M2 June 2012 Q3
3.
A uniform rod \(AB\), of mass 5 kg and length 4 m, has its end \(A\) smoothly hinged at a fixed point. The rod is held in equilibrium at an angle of 25\(^\circ\) above the horizontal by a force of magnitude \(F\) newtons applied to its end \(B\). The force acts in the vertical plane containing the rod and in a direction which makes an angle of 40\(^\circ\) with the rod, as shown in Figure 1.
| Scheme | Marks |
|---|---|
| \(M(A),\ F.4\sin 40^\circ = 5g.2\cos 25^\circ\) | M1 A1 A1 |
| \(F = 35\) | A1 |
| (4) |
Notes
M1 A complete method to find \(F\), e.g. take moments about \(A\). Condone sin/cos confusion. Requires correct ratio of lengths.
A1 Correct terms with at most one slip
A1 All correct
A1 35 or 34.5 (>3sf not acceptable due to use of 9.8, but only penalise once in a question)
| Scheme | Marks |
|---|---|
| \(F\cos 75^\circ \pm Y = 5g\) | M1 A1 |
| \(Y = 40\); | A1 |
| UP | A1 |
| (4) | |
| (8 marks) |
Notes
M1 Resolve vertically. Need all three terms but condone sign errors. Must be attempting to work with their 75\(^\circ\) or 15\(^\circ\).
A1 Correct equation (their \(F\))
A1 40 or 40.1 Apply ISW if the candidate goes on to find \(R\).
A1 cso (the Q does specifically ask for the direction, so this must be clearly stated)
OR1 (b)
| \(4\text{m}\cos 25 \times Y = 5g \times 2\text{m}\cos 25 + F\cos 15 \times 4\text{m}\sin 25\) etc. | M1 A1 |
M1 A1 Taking moments about the point vertically below \(B\) and on the same horizontal level as \(A\). (Their \(F\))
OR2 (b)
| \(R\cos\alpha = F\cos 40 + 5g\cos 65\) | |
| \(R\sin\alpha + F\sin 40 = 5g\cos 25\) | |
| \(R = 52.1,\ \alpha = 25.3^\circ\) | |
| \(Y = R\sin(25 + \alpha)\) Etc. | M1A1 |
Resolve parallel & perpendicular to the rod
Solve for \(R\), \(\alpha\)
M1A1 Need a complete strategy to find \(Y\) for M1.