Two small kids of masses 10 kg and 15 kg are trying to balance a seesaw of total length 5.0 m, with the fulcrum at the centre. If lighter kid is sitting at an end, where should the other sit?

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Solution

Given : $W_{1} = 10 \times 10 = 100 N$ $W_{2} = 15 \times 10 = 150 N$
Position of $W_{1}$ from centre $r_{1} = 2.5 m$
Using $W_{1} r_{1} = W_{2} r_{2}$
Or $100 \times 2.5 = 150 \times r_{2}$
$⟹ r_{2} = 1.67 m$

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Two small kids of masses 10 kg and 15 kg are trying to balance a seesaw of total length 5.0 m, with the fulcrum at the centre. If lighter kid is sitting at an end, where should the other sit?

Given : W1=10×10=100 N W2=15×10=150 NPosition of W1 from centre r1=2.5 mUsing W1r1=W2r2Or 100×2.5=150×r2⟹ r2=1.67 m

Given : $W_{1} = 10 \times 10 = 100 N$ $W_{2} = 15 \times 10 = 150 N$
Position of $W_{1}$ from centre $r_{1} = 2.5 m$
Using $W_{1} r_{1} = W_{2} r_{2}$
Or $100 \times 2.5 = 150 \times r_{2}$
$⟹ r_{2} = 1.67 m$