M1 June 2014 Q4
4.
A beam \(AB\) has weight \(W\) newtons and length 4 m. The beam is held in equilibrium in a horizontal position by two vertical ropes attached to the beam. One rope is attached to \(A\) and the other rope is attached to the point \(C\) on the beam, where \(AC = d\) metres, as shown in Figure 3. The beam is modelled as a uniform rod and the ropes as light inextensible strings. The tension in the rope attached at \(C\) is double the tension in the rope attached at \(A\).
A small load of weight \(kW\) newtons is attached to the beam at \(B\). The beam remains in equilibrium in a horizontal position. The load is modelled as a particle. The tension in the rope attached at \(C\) is now four times the tension in the rope attached at \(A\).
| Scheme | Marks |
|---|---|
| Resolving vertically: \(T + 2T\ (= 3T) = W\) | M1A1 |
| Moments about \(A\): \(2W = 2T \times d\) | M1A1 |
| Substitute and solve: \(2W = 2\dfrac{W}{3}d\) | DM1 |
| \(d = 3\) | A1 |
| (6) |
Notes
N.B. In moments equations, for the M mark, all terms must be force x distance but take care in the cases when the distance is 1.
N.B. If \(Wg\) is used, mark as a misread. If \(T\) and \(2T\) are reversed, mark as per scheme NOT as a misread.
First M1 for an equation in \(W\) and \(T\) and possibly \(d\) (either resolve vertically or moments about any point other than the mid-pt), with usual rules.
First A1 for a correct equation.
Second M1 for an equation in \(W\) and \(T\) and possibly \(d\) (either resolve vertically or moments about any point other than the mid-pt), with usual rules.
Second A1 for a correct equation.
Third M1, dependent on first and second M marks, for solving for \(d\)
Third A1 for \(d = 3\) cso
N.B. If a single equation is used (see below) by taking moments about the mid-point of the rod, \(2T = 2T(d - 2)\), this scores M2A2 (-1 each error)
Third M1, dependent on first and second M marks, for solving for \(d\)
Third A1 for \(d = 3\) cso
| Scheme | Marks |
|---|---|
| Resolving vertically: \(T + 4T = W + kW\) \(\left(5T = W(1 + k)\right)\) | M1A1 ft |
| Moments about A: \(2W + 4kW = 3 \times 4T\) | M1A1 ft |
| Substitute and solve: \(2W + 4kW = \dfrac{12}{5}W(1 + k)\) \(2 + 4k = \dfrac{12}{5} + \dfrac{12}{5}k\) | DM1 |
| \(\dfrac{8}{5}k = \dfrac{2}{5}\), \(k = \dfrac{1}{4}\) | A1 |
| (6) | |
| (12 marks) |
Notes
N.B. If \(Wg\) and \(kWg\) are used, mark as a misread.
If they use any results from (a), can score max M1A1 in (b) for one equation.
If \(T\) and \(4T\) are reversed, mark as per scheme NOT as a misread.
First M1 for an equation in \(W\) and a tension \(T_1\) and possibly their \(d\) or their \(d\) and \(k\) (either resolve vertically or moments about any point), with usual rules.
First A1 ft on their \(d\), for a correct equation.
Second M1 for an equation in \(W\) and the same tension \(T_1\) and possibly their \(d\) or their \(d\) and \(k\) (either resolve vertically or moments about any point), with usual rules.
Second A1 ft on their \(d\), for a correct equation.
Third M1, dependent on first and second M marks, for solving to give a numerical value of \(k\)
Third A1 for \(k = 1/4\) oe cso