CAIE M2 2007 November — Question 3 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2007
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod hinged to wall with string support
DifficultyStandard +0.3 This is a standard moments problem requiring taking moments about the hinge, resolving forces horizontally and vertically, and using Pythagoras to find the angle. The setup is straightforward with perpendicular string, and all calculations follow routine mechanics procedures with no conceptual challenges beyond typical M2 level.
Spec3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces

3 \includegraphics[max width=\textwidth, alt={}, center]{b9080e9f-2c23-43ce-b171-bd68648dc56b-3_764_627_274_758} A uniform beam \(A B\) has length 2 m and mass 10 kg . The beam is hinged at \(A\) to a fixed point on a vertical wall, and is held in a fixed position by a light inextensible string of length 2.4 m . One end of the string is attached to the beam at a point 0.7 m from \(A\). The other end of the string is attached to the wall at a point vertically above the hinge. The string is at right angles to \(A B\). The beam carries a load of weight 300 N at \(B\) (see diagram).
  1. Find the tension in the string. The components of the force exerted by the hinge on the beam are \(X \mathrm {~N}\) horizontally away from the wall and \(Y \mathrm {~N}\) vertically downwards.
  2. Find the values of \(X\) and \(Y\).

Question 3:
Part (i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
M1For taking moments about A (3 terms)
\(100x(1\cos\alpha) + 300x(2\cos\alpha) = T \times 0.7\)A1 \(\alpha\) is the angle made by the string with the vertical
where \(\cos\alpha = 0.96\)A1
Tension is 960NA1ft [4 marks] ft \(1000\cos\alpha\)
Part (ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(X = 268.8\) (269)B1ft ft \(1000\sin\alpha\cos\alpha\)
\(Y + 10g + 300 = 960\cos\alpha\)M1 For resolving forces vertically (4 terms)
\(Y = 521.6\) (522)A1 [3 marks] [Total: 7]
## Question 3:

### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For taking moments about A (3 terms) |
| $100x(1\cos\alpha) + 300x(2\cos\alpha) = T \times 0.7$ | A1 | $\alpha$ is the angle made by the string with the vertical |
| where $\cos\alpha = 0.96$ | A1 | |
| Tension is 960N | A1ft | **[4 marks]** ft $1000\cos\alpha$ |

### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X = 268.8$ (269) | B1ft | ft $1000\sin\alpha\cos\alpha$ |
| $Y + 10g + 300 = 960\cos\alpha$ | M1 | For resolving forces vertically (4 terms) |
| $Y = 521.6$ (522) | A1 | **[3 marks]** **[Total: 7]** |

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\includegraphics[max width=\textwidth, alt={}, center]{b9080e9f-2c23-43ce-b171-bd68648dc56b-3_764_627_274_758}

A uniform beam $A B$ has length 2 m and mass 10 kg . The beam is hinged at $A$ to a fixed point on a vertical wall, and is held in a fixed position by a light inextensible string of length 2.4 m . One end of the string is attached to the beam at a point 0.7 m from $A$. The other end of the string is attached to the wall at a point vertically above the hinge. The string is at right angles to $A B$. The beam carries a load of weight 300 N at $B$ (see diagram).\\
(i) Find the tension in the string.

The components of the force exerted by the hinge on the beam are $X \mathrm {~N}$ horizontally away from the wall and $Y \mathrm {~N}$ vertically downwards.\\
(ii) Find the values of $X$ and $Y$.

\hfill \mbox{\textit{CAIE M2 2007 Q3 [7]}}