(c9p30) A 11.89 -kg block of metal is suspended from a scale and immersed in wat
ID: 1261573 • Letter: #
Question
(c9p30) A 11.89 -kg block of metal is suspended from a scale and immersed in water as in Figure P9.30. The dimensions of the block are 14.0 cm x 12.0 cm x 12.0 cm. The 14.0-cm dimension is vertical, and the top of the block is 13.0 cm below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? (Take P0 = 1.013
Explanation / Answer
(a)
the absolute pressure at the level of the top of theblock will be
Ptop = Po +?water g htop
=(1.0130 105 N/m2) + (103 kg /m3) (9.80 m / s2) (13x 10-2m)
= 102574 Pa
at the level of the bottom of the block will be
Pbottom = Po +?water g hbottom
=(1.0130 105 N/m2) + (103 kg /m3) (9.80 m / s2) (27.0 x 10-2m)
= 103946 Pa
so the downward force exerted on the top by the waterwill be
Ftop = Ptop A = (102574)(0.12 m)2 = 1477.0 N
the upward force the water exerts on the bottom of theblock is
Fbottom = Pbottom A = (103946)(0.12 m)2 = 1496.8 N
(b)
so the scale reads the tension in the T in the cordsupporting the block
as the block is in equilibrium we can write
? Fy = T + Fbottom - Ftop -m g
= 0
T = (11.98)(9.8)+1477.0 -1496.8 N
= 97.58 N
(c)
According to the archimedes principle the buoyantforce on the block equals the weight of the
displaced water so we can write
B = (?water Vblock) g = 1000*0.002016*9.8 = 19.8 N
From part (a), F = 1496.8 - 1477 = 19.8 N
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