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.
(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.
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Correct Answer: Option n
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.
(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.