June 2022 Paper 3 Mechanics Q4
4.
A uniform rod \(AB\) has mass \(M\) and length \(2a\)
A particle of mass \(2M\) is attached to the rod at the point \(C\), where \(AC = 1.5a\)
The rod rests with its end \(A\) on rough horizontal ground.
The rod is held in equilibrium at an angle \(\theta\) to the ground by a light string that is attached to the end \(B\) of the rod.
The string is perpendicular to the rod, as shown in Figure 2.
The tension in the string is \(T\)
Given that \(\cos\theta = \dfrac{3}{5}\)
The coefficient of friction between the rod and the ground is \(\mu\)
Given that the rod is in limiting equilibrium,
| Scheme | Marks | AO |
|---|---|---|
| The horizontal component of \(T\) acts to the left and since the only other horizontal force is friction, it must act to the right oe | B1 | 2.4 |
| (1) |
Notes
B1: Any equivalent explanation
| Scheme | Marks | AO |
|---|---|---|
| Take moments about \(A\) or any other complete method to obtain an equation in T, M and \(\theta\) only. (see possible equations below that they may use) | M1 | 3.1b |
| \(T.2a = Mga\cos\theta + 2Mg \times 1.5a\cos\theta\) (A0 if \(a\)’s missing) | A1 | 1.1b |
| Other possible equations but \(F\) and \(R\) would need to be eliminated. \((\nwarrow)\), \(R\cos\theta + T = F\sin\theta + Mg\cos\theta + 2Mg\cos\theta\) \((\nearrow)\), \(R\sin\theta + F\cos\theta = Mg\sin\theta + 2Mg\sin\theta\) \((\rightarrow)\), \(F = T\sin\theta\) M(\(B\)), \(R.2a\cos\theta = Mga\cos\theta + 2Mg \times 0.5a\cos\theta + F.2a\sin\theta\) M(\(G\)), \(Fa\sin\theta + Ta = Ra\cos\theta + 2Mg \times 0.5a\cos\theta\) M(\(C\)), \(R \times 1.5a\cos\theta = T \times 0.5a + Mg \times 0.5a\cos\theta + F \times 1.5a\sin\theta\) | ||
| \(T = 2Mg\cos\theta\) * | A1* | 1.1b |
| (3) |
Notes
M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors
A1: Correct equation, trig does not need to be substituted
(Allow: \(T.2a = Mga\cos\theta + 3Mga\cos\theta\))
A1*: Given answer correctly obtained with no wrong working seen.
Allow \(2Mg\cos\theta = T\)
But not \(T = 2\cos\theta Mg\)
| Scheme | Marks | AO |
|---|---|---|
| e.g. Resolve vertically | M1 | 3.4 |
| \((\uparrow)\), \(R + T\cos\theta = Mg + 2Mg\) | A1 | 1.1b |
| \(R = \dfrac{57Mg}{25}\) * | A1* | 1.1b |
| Other possible equations but \(F\) would need to be eliminated. \((\nwarrow)\), \(R\cos\theta + T = F\sin\theta + Mg\cos\theta + 2Mg\cos\theta\) \((\nearrow)\), \(R\sin\theta + F\cos\theta = Mg\sin\theta + 2Mg\sin\theta\) \((\rightarrow)\), \(F = T\sin\theta\) M(\(B\)), \(R.2a\cos\theta = Mga\cos\theta + 2Mg \times 0.5a\cos\theta + F.2a\sin\theta\) M(\(G\)), \(Fa\sin\theta + Ta = Ra\cos\theta + 2Mg \times 0.5a\cos\theta\) M(\(C\)), \(R \times 1.5a\cos\theta = T \times 0.5a + Mg \times 0.5a\cos\theta + F \times 1.5a\sin\theta\) | ||
| (3) |
Notes
M1: For an equation in \(R\), \(M\), \(T\) and \(\theta\) only
Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved
A1: Correct equation, \(T\) and trig do not need to be substituted
A1*: Given answer correctly obtained with no wrong working seen
| Scheme | Marks | AO |
|---|---|---|
| Find an equation containing \(F\) e.g. Resolve horizontally | M1 | 3.4 |
| \((\rightarrow)\), \(F = T\sin\theta\) | A1 | 1.1b |
| Other possible equations \((\nwarrow)\), \(R\cos\theta + T = F\sin\theta + Mg\cos\theta + 2Mg\cos\theta\) \((\nearrow)\), \(R\sin\theta + F\cos\theta = Mg\sin\theta + 2Mg\sin\theta\) \((\rightarrow)\), \(F = T\sin\theta\) M(\(B\)), \(R.2a\cos\theta = Mga\cos\theta + 2Mg \times 0.5a\cos\theta + F.2a\sin\theta\) M(\(G\)), \(Fa\sin\theta + Ta = Ra\cos\theta + 2Mg \times 0.5a\cos\theta\) M(\(C\)), \(R \times 1.5a\cos\theta = T \times 0.5a + Mg \times 0.5a\cos\theta + F \times 1.5a\sin\theta\) | ||
| \(F = \mu R\) used i.e. both \(F\) and \(R\) are substituted. | M1 | 3.1b |
| \(\mu = \dfrac{8}{19}\) * | A1* | 2.2a |
| (4) | ||
| (11 marks) |
Notes
M1: For any equation with \(F\) in it
Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved
A1: Correct equation, trig does not need to be substituted
M1: Must be used i.e M0 if merely quoting it.
A1*: Given answer correctly obtained with no wrong working seen