June 2023 Paper 3 Q11
11
A uniform rod \(AB\), of weight \(20\,\mathrm{N}\) and length \(2.8\,\mathrm{m}\), rests in equilibrium with the end \(A\) in contact with rough horizontal ground and the end \(B\) resting against a smooth wall inclined at \(55^\circ\) to the horizontal. The rod, which rests in a vertical plane that is perpendicular to the wall, is inclined at \(30^\circ\) to the horizontal (see diagram).
| Scheme | Marks | AO |
|---|---|---|
| M1 | 3.1b | |
| \(1.4(20\cos 30) = R_B(2.8\cos 25)\) | M1 | 1.1 |
| \(R_B = \dfrac{28\cos 30}{2.8\cos 25} = 9.5555330\ldots\) \(= 9.56\) (N) (correct to 3 significant figures) | A1 | 1.1 |
| [3] |
Notes
M1: Moments about \(A\) to form an equation with the correct number of terms and both the weight and contact force at \(B\) resolved - condone \(20g\) for the weight of the rod. All relevant values must have been substituted. May take moments about another point (and resolve) but must end up after elimination with an equation in \(R_B\) only
Resolved for the weight means either sin or cos with an angle of either 30 or 60 only, and for the contact force at \(B\) either sin or cos with an angle of 25 or 35 or 55 or 65 only
M1: One of the two moment terms correct
Must be as part of a two-term moment equation (so dimensionally correct so M0 if \(20g\) for the weight) but the other term need not be resolved (or resolved correctly) e.g. \(1.4(20\cos 30) = 2.8R_B\) scores M0 M1
A1: AG - correct equation followed by 9.56 scores all 3 marks
Allow awrt 9.56
For reference in parts (a) and (b):
Moments about \(B\): \(R_A(2.8\cos 30) = F_A(2.8\sin 30) + 20(1.4\cos 30)\)
Moments about the mid-point: \(R_A(1.4\cos 30) = F_A(1.4\sin 30) + R_B(1.4\cos 25)\)
| Scheme | Marks | AO |
|---|---|---|
| M1* | 3.3 | |
| \(R_A + 9.56\sin 35 = 20\) \(F_A = 9.56\cos 35\) | A1 A1 | 1.1 1.1 |
| \(\sqrt{(9.56\cos 35)^2 + (20 - 9.56\sin 35)^2}\) | M1dep* | 1.1 |
| 16.5 (N) | A1 | 2.2a |
| [5] |
Notes
M1*: Attempt at resolving vertically or horizontally – correct number of terms with the contact force at \(B\) resolved (angle must be one of 25, 35, 55 or 65 only) – or taking moments again (see below) (same conditions for taking moments as in part (a)). M0 if using \(W\) or \(mg\) only for the weight of the rod (must substitute in the given value before awarding this mark – see second guidance column if using \(20g\))
Allow sign errors, sin/cos confusion and \(20g\) used as the weight for the M mark. Allow \(R_B\) instead of the given value of 9.56 for the M mark
A1: Correct expression/equation for \(R_A\)
\(R_A = 20 - 5.48\ldots = 14.5\ldots\)
A1: Correct expression/equation for \(F_A\)
\(F_A = 7.83\ldots\)
M1dep*: Correct method for calculating the magnitude of the contact force at \(A\) where \(R_A\) and \(F_A\) are a linear combination of the correct number of terms with 9.56 resolved in both equations (angle must be one of 25, 35, 55 or 65 only)
A1: www awrt 16.5. For reference if using 9.56 for \(R_B\) then \(16.494179\ldots\)
If using more accurate value for \(R_B\) then \(16.494698\ldots\)
For reference in parts (a) and (b):
Moments about \(B\): \(R_A(2.8\cos 30) = F_A(2.8\sin 30) + 20(1.4\cos 30)\)
Moments about the mid-point: \(R_A(1.4\cos 30) = F_A(1.4\sin 30) + R_B(1.4\cos 25)\)