October 2021 Paper 3 Q12
12
A beam, \(AB\), has length \(4\,\mathrm{m}\) and mass \(20\,\mathrm{kg}\). The beam is suspended horizontally by two vertical ropes. One rope is attached to the beam at \(C\), where \(AC = 0.5\,\mathrm{m}\). The other rope is attached to the beam at \(D\), where \(DB = 0.7\,\mathrm{m}\) (see diagram).
The beam is modelled as a non-uniform rod and the ropes as light inextensible strings.
It is given that the tension in the rope at \(C\) is three times the tension in the rope at \(D\).
A particle of mass \(m\,\mathrm{kg}\) is now placed on the beam at a point where the magnitude of the moment of the particle's weight about \(C\) is \(3.5mg\,\mathrm{N\,m}\). The beam remains horizontal and in equilibrium.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{array}{ll}0.5T_C + (4 - 0.7)T_D = 20g\bar{x} & \text{(moments about } A\text{)}\\ (\bar{x} - 0.5)(20g) = (4 - 1.2)T_D & \text{(moments about } C\text{)}\\ (4 - 1.2)T_C = 20g(4 - 0.7 - \bar{x}) & \text{(moments about } D\text{)}\\ 0.7T_D + (4 - 0.5)T_C = 20g(4 - \bar{x}) & \text{(moments about } B\text{)}\\ (\bar{x} - 0.5)T_C = (4 - 0.7 - \bar{x})T_D & \text{(moments about com)}\end{array}\) | M1* A1FT | 3.3 1.1 |
| \(T_C = 3T_D\) or \(T_C = 3T\) and \(T_D = T\) | B1 | 3.3 |
| \(T_C = 147\,(= 15g)\) and/or \(T_D = 49\,(= 5g)\) (from \(T_C + T_D = 20g\)) used in relevant moment equation(s) | M1dep* | 1.1 |
| \(\bar{x} = 1.2\ (\mathrm{m})\) | A1 | 2.2a |
| [5] |
Notes
M1*: Moments (with correct number of terms) about \(A\), \(C\), \(D\), \(B\) or com
\(\bar{x}\) is the centre of mass of the rod from \(A\)
Allow \(g\) to be absent
A1FT: Follow though their \(T_C\) and \(T_D\) (allow just \(T\)) and allow \(W\) for \(20g\)
B1: Correct relationship(s) for the tensions at \(C\) and \(D\) (soi)
M1dep*: Equation in \(\bar{x}\) only, e.g.,
• \(W = 4T\) and any one mom. eq.
• \(4T = 20g\) and any one mom. eq.
• \(W = 20g\) and any two mom. eq.
M0 if tension used are the same in both ropes
A1: Must be 1.2 as question asks for com from \(A\)
Alternative: B1 as main scheme (soi), then M1 for splitting 2.8 in the ratio 1: 3 or 3:1 then A1 for correct 1:3 then B2 for \(0.5 + 0.25(2.8) = 1.2\,(\mathrm{m})\)
| Scheme | Marks | AO |
|---|---|---|
| \(0.7mg = 20g(4 - \bar{x} - 0.7)\) (moments about \(D\)) \(20g(\bar{x} - 0.5) + 3.5mg = (4 - 0.5 - 0.7)(20g + mg)\) (moments about \(C\)) | M1 | 3.1b |
| \(m = 60\) | A1 | 2.2a |
| [2] |
Notes
M1: Moments about \(D\) (oe) – correct number of terms – if taking moments about another point then must set \(T_C = 0\) and \(T_D = 20g + mg\)
Allow \(g\) to be absent