M2 June 2017 Q5
5.
A uniform rod \(AB\), of mass 5 kg and length 8 m, has its end \(B\) resting on rough horizontal ground. The rod is held in limiting equilibrium at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{3}{4}\), by a rope attached to the rod at \(C\). The distance \(AC = 1\) m. The rope is in the same vertical plane as the rod. The angle between the rope and the rod is \(\beta\) and the tension in the rope is \(T\) newtons, as shown in Figure 3. The coefficient of friction between the rod and the ground is \(\dfrac{2}{3}\). The vertical component of the force exerted on the rod at \(B\) by the ground is \(R\) newtons.
| Scheme | Marks |
|---|---|
| \(F = \dfrac{2}{3}R\) seen or implied | B1 |
| M\((C)\): \(5g \times 3\cos\alpha + F \times 7\sin\alpha = 7\cos\alpha \times R\) | M1 A1 A1 |
| \(15g\cos\alpha = R\left(7\cos\alpha - \dfrac{14}{3}\sin\alpha\right)\) | |
| \(15g \times \dfrac{4}{5} = R\left(7 \times \dfrac{4}{5} - \dfrac{14}{3} \times \dfrac{3}{5}\right) = \dfrac{14}{5}R\) | dM1 |
| \(R = \dfrac{30}{7}g = 42\) (N) | A1 |
| (6) |
Notes
B1 Use of \(F = \mu R\). Could be on diagram. Allow in (b) if not seen before
M1 Moments about \(C\) or alternative complete method to find equation in \(F\) and \(R\) or \(R\) only. Dimensionally correct and all terms needed. Condone sin/cos confusion and sign error(s).
A1 At most one error
A1 Correct unsimplified equation
dM1 Substitute for \(F\) and trig and solve for \(R\). Dependent on previous M1
e.g. of alternative for M1A1A1:
| M\((A)\): \(T\sin\beta + 8R\cos\alpha = 8F\sin\alpha + 20g\cos\alpha\) and M\((B)\): \(\ 7T\sin\beta = 20g\cos\alpha\) | (M1) (A1) |
| \(\dfrac{20g}{7}\cos\alpha + 8R\cos\alpha = 8F\sin\alpha + 20g\cos\alpha\) | (A1) |
(A1) At most 1 error
(A1) Correct unsimplified equation in \(F\) and \(R\) or \(R\) only
| Scheme | Marks |
|---|---|
| Resolve \(\updownarrow\): \(\ \ T\cos\theta + R = 5g\) \(R + T\sin(\beta - \alpha) = 5g\) | M1 A1 |
| Resolve \(\leftrightarrow\): \(\ \ T\sin\theta = F\ (= 28)\) \(F\left(= \dfrac{2}{3}R\right) = T\cos(\beta - \alpha)\) | M1 A1 |
| Solve simultaneous equations for \(\beta - \alpha\) | |
| \(\tan(\beta - \alpha) = 4,\ \ \beta = 50.9^\circ\ \ (51^\circ)\) | A1 |
| (5) | |
| (11 marks) |
Notes
M1 Need all terms. Condone sin/cos confusion and sign error(s).
A1 Correct in \(R\) or their \(R\)
M1 Need both terms. Condone sin/cos confusion
A1 Correct in \(R\) or their \(R\)
A1 cso . Max 3 s.f.
Alt 5b
| M\((B)\): \(\ 7 \times T\sin\beta = 5g\cos\alpha \times 4\) | M1 |
| \(\left(T\sin\beta = \dfrac{16}{7}g\right)\) | A1 |
| OR: resolve perpendicular to the rod: \(T\sin\beta + R\cos\alpha = 5g\cos\alpha + \dfrac{2}{3}R\sin\alpha\) | (M1) (A1) |
| Resolve parallel to rod: \(T\cos\beta + 5g\sin\alpha = F\cos\alpha + R\sin\alpha\) \(\left(= \dfrac{2}{3}R\cos\alpha + R\sin\alpha\right)\) | M1 |
| \(\left(T\cos\beta = \dfrac{13}{7}g\right)\) | A1 |
| Solve simultaneous equations for \(\beta\) | |
| \(\tan\beta = \dfrac{16}{13},\ \ \beta = 50.9^\circ\ \ (51^\circ)\) | A1 |
M1 Moments equation. Dimensionally correct. Condone sin/cos confusion and sign error(s).
M1 All terms needed. Condone sin/cos confusion and sign error(s).
A1 cso. Max 3 s.f.