OCR MEI M1 2012 January — Question 2 7 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2012
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeTriangle of forces method
DifficultyModerate -0.3 This is a standard M1 equilibrium problem using triangle of forces. Part (i) requires basic angle geometry, part (ii) is routine diagram drawing, and part (iii) involves straightforward trigonometry on a right-angled triangle with given angles (30°, 60°, 90°) and one known force (20N). The question is slightly easier than average because the right angle simplifies calculations and the special angles make the arithmetic clean.
Spec3.03a Force: vector nature and diagrams3.03m Equilibrium: sum of resolved forces = 0
2 Fig. 2 shows a small object, P , of weight 20 N , suspended by two light strings. The strings are tied to points A and B on a sloping ceiling which is at an angle of \(60 ^ { \circ }\) to the upward vertical. The string AP is at \(60 ^ { \circ }\) to the downward vertical and the string BP makes an angle of \(30 ^ { \circ }\) with the ceiling. The object is in equilibrium. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0330185f-d79d-4a78-9fa2-29ec345c2856-2_430_670_1546_699} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. Show that \(\angle \mathrm { APB } = 90 ^ { \circ }\).
  2. Draw a labelled triangle of forces to represent the three forces acting on P .
  3. Hence, or otherwise, find the tensions in the two strings.
Question 2(i):
AnswerMarks Guidance
AnswerMark Guidance
\(\angle BAD = 60°\) and \(\angle PAE = 60° \Rightarrow \angle PAB = 60°\) (Angles on straight line); \(\angle PAB = 60°\) and \(\angle ABP = 30° \Rightarrow \angle APB = 90°\) (Angles in triangle)B1 Any valid argument. Allow an "argument" containing only numbers and no words
[1]
Question 2(ii):
AnswerMarks Guidance
AnswerMark Guidance
Triangle with angles 90°, 60°, 30° with sides labelled \(T_1\), \(T_2\), 20B1 A triangle with angles approximately 90°, 60° and 30°. No mark for isosceles or equilateral triangle. No extra forces. Do not award to candidates who draw a force diagram showing 3 concurrent forces
Arrows following cycle round triangleB1 Triangle: sides marked with arrows following a cycle. Force diagram: directions of three forces indicated by arrows. No extra forces
All sides labelled consistently: 20 opposite 90°, \(T_{AP}\) opposite 30°, \(T_{BP}\) opposite 60°B1 Triangle: all sides labelled consistently and unambiguously with the angles. Force diagram: all forces labelled. Condone \(20g\) or \(W\)
[3]
Question 2(iii):
AnswerMarks Guidance
AnswerMark Guidance
\(T_2 = 20\cos 30°\)M1 An attempt to apply trigonometry to a triangle of forces to find the tension in either string
Tension in string BP is 17.3 NA1 No mark if derived from "weight \(= 20g\)" (giving 169.7)
Tension in string AP is 10 NA1 FT for "weight \(= 20g\)" (giving 98)
[3]
Alternative (horizontal/vertical equilibrium):
AnswerMarks Guidance
AnswerMark Guidance
Horizontal: \(T_2\cos 60° = T_1\cos 30°\); Vertical: \(T_2\sin 60° + T_1\sin 30° = 20\)(M1) An attempt at both horizontal and vertical equilibrium equations
Tension in BP is 17.3 N(A1) No mark if derived from "weight \(= 20g\)"
Tension in AP is 10 N(A1) FT from "weight \(= 20g\)" (giving 98)
([3])
Alternative (resolving along PA and PB):
AnswerMarks Guidance
AnswerMark Guidance
\(T_1 = 20\cos 60°\), \(T_2 = 20\cos 30°\)(M1) An attempt to resolve the weight in directions PA and PB. Method may be implied by subsequent work
Tension in BP is 17.3 N(A1) No mark if derived from "weight \(= 20g\)"
Tension in AP is 10 N(A1) FT from "weight \(= 20g\)" (giving 98)
([3]) 
## Question 2(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\angle BAD = 60°$ and $\angle PAE = 60° \Rightarrow \angle PAB = 60°$ (Angles on straight line); $\angle PAB = 60°$ and $\angle ABP = 30° \Rightarrow \angle APB = 90°$ (Angles in triangle) | B1 | Any valid argument. Allow an "argument" containing only numbers and no words |
| | **[1]** | |

---

## Question 2(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Triangle with angles 90°, 60°, 30° with sides labelled $T_1$, $T_2$, 20 | B1 | A triangle with angles approximately 90°, 60° and 30°. No mark for isosceles or equilateral triangle. No extra forces. Do not award to candidates who draw a force diagram showing 3 concurrent forces |
| Arrows following cycle round triangle | B1 | Triangle: sides marked with arrows following a cycle. Force diagram: directions of three forces indicated by arrows. No extra forces |
| All sides labelled consistently: 20 opposite 90°, $T_{AP}$ opposite 30°, $T_{BP}$ opposite 60° | B1 | Triangle: all sides labelled consistently and unambiguously with the angles. Force diagram: all forces labelled. Condone $20g$ or $W$ |
| | **[3]** | |

---

## Question 2(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $T_2 = 20\cos 30°$ | M1 | An attempt to apply trigonometry to a triangle of forces to find the tension in either string |
| Tension in string BP is 17.3 N | A1 | No mark if derived from "weight $= 20g$" (giving 169.7) |
| Tension in string AP is 10 N | A1 | FT for "weight $= 20g$" (giving 98) |
| | **[3]** | |

**Alternative (horizontal/vertical equilibrium):**

| Answer | Mark | Guidance |
|--------|------|----------|
| Horizontal: $T_2\cos 60° = T_1\cos 30°$; Vertical: $T_2\sin 60° + T_1\sin 30° = 20$ | (M1) | An attempt at both horizontal and vertical equilibrium equations |
| Tension in BP is 17.3 N | (A1) | No mark if derived from "weight $= 20g$" |
| Tension in AP is 10 N | (A1) | FT from "weight $= 20g$" (giving 98) |
| | **([3])** | |

**Alternative (resolving along PA and PB):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $T_1 = 20\cos 60°$, $T_2 = 20\cos 30°$ | (M1) | An attempt to resolve the weight in directions PA and PB. Method may be implied by subsequent work |
| Tension in BP is 17.3 N | (A1) | No mark if derived from "weight $= 20g$" |
| Tension in AP is 10 N | (A1) | FT from "weight $= 20g$" (giving 98) |
| | **([3])** | |

---
2 Fig. 2 shows a small object, P , of weight 20 N , suspended by two light strings. The strings are tied to points A and B on a sloping ceiling which is at an angle of $60 ^ { \circ }$ to the upward vertical. The string AP is at $60 ^ { \circ }$ to the downward vertical and the string BP makes an angle of $30 ^ { \circ }$ with the ceiling.

The object is in equilibrium.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{0330185f-d79d-4a78-9fa2-29ec345c2856-2_430_670_1546_699}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

(i) Show that $\angle \mathrm { APB } = 90 ^ { \circ }$.\\
(ii) Draw a labelled triangle of forces to represent the three forces acting on P .\\
(iii) Hence, or otherwise, find the tensions in the two strings.

\hfill \mbox{\textit{OCR MEI M1 2012 Q2 [7]}}
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