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 plank starting from P. If the plank is in equilibrium, calculate the
(a) reactions at P and R when he walked 1.5 metres;
(b) distance he had walked when the two reactions are equal;
(c) distance he walked before the plank tips over.
Explanation
(a)
Take moments about P
Clockwise moment = (200 times 1.5 + 100 times 4)
= (300 + 400 = 700N)
Anti-clockwise moment = (5R_{2} N)
(5R_{2} = 700 ; R_{2} = 140N)
For vertical equilibrium, (R_{1} + R_{2} = 300)
(R_{1} = 300 – 140 = 160N)
(b)
Suppose he walked x m from P
Let each reaction be R N
Equating forces, 2R = 300
(R = 150N)
We take moments about P.
Clockwise moments =( 200x + 100 times 4)
= (200x + 400 Nm)
Anti-clockwise moments = 5R = 750 Nm
(200x + 400 = 750)
(200x = 350 ; x =1.75m)
(c) Plank tips over when he is between S and Q.
Let boy be y m from S.
We take moments about S.
(200y = 100(5 – 4) = 100)
(y = 0.5m)
The boy must be 5.5m from P.