Question

A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto the barge.
Part A) How much deeper does the barge float?

Answer 1

Answer:

The depth of barge float=3 cm

Explanation:

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float=$l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water=$10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$cm

Hence, the depth of barge float=3 cm

Answer 2

Answer:

Explanation:

length, l = 5.2 m

width, w = 2.4 m

density of water = 1000 kg/m³

Let the depth immersed in water is h.

So, According to the principle of flotation

Weight of the auto parts = Buoyant force acting on the barge

410 x g = 5.2 x 2.4 x h x 1000 x g

410 = 12480 h

h = 3.3 cm