A simple \"box-like\" canoe has outer dimensions of 15.0 ft in length, 2.0 ft in

Question

A simple "box-like" canoe has outer dimensions of 15.0 ft in length, 2.0 ft in width, and 1.0 ft in depth. The canoe shell is made of concrete with a specific weight of 58.0 lbs/ft^3 and contains 15.0 lbs of steel reinforcement with negligible volume. The volume of concrete in the canoe shell can be estimated by the following equation: V=tx64 ft^2, where t is the hull thickness in feet.

a four inch freeboard is desired. for a hull thickness of .5 in what is the maximum weight of the paddlers and other gear?

Explanation / Answer

length=15.0 ft width=2.0 ft depth=1.0 ft volume of canoe=15*2*1=30ft^3 volume of concrete in the canoe shell=tx64 ft^2 for .5 hull thickness volume of concrete in the canoe shell=32ft^3 AS specific weight of concrete is 58.0 lbs/ft^ so for 32ft^3 weight will 32*58=1856lbs secondly free board is of 4 inch so so total depht will =1+4/3=1.333ft volume 15*2*1.333=39.99 ft^3 for this weight will =39.99*582319.42lb

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