A Mechanics Problem Involving the Centre of Mass of a Uniform Trapezoidal Lamina in Equilibrium

A mathematically inclined sixth form pupil brought me a Mechanics 2 type problem involving the centre of mass of a uniform trapezoidal lamina in equilibrium, with which he was having difficulty. The problem involves a number of useful issues so I am recording it here along with my suggested solution.

The Problem

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Suggested solution to parts (a) and (b)

The original situation is shown in the diagram below:

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After folding, we observe that the mass of the triangle FBE is the same as the mass of the square FBCE. Since the lamina is uniform, the mass of the total trapezium is proportional to 3a^2, the mass of AFED is proportional to 2a^2 and the mass of FBE is proportional to a^2.

By symmetry, the centre of mass of AFED is at \big(a, \frac{a}{2}\big) (assuming the origin is at A). Using the coordinates of the vertices of the triangle FBE, namely (2a, 0), (3a, 0), (2a, a), we use the formula for the centre of mass of a uniform triangular lamina to find the location of the triangle’s centre of mass:

\big(\frac{2a+3a+2a}{3}, \frac{a}{3}\big) = \big(\frac{7a}{3}, \frac{a}{3}\big)

Therefore the centre of mass of the trapezium has coordinates

\frac{2a^2}{3a^2} \begin{pmatrix} a \\ \frac{a}{2} \end{pmatrix} + \frac{a^2}{3a^2} \begin{pmatrix} \frac{7a}{3} \\ \frac{a}{3} \end{pmatrix} = \begin{pmatrix} \frac{13a}{9} \\ \frac{4a}{9} \end{pmatrix}

The distance of the centre of mass from AD is then \frac{13a}{9} and the distance from AB is \frac{4a}{9}. (Answer)

Suggested solution to part (c)

The centre of mass will be vertically below the point of suspension. The situation is shown in the following diagram:

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Using simple trigonometry, the angle between DE and the vertical line going through the centre of mass is \arctan \frac{5a/9}{5a/9} = 45^{\circ}, and this implies that the angle between DE and the horizontal is also 45^{\circ}. (Answer)

Suggested solution to part (d)

The situation now is shown in the following diagram:

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The centre of mass is now

\frac{M}{M+m} \begin{pmatrix} \frac{13a}{9} \\ \frac{4a}{9} \end{pmatrix} + \frac{m}{M+m} \begin{pmatrix} 3a \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{M(13a/9) + 3am}{M+m} \\ \frac{M(4a/9)}{M+m} \end{pmatrix}

But the centre of mass must be below the point of suspension. Therefore we must have

\frac{M(13a/9) + 3am}{M+m} = 2a

\Longrightarrow

m = \frac{5M}{9}   (Answer)