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Question

A uniform wooden plank $5 m$ long and weighting $40 k g$ is resting on two supports $0.5 m$ from each end. A boy weighting $45 k g$ stands $1.5 m$ from one end of the wooden plank. Find the reactions at the supports.

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Solution

Now from the laws of equilibrium

$Σ F$ = 0

$R_{1} - 40 g - 45 g + R_{2} = 0$

$R_{1} + R_{2} = 85 g$
850N

and
$Σ τ = 0$

$R_{1} \times 0.5 - 40 g$ $\times 2.5 - 45 g$ $\times 3.5 + R_{2}$ $\times 4.5$ = 0

$R_{1} + 9 R_{2} = 5150$

solving for $R_{1}$ and $R_{2}$

$R_{1}$ = 312.5 N

$R_{2}$ = 537.5 N

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