In a SeeSaw a child of mass 25 kg is at its one end and another child has mass of 20 kg at the other end. Find the distance of second child to balance the seesaw, if first child is at 2 metre from the fix it.. take G is equal to 10 metre per second square

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Solution

net torque about the centre of seesaw should be zero for equilibrium
Weight of first child x distance from fixit = weight of second child x distance from fixit
25 x g x 2 = 20 x g x Y
Y = 2.5 metres

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In a SeeSaw a child of mass 25 kg is at its one end and another child has mass of 20 kg at the other end. Find the distance of second child to balance the seesaw, if first child is at 2 metre from the fix it.. take G is equal to 10 metre per second square

net torque about the centre of seesaw should be zero for equilibrium Weight of first child x distance from fixit = weight of second child x distance from fixit 25 x g x 2 = 20 x g x Y Y = 2.5 metres

net torque about the centre of seesaw should be zero for equilibrium
Weight of first child x distance from fixit = weight of second child x distance from fixit
25 x g x 2 = 20 x g x Y
Y = 2.5 metres