CAIE M1 2016 November — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2016
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeParticle suspended by strings
DifficultyModerate -0.3 This is a standard two-rope equilibrium problem requiring resolution of forces in two directions and solving simultaneous equations. While it involves trigonometry and careful angle work, it's a routine mechanics exercise with a well-established method (resolve horizontally and vertically, solve for two unknowns). Slightly easier than average due to being a straightforward application of equilibrium principles without any conceptual complications.
Spec3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces

2 \includegraphics[max width=\textwidth, alt={}, center]{94c11160-a718-4de5-867a-27c755051fa6-2_342_629_616_756} The diagram shows a small object \(P\) of mass 20 kg held in equilibrium by light ropes attached to fixed points \(A\) and \(B\). The rope \(P A\) is inclined at an angle of \(50 ^ { \circ }\) above the horizontal, the rope \(P B\) is inclined at an angle of \(10 ^ { \circ }\) below the horizontal, and both ropes are in the same vertical plane. Find the tension in the rope \(P A\) and the tension in the rope \(P B\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
M1For resolving horizontally
M1For resolving vertically
\(T_A \cos 50° - T_B \cos 10° = 0\) and \(T_A \sin 50° - T_B \sin 10° - 20g = 0\)A1
M1For solving equations to find \(T_A\) and \(T_B\)
Tension in PA is \(306\text{ N}\), Tension in PB is \(200\text{ N}\)A1 [5]
Alternative (Lami's Theorem):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\([T_A/\sin 80° = T_B/\sin 140° = 20g/\sin 140°]\)M1 For applying Lami's Theorem
\([T_A = 20g \sin 80°/\sin 140°]\)M1 For solving for \(T_A\)
Tension in PA is \(306\text{ N}\)A1
\([T_B = 20g \sin 140°/\sin 140°]\)M1 For solving for \(T_B\)
Tension in PB is \(200\text{ N}\)A1 [5]
# Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving horizontally |
| | M1 | For resolving vertically |
| $T_A \cos 50° - T_B \cos 10° = 0$ **and** $T_A \sin 50° - T_B \sin 10° - 20g = 0$ | A1 | |
| | M1 | For solving equations to find $T_A$ and $T_B$ |
| Tension in PA is $306\text{ N}$, Tension in PB is $200\text{ N}$ | A1 | [5] |

**Alternative (Lami's Theorem):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T_A/\sin 80° = T_B/\sin 140° = 20g/\sin 140°]$ | M1 | For applying Lami's Theorem |
| $[T_A = 20g \sin 80°/\sin 140°]$ | M1 | For solving for $T_A$ |
| Tension in PA is $306\text{ N}$ | A1 | |
| $[T_B = 20g \sin 140°/\sin 140°]$ | M1 | For solving for $T_B$ |
| Tension in PB is $200\text{ N}$ | A1 | [5] |

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2\\
\includegraphics[max width=\textwidth, alt={}, center]{94c11160-a718-4de5-867a-27c755051fa6-2_342_629_616_756}

The diagram shows a small object $P$ of mass 20 kg held in equilibrium by light ropes attached to fixed points $A$ and $B$. The rope $P A$ is inclined at an angle of $50 ^ { \circ }$ above the horizontal, the rope $P B$ is inclined at an angle of $10 ^ { \circ }$ below the horizontal, and both ropes are in the same vertical plane. Find the tension in the rope $P A$ and the tension in the rope $P B$.

\hfill \mbox{\textit{CAIE M1 2016 Q2 [5]}}