A uniform see-saw has a length of $8 m$ . A boy of weight $50 k g$ sits at a distance $2 m$ from one end and a girl of the weight $40 k g$ sits at a distance $1 m$ from the other end. Is the see-saw balanced? How much should the girl shift to make the see-saw balanced?

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Solution

Step 1: Given data
Weight of boy $= 50 k g$
Boy's distance $= 2 m$
Weight of girl $= 40 k g$
Girl's distance $= 1 m$
Step 2: Formula used
Moment of force $= F o r c e \times M o m e n t s a r m$
If see-saw to be equilibrium, $C l o c k w i s e m o m e n t s o f f o r c e = A n t i c l o c k w i s e m o m e n t s o f f o r c e$
Here, weight is taken as a force.
$50 k g \times 2 m = 40 k g \times 1 m 100 N m \neq 40 N m$
Hence, the see-saw is unbalanced.
Step 4: Finding the girl's distance to keep see-saw in equilibrium:
Let x be the distance of girl to keep the sea-saw in equilibrium.
$50 k g \times 2 m = 40 k g \times x m x = \frac{100}{40} = 2.5 m$
Hence, to make the see-saw balanced girl should sit at $2.5 m$ .

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A uniform see-saw has a length of 8m. A boy of weight 50kg sits at a distance 2m from one end and a girl of the weight 40kg sits at a distance 1m from the other end. Is the see-saw balanced? How much should the girl shift to make the see-saw balanced?

A uniform see-saw has a length of $8 m$ . A boy of weight $50 k g$ sits at a distance $2 m$ from one end and a girl of the weight $40 k g$ sits at a distance $1 m$ from the other end. Is the see-saw balanced? How much should the girl shift to make the see-saw balanced?