M1 June 2013 (R) Q2
2.
A particle of weight 8 N is attached at \(C\) to the ends of two light inextensible strings \(AC\) and \(BC\). The other ends, \(A\) and \(B\), are attached to a fixed horizontal ceiling. The particle hangs at rest in equilibrium, with the strings in a vertical plane. The string \(AC\) is inclined at 35\(^\circ\) to the horizontal and the string \(BC\) is inclined at 25\(^\circ\) to the horizontal, as shown in Figure 1. Find
| Scheme | Marks |
|---|---|
| Resolve horizontally: \(\mathrm{T}_A\cos 35^\circ = \mathrm{T}_B\cos 25^\circ\) | M1A1 |
| Resolve vertically: \(\mathrm{T}_A\sin 35^\circ + \mathrm{T}_B\sin 25^\circ = 8\) | M1A1 |
| Equation in one unknown: \(\mathrm{T}_B\dfrac{\cos 25^\circ}{\cos 35^\circ}\sin 35^\circ + \mathrm{T}_B\sin 25^\circ = 8\) or \(\mathrm{T}_A\sin 35^\circ + \mathrm{T}_A\dfrac{\cos 35^\circ}{\cos 25^\circ}\sin 25^\circ = 8\) | DM1A1 |
| \(\mathrm{T}_A = 8.4\), 8.37, 8.372 (N) or better | A1 |
| \(\mathrm{T}_B = 7.6\), 7.57, 7.567 (N) or better | A1 |
| (8 marks) |
Notes
First M1 for resolving horizontally with correct no. of terms and both \(T_A\) and \(T_B\) terms resolved.
First A1 for a correct equation.
Second M1 for resolving vertically with correct no. of terms and both \(T_A\) and \(T_B\) terms resolved.
Second A1 for a correct equation.
Third M1, dependent on first two M marks, for eliminating \(T_A\) or \(T_B\)
Third A1 for a correct equation in one unknown
Fourth A1 for \(T_A = 8.4\) (N) or better.
Fifth A1 for \(T_B = 7.6\) (N) or better.
N.B. The first two M marks can be for two resolutions in any two directions.
N.B. If the two tensions are taken to be equal, can score max M1A0 for vertical resolution.
2alt
| Using Sine Rule on triangle of forces: \(\dfrac{8}{\sin 60^\circ} = \dfrac{\mathrm{T}_A}{\sin 65^\circ} = \dfrac{\mathrm{T}_B}{\sin 55^\circ}\) | M1A1 |
| \(\dfrac{8 \times \sin 65^\circ}{\sin 60^\circ} = \mathrm{T}_A\), \(= 8.4\), 8.37, 8.372 (N) or better | M1A1, A1 |
| \(\dfrac{8 \times \sin 55^\circ}{\sin 60^\circ} = \mathrm{T}_B\), \(= 7.6\), 7.57, 7.567 (N) or better | M1A1, A1 |
2 alt 1
See Alternative 1 using a Triangle of Forces and the Sine Rule.
2 alt 2
Alternative 2 is to resolve perpendicular to each string:
The scheme is similar to Alt 1 and gives the same expressions for \(T_A\) and \(T_B\)
M1A1 resolving perp to both strings as a complete method.
M1A1A1 for finding \(T_A\)
M1A1A1 for finding \(T_B\)