A uniform plank PQ of length 8m and mass 10kg is supported horizontally at the end P and at point R, 3 metres from Q. A boy of mass 20 kg walks along the pl...

Answer Details

The plank \(PQ\) is \(8\ \text{m}\) long, mass \(10\ \text{kg}\), supported at \(P\) (left end) and at \(R\), which is \(3\ \text{m}\) from \(Q\), i.e. \(5\ \text{m}\) from \(P\). The plank weight \(10g\) acts at the midpoint, \(4\ \text{m}\) from \(P\). A boy of weight \(20g\) stands at distance \(d\) from \(P\). (Take \(g = 10\ \text{m s}^{-2}\), so weights are \(100\ \text{N}\) and \(200\ \text{N}\).)

(a) When \(d = 1.5\ \text{m}\). Taking moments about \(P\) (so \(R_P\) has no moment):

\[R_R \times 5 = 10g(4) + 20g(1.5) = 40g + 30g = 70g \Rightarrow R_R = 14g = 140\ \text{N}.\]

Vertical equilibrium: \(R_P + R_R = 30g = 300\ \text{N} \Rightarrow R_P = 160\ \text{N}\).

So \(R_P = 160\ \text{N},\ R_R = 140\ \text{N}\).

(b) When the reactions are equal. Each reaction \(= \tfrac12(30g) = 15g = 150\ \text{N}\). Moments about \(P\):

\[15g \times 5 = 40g + 20g\,d \Rightarrow 75g = 40g + 20g\,d \Rightarrow d = \frac{35}{20} = 1.75\ \text{m}.\]

(c) When the plank tips. The plank tips about \(R\) when \(R_P = 0\). Taking moments about \(R\), the boy (on the \(Q\) side, \(d - 5\) from \(R\)) balances the plank weight (\(5 - 4 = 1\ \text{m}\) on the \(P\) side):

\[20g(d - 5) = 10g(1) \Rightarrow d - 5 = 0.5 \Rightarrow d = 5.5\ \text{m}.\]

The boy can walk \(5.5\ \text{m}\) from \(P\) before the plank tips over.