A person is sitting at the very back of a canoe of length L, when the front just

Question

A person is sitting at the very back of a canoe of length L, when the front just bumps into the dock. If the canoeist gets up and walks to the front of the canoe and there is no friction between the boat and the water, how far will they be from the dock? Express your answer in terms of the length of the canoe, the mass of the person m_p, and the mass of the canoe m_c. What will this distance be, in meters, if the length of canoe is 3.1 m, the person's mass is 56 kg, and the canoe's mass is 74 kg?

Explanation / Answer

m1 = 56 cm        x1 = L

m2 = 74 kg        x2 = L/2

xCM = (m1*x1+m2*x2)/(m1+m2)

Xcm = L(56+37)/(56+74)

Xcm = 0.72 L

new position

x1' = d

x2' = d+L/2

Xcm = 0.72L

xCM = (m1*x1+m2*x2)/(m1+m2)

0.72L = (56*d + 74*(d+0.5L))/(56+74)

part(b)

L = 3.1

0.72*3.1 = (56*d + 74*(d+0.5*3.1))/(56+74)

d = 1.35 m <<---------answer

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