve\) \(y\) direction | B1 | 1 mark total |
## Question 2:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $[X = 25 \times 0.96 - 30 \times 0.8 = 0]$ | M1 | For resolving forces in the $x$ direction |
| Component in $x$-direction is zero | A1 | 2 marks total; AG |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $[Y = 25 \times 0.28 - 20 + 30 \times 0.6 = 5]$ | M1 | For resolving forces in the $y$ direction |
| Resultant has magnitude 5 N and acts in the positive $y$ direction | A1 | 2 marks total |
### Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| Replacement has magnitude 30 N and acts in the $-ve$ $y$ direction | B1 | 1 mark total |
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\includegraphics[max width=\textwidth, alt={}, center]{c7133fc4-9a14-43fd-b5ed-788da72291cd-2_666_953_662_596}
Three coplanar forces act at a point. The magnitudes of the forces are $20 \mathrm {~N} , 25 \mathrm {~N}$ and 30 N , and the directions in which the forces act are as shown in the diagram, where $\sin \alpha = 0.28$ and $\cos \alpha = 0.96$, and $\sin \beta = 0.6$ and $\cos \beta = 0.8$.\\
(i) Show that the resultant of the three forces has a zero component in the $x$-direction.\\
(ii) Find the magnitude and direction of the resultant of the three forces.\\
(iii) The force of magnitude 20 N is replaced by another force. The effect is that the resultant force is unchanged in magnitude but reversed in direction. State the magnitude and direction of the replacement force.
\hfill \mbox{\textit{CAIE M1 2014 Q2 [5]}}