A(40kg) and B(46kg). They went to a playground and wanted to play on centrally pivoted seesaw. "A" sat on the left of the seesaw at 2m from the pivot. If the 4.4m long board must be balanced horizontally,where must "B" sit?

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A(40kg) and B(46kg). They went to a playground and wanted to play on centrally pivoted seesaw. "A" sat on the left of the seesaw at 2m from the pivot. If the 4.4m long board must be balanced horizontally,where must "B" sit?

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Answer 1 · 2015-08-07T14:08:46

I found: $1.7 m$ $1.7m$ (to the right).

Explanation:

At equilibrium the torques with respect to the pivot point should cancel:
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Where $torque = weight \cdot distance$ $"torque"="weight"*"distance"$
Giving:
$[40 \cdot 9.8 \cdot (- 2)] + [46 \cdot 9.8 \cdot x] = 0$ $[40*9.8*(-2)]+[46*9.8*x]=0$
$x = \frac{784}{451} = 1.7 m$ $x=784/451=1.7m$

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A(40kg) and B(46kg). They went to a playground and wanted to play on centrally pivoted seesaw. "A" sat on the left of the seesaw at 2m from the pivot. If the 4.4m long board must be balanced horizontally,where must "B" sit?

1 Answer

Explanation:

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