need help with physics

Question

two children are playing on a see saw. the older child has a mass of 40 kg. the younger child has a mass of 25 kg. assume that the see saw is massless and the two seats are each 1.5 m from the fulcrum. assume the see saw starts out perfectly horizontal with the children standing over their respective seats. The children then sit down at the same instant.

A) Calculate the net torque the children exert on the see-saw the moment that they are seated.

B) determine the magnitude of the linear acceleration the children should expect to experience the instant the see saw begins to move.

C) determine the length the younger child's side would need to be in order for the see saw to not move as they sat down.

any help would be appreciated

Answer 1

A) Using g = 9.8 m/s² to find weight force of each child, and torque τ = F*r, Στ = (40)(9.8)(1.5) - (25)(9.8)(1.5) = 221 N*m for the magnitude of the difference of torques in opposite directions.

B) For this we have the relationships τ = I*α where I is moment of inertia and α is angular acceleration, I = Σm*r², and linear a = r * α. I = (40)(1.5)² + (25)(1.5)² = 146 kg m², α = τ/I = 221/146 = 1.51 rad/s², and linear a = 2.26 m/s².

C) To balance the torques, (40)(9.8)(1.5) = (25)(9.8) x, so the younger child's distance x = (40)(1.5)/25 = 2.4 m.