. Two children of mass m1 = 35kg and m2 = 46kg sit on opposite ends of a homogeneous seesaw of length L = 2m, which has a fulcrum in the middle, as shown. At what position on the seesaw should a downward 200N force be applied so that the seesaw is in equilibrium? Give your answer as a position :13, with :1: = 0 at the fulcrum, and positive to the right. A- 0m 2.0m lI B. 0.056m :nl m2 C. 0.54m , f" D. 0.62111\" Firs":- E. It is impossible to answer this ques 35k 46kg tion without knowing the mass of 0-+Qj the seesaw. Please explain why net torque equation is m1*g*(L/2] - m2*g*(L/2) + F*d. I don't understand why torque on the right is treated as negative, also where did F*d came from