| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.3 This is a standard M1 equilibrium problem requiring resolution of forces in two perpendicular directions and application of friction laws. While it involves multiple forces at angles and requires solving simultaneous equations, the techniques are routine for mechanics students with no novel problem-solving insight needed. The constant speed condition making it an equilibrium problem is explicitly stated, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For resolving the applied forces on the box in the \(x\)-direction or the \(y\)-direction | |
| \(100\cos 30° + 120\cos 60° - F\cos\alpha = 136\) \((F\cos\alpha = 10.6025...)\) or \(100\sin 30° - 120\sin 60° + F\sin\alpha = 0\) \((F\sin\alpha = 53.9230...)\) | A1 | |
| \(100\sin 30° - 120\sin 60° + F\sin\alpha = 0\) \((F\sin\alpha = 53.9230...)\) or \(100\cos 30° + 120\cos 60° - F\cos\alpha = 136\) \((F\cos\alpha = 10.6025...)\) | B1 | |
| M1 | For using \(F^2 = (F\cos\alpha)^2 + (F\sin\alpha)^2\) or \(\tan\alpha = F\sin\alpha \div F\cos\alpha\) | |
| \(F = 55.0\) or \(\alpha = 78.9\) | A1 | |
| \(\alpha = 78.9\) or \(F = 55.0\) | B1 [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Magnitude is \(136\ \text{N}\) | B1 | |
| \(R = 40g\) | B1 | |
| Coefficient is \(0.34\) | B1 [3] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving the applied forces on the box in the $x$-direction or the $y$-direction |
| $100\cos 30° + 120\cos 60° - F\cos\alpha = 136$ $(F\cos\alpha = 10.6025...)$ or $100\sin 30° - 120\sin 60° + F\sin\alpha = 0$ $(F\sin\alpha = 53.9230...)$ | A1 | |
| $100\sin 30° - 120\sin 60° + F\sin\alpha = 0$ $(F\sin\alpha = 53.9230...)$ or $100\cos 30° + 120\cos 60° - F\cos\alpha = 136$ $(F\cos\alpha = 10.6025...)$ | B1 | |
| | M1 | For using $F^2 = (F\cos\alpha)^2 + (F\sin\alpha)^2$ or $\tan\alpha = F\sin\alpha \div F\cos\alpha$ |
| $F = 55.0$ or $\alpha = 78.9$ | A1 | |
| $\alpha = 78.9$ or $F = 55.0$ | B1 [6] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Magnitude is $136\ \text{N}$ | B1 | |
| $R = 40g$ | B1 | |
| Coefficient is $0.34$ | B1 [3] | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{ceb367ee-4e12-4cb2-9020-078ea5724d6e-3_703_700_255_721}
A small box of mass 40 kg is moved along a rough horizontal floor by three men. Two of the men apply horizontal forces of magnitudes 100 N and 120 N , making angles of $30 ^ { \circ }$ and $60 ^ { \circ }$ respectively with the positive $x$-direction. The third man applies a horizontal force of magnitude $F \mathrm {~N}$ making an angle of $\alpha ^ { \circ }$ with the negative $x$-direction (see diagram). The resultant of the three horizontal forces acting on the box is in the positive $x$-direction and has magnitude 136 N .\\
(i) Find the values of $F$ and $\alpha$.\\
(ii) Given that the box is moving with constant speed, state the magnitude of the frictional force acting on the box and hence find the coefficient of friction between the box and the floor.
\hfill \mbox{\textit{CAIE M1 2013 Q6 [9]}}