Problem 01 - Buoyancy

$V_{wood} = (0.305^2)(3)$

$V_{wood} = 0.279\,075 \, \text{m}^3$
 

$W = \gamma_{wood} V_{wood} = 6288.46(0.279\,075)$

$W = 1754.95 \, \text{N}$
 

 

$BF = \gamma_{water} V_{wood} = 9810(0.279\,075)$

$BF = 2737.72 \, \text{N}$
 

$\Sigma F_v = 0$

$F + W = BF$

$F + 1754.95 = 2737.72$

$F = 982.77 \, \text{N}$           answer

$V_{wood} = 1(1)(10)$

$V = 10 \, \text{ft}^3$
 

$W = \gamma_{wood} V_{wood} = 40(10)$

$W = 400 \, \text{lb}$
 

 

$BF = \gamma_{water} V_{wood} = 62.4(10)$

$BF = 624 \, \text{lb}$
 

$\Sigma F_v = 0$

$F + W = BF$

$F + 400 = 624$

$F = 224 \, \text{lb}$           answer