A uniform rectangular lamina ABCD of mass 10 kg has AD = BC = 2a and AB = DC = a. It rests in a vertical plane with A on a rough horizontal table and AD inclined at 30^° to the horizontal. A force P along DC maintains equilibrium. Taking g = 9.81 mathrmms^-2, find P in newtons, correct to 2 decimal places.

Paper 2 · Specimen 2026 · Mechanics

A uniform rectangular lamina ABCD of mass $10$ kg has $AD = BC = 2a$ and $AB = DC = a$ . It rests in a vertical plane with A on a rough horizontal table and AD inclined at $30^\circ$ to the horizontal. A force $P$ along DC maintains equilibrium. Taking $g = 9.81\,\mathrm{ms^{-2}}$ , find $P$ in newtons, correct to 2 decimal places.

Model answer

30.22

Also accepted: 30.22 N, 30.216, 30.2

Explanation

Take moments about A so the unknown reaction and friction there drop out. The centre of mass of a uniform rectangle is at its centre, a horizontal distance

a\left(\dfrac{\sqrt3}{2} - \dfrac14\right) = 0.61603a

from A, so the weight has moment

10g(0.61603a)

. The force

P

acts along DC, a perpendicular distance

2a

from A, so its moment is

2aP

. Equating and cancelling

a

P = 5g(0.61603) = 5 \times 9.81 \times 0.61603 = 30.216

, that is

30.22

N. With

g = 9.8

the figure would be

30.19

N, so the value of

g

matters at the second decimal place asked for.

Derived from ZIMSEC Additional Mathematics Paper 2, Specimen Paper, Q2

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