Statics of rigid bodies — WME03 IAL Further Mathematics practice questions · Study-IAL

Statics of rigid bodies

WME03Topic 3.53 questions

Centres of mass of solids and laminas by integration, and toppling and sliding.

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Get a plain-language explanation, a worked example and the usual mistake.

Original practice questions written in the style of the exam.

\bar{x} = \dfrac{\int_0^2 x \cdot x^2\,dx}{\text{area}}

(1)

\bar{x} = \dfrac{4}{8/3} = \dfrac{3}{2}

(1)

\bar{y} = \dfrac{\int_0^2 \tfrac{1}{2}y^2\,dx}{\text{area}} = \dfrac{\tfrac{1}{2}\int_0^2 x^4\,dx}{8/3}

(1)

\tfrac{1}{2}\displaystyle\int_0^2 x^4\,dx = \dfrac{16}{5}

(1)

\bar{y} = \dfrac{6}{5}

, so the centre of mass is

\left(\tfrac{3}{2}, \tfrac{6}{5}\right)

(1)

=

2

cm

\tfrac{8}{4} = 2\,\text{cm}

\frac{8}{4} = 2 cm

M1 Topples when the centre of mass is vertically above the lowest point of the base:

\tan\alpha = \dfrac{3}{2}

(1)

\alpha = 56^\circ

(1)

B1 Slides when

\tan\alpha = \mu

, so it topples first if

\mu > 1.5

(1)