Question

How much extra water does a 147-lb concrete canoe displace compared to an ultra -lightweight 40-lb Kevlar canoe of the same size carrying the same load?

Answer 1

F.B.D: Load M FbT w Start by writing an equilibrium of forces equation: Erro Fb - W = 0 Where Fb is the buoyancy force and W the weight of the canoe and the load combined. Knowing that the weight is W = m*g, the previous equation is: Fb = W = (Mc + M)g (Eq. I) And the buoyancy force is calculated using the following equation: Fb = V *p*g (Eq. 2) Where V is the volume displaced of water by the canoe and p is the density of the fluid (in this case, water at room temperature: p = 62,4 ).|| Matching Eq. 1 and 2: V*p = Mc + MI

Mc + MI V =- Which would give us the total volume displace by each canoe, however, the problem asks us for the difference between volumes displaced, and thus the difference is: av = (Mc M'). - (Me + M"), Where the sub index"c" represents concrete and "K" Kevlar. Since both the mass of the load and the density of the fluid is the same for both cases, the previous equation becomes: Moc - MCK AV = 1471b - 401b AV = - AV = 1,715ft3