October 2020 Paper 3 Mechanics Q4
4.
A ladder \(AB\) has mass \(M\) and length \(6a\).
The end \(A\) of the ladder is on rough horizontal ground.
The ladder rests against a fixed smooth horizontal rail at the point \(C\).
The point \(C\) is at a vertical height \(4a\) above the ground.
The vertical plane containing \(AB\) is perpendicular to the rail.
The ladder is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{4}{5}\), as shown in Figure 1.
The coefficient of friction between the ladder and the ground is \(\mu\).
The ladder rests in limiting equilibrium.
The ladder is modelled as a uniform rod.
Using the model,
| Scheme | Marks | AO |
|---|---|---|
| Take moments about \(A\) | M1 | 3.3 |
| \(N \times \dfrac{4a}{\sin\alpha} = Mg \times 3a\cos\alpha\) | A1 | 1.1b |
| \(\dfrac{9Mg}{25}\) * | A1* | 1.1b |
| (3) |
Notes
M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors for an equation in \(N\) and \(Mg\) only.
For perp distance allow any of: \(\dfrac{4a}{\sin\alpha}, \dfrac{4a}{\cos\alpha}, 5a\) but
use of any of: \(6a, 5a\sin\alpha, 4a\cos\alpha, \ldots\) or anything involving \(\tan\alpha\) is M0
Also M0 if no \(a\)’s in their first equation.
A1: Correct equation, trig does not need to be substituted
A1*: Given answer correctly obtained.
| Scheme | Marks | AO |
|---|---|---|
| Resolve horizontally | M1 | 3.4 |
| \((\rightarrow)\ F = \dfrac{9Mg}{25}\sin\alpha\) | A1 | 1.1b |
| Resolve vertically | M1 | 3.4 |
| \((\uparrow)\ R + \dfrac{9Mg}{25}\cos\alpha = Mg\) | A1 | 1.1b |
| Other possible equations: \((\nwarrow),\ R\cos\alpha + \dfrac{9Mg}{25} = Mg\cos\alpha + F\sin\alpha\) \((\nearrow),\ Mg\sin\alpha = F\cos\alpha + R\sin\alpha\) M\((C)\), \(Mg.2a\cos\alpha + F.5a\sin\alpha = R.5a\cos\alpha\) M\((G)\), \(\dfrac{9Mg}{25}.2a + F.3a\sin\alpha = R.3a\cos\alpha\) M\((B)\), \(Mg.3a\cos\alpha + F.6a\sin\alpha = R.6a\cos\alpha + \dfrac{9Mg}{25}a\) \(\left(F = \dfrac{36Mg}{125},\ R = \dfrac{98Mg}{125}\right)\) | ||
| \(F = \mu R\) used | M1 | 3.4 |
| Eliminate \(R\) and \(F\) and solve for \(\mu\) | M1 | 3.1b |
| Alternative equations if they have at \(A\): \(X\) horizontally and \(Y\) perpendicular to the rod. \((\nwarrow),\ Y + \dfrac{9Mg}{25} = Mg\cos\alpha + X\sin\alpha\) \((\nearrow),\ Mg\sin\alpha = X\cos\alpha\) \((\uparrow),\ \dfrac{9Mg}{25}\cos\alpha + Y\cos\alpha = Mg\) \((\rightarrow),\ Y\sin\alpha + \dfrac{9Mg}{25}\sin\alpha = X\) M\((C)\), \(Mg.2a\cos\alpha + X.5a\sin\alpha = Y.5a\) M\((G)\), \(\dfrac{9Mg}{25}.2a + X.3a\sin\alpha = Y.3a\) M1A1 M1A1 M\((B)\), \(Mg.3a\cos\alpha + X.6a\sin\alpha = Y.6a + \dfrac{9Mg}{25}a\) \(\left(X = \dfrac{4Mg}{3},\ Y = \dfrac{98Mg}{75}\right)\) Then \(F = \mu R\) becomes: \(X - Y\sin\alpha = \mu Y\cos\alpha\) M1 Eliminate \(X\) and \(Y\) and solve for \(\mu\) M1 | ||
| \(\mu = \dfrac{18}{49}\) (0.3673…..accept 0.37 or better) | A1 | 2.2a |
| (7) | ||
| (10 marks) |
Notes
M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors
A1: Correct equation, trig does not need to be substituted but \(N\) does.
M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors
A1: Correct equation, trig does not need to be substituted but \(N\) does.
N.B. The above 4 marks are for any two equations, either resolutions or moments or one of each. Mark best two equations.
Equations may appear in part (a) but must be used in (b) to earn marks.
M1: Must be used, e.g. seen on the diagram. i.e. M0 if merely quoting it.
(M0 if \(F = \mu \times \dfrac{9Mg}{25}\) used)
M1: Must have 3 equations (and all 3 previous M marks)
A1: Accept 0.37 or better