In figure (a), a 5.4 kg dog stands on a 15 kg flatboat at distance D = 6.1 m fro
ID: 2175900 • Letter: I
Question
In figure (a), a 5.4 kg dog stands on a 15 kg flatboat at distance D = 6.1 m from the shore. It walks 2.1 m along the boat toward shore and then stops. Assuming no friction between the boat and the water, find how far the dog is then from the shore. m The canoe and dog form a system; treat each as a particle located at its own com. The system's com cannot move because of internal forces, such as due to the dog's feet pushing on the canoe as the dog moves. Read the eBook Section 9-3 Newton's Second Law for a System of ParticlesExplanation / Answer
let the center of mass of boat be at x m from shore. com of system, com1 = (x*15+6.1*5.4)/(20.4) now, let the boat move y m away from shore, now, com2 = ((x+y)15 + (4+y)*5.4)/20.4 com1 = com2 (x*15+6.1*5.4)/(20.4) = ((x+y)15 + (4+y)*5.4)/(20.4) (x*15+6.1*5.4)=((x+y)15 + (4+y)*5.4) 6.1*5.4 = 15y + 21.6 + 5.4y 20.4y = 11.34 y = 11.34/20.4 y = 0.56 m distance of dog from shore = (4+y) = 4.56 m