Unreasonable Results Suppose two children are using a uniform seesaw that is 3.00 m long and has its center of mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from the pivot. (a) Calculate where the second 18.0 kg child must sit to balance the seesaw. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?

Solution 39PE Step-by-step solution Step 1 of 4 (a) Since that two children are using a uniform seesaw and it has center of mass over the pivot. Draw the free body of diagram as given below, Step 2 of 4 Apply the second condition of equilibrium that is net applied torque around pivot point, A is zero. Here, is mass of first children, is mass of second children, is position of first children from pivot point A and is position of second children from pivot point A. Since weight of first children is, Here, is mass of first children and g is acceleration due to gravity. Since weight of second children is, Here, is mass of second children and g is acceleration due to gravity. Substitute and , Substitute 30 kg for , 18.0 kg for , 9.8 m/s2 for g and 1.40 m for solve for , Hence, Second child must sit at a distance of form the pivot point.