| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Maximum/minimum force for equilibrium |
| Difficulty | Standard +0.3 This is a standard connected particles equilibrium problem requiring resolution of forces and solving simultaneous equations. Part (a) involves straightforward force resolution with a given F value, while part (b) requires understanding that equilibrium breaks when tension becomes zero—a common textbook scenario that's slightly above routine but well within standard M1 expectations. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_A \times 0.8 - T_B \times 0.6 - 20 = 0\) or \(T_A \times 0.6 + T_B \times 0.8 - 10g = 0\) | M1 | Resolving horizontally or vertically |
| \(T_A \times 0.8 - T_B \times 0.6 - 20 = 0\) | A1 | |
| \(T_A \times 0.6 + T_B \times 0.8 - 10g = 0\) | A1 | |
| \(0.8T_A - \dfrac{0.6(10g - 0.6T_A)}{0.8} = 20 \rightarrow T_A = \ldots\) | M1 | Attempt to solve simultaneously |
| \(T_A = 76 \text{ N}, T_B = 68 \text{ N}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_A \times 0.6 - 10g = 0 \Rightarrow T_A = \dfrac{500}{3}\) | B1 | From using \(T_B = 0\) |
| \(T_A \times 0.8 - F = 0\) | M1 | |
| \(F = \dfrac{400}{3}\) | A1 | Allow \(F = 133\) to 3 s.f. |
## Question 4:
### Part (a):
$T_A \times 0.8 - T_B \times 0.6 - 20 = 0$ or $T_A \times 0.6 + T_B \times 0.8 - 10g = 0$ | M1 | Resolving horizontally or vertically
$T_A \times 0.8 - T_B \times 0.6 - 20 = 0$ | A1 |
$T_A \times 0.6 + T_B \times 0.8 - 10g = 0$ | A1 |
$0.8T_A - \dfrac{0.6(10g - 0.6T_A)}{0.8} = 20 \rightarrow T_A = \ldots$ | M1 | Attempt to solve simultaneously
$T_A = 76 \text{ N}, T_B = 68 \text{ N}$ | A1 |
### Part (b):
$T_A \times 0.6 - 10g = 0 \Rightarrow T_A = \dfrac{500}{3}$ | B1 | From using $T_B = 0$
$T_A \times 0.8 - F = 0$ | M1 |
$F = \dfrac{400}{3}$ | A1 | Allow $F = 133$ to 3 s.f.
---\begin{enumerate}[label=(\alph*)]
\item In the case where $F = 20$, find the tensions in each of the strings.
\item Find the greatest value of $F$ for which the block remains in equilibrium in the position shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q4 [8]}}