# Solucionario de fisica

Solo disponible en BuenasTareas
• Páginas : 67 (16736 palabras )
• Descarga(s) : 0
• Publicado : 8 de febrero de 2011

Vista previa del texto
IDENTIFY: For a floating object, the weight of the object equals the upward buoyancy force, B, exerted by the
fluid.
SET UP: fluid submerged B = ρ V g . The weight of the object can be written as object object w = ρ V g . For seawater,
ρ =1.03×103 kg/m3 .
EXECUTE: (a) The displaced fluid must weigh more than the object, so fluid ρ < ρ .
(b) If the ship does not leak, much of the waterwill be displaced by air or cargo, and the average density of the
floating ship is less than that of water.
(c) Let the portion submerged have volume V, and the total volume be 0 V . Then 0 fluid ρV = ρ V , so
0 fluid
V ρ
V ρ
= The
fraction above the fluid surface is then
fluid
1 ρ . If ρ 0,
ρ
− → the entire object floats, and if fluid ρ →ρ , none of the
object is above the surface.(d) Using the result of part (c),
6 3
3
fluid
1 1 (0.042 kg) ([5.0][4.0][3.0] 10 m ) 0.32 32%.
1030kg m
ρ
ρ
× −
− =− = =
EVALUATE: For a given object, the fraction of the object above the surface increases when the density of the
fluid in which it floats increases.
14.30. IDENTIFY: water obj B = ρ V g . The net force on the sphere is zero.
SET UP: The density of water is 1.00×103kg/m3 .
EXECUTE: (a) B = (1000 kg/m3)(0.650 m3 )(9.80 m/s2 ) = 6.37×103 N
(b) B = T + mg and
3
2
6.37 10 N 900 N 558 kg
9.80 m/s
m B T
g
− × −
= = = .
(c) Now water sub B = ρ V g , where sub V is the volume of the sphere that is submerged. B = mg . water sub ρ V = mg and
3
sub 3
water
558 kg 0.558 m
1000 kg/m
V m
ρ
= = = .
3
sub
3
obj
0.558 m 0.858 85.8%
0.650 m
V
V
= == .
EVALUATE: The average density of the sphere is 3
sph 3
558 kg 858 kg/m
0.650 m
m
V
ρ = = = . sph water ρ < ρ , and that is why
it floats with 85.8% of its volume submerged.
Fluid Mechanics 14-7
14.31. IDENTIFY and SET UP: Use Eq.(14.8) to calculate the gauge pressure at the two depths.
(a) The distances are shown in Figure 14.31a.

EXECUTE: p − p0 = ρ gh
The upper face is1.50 cm below the top of the oil, so
3 2
0 p − p = (7.90 kg/m )(9.80 m/s )(0.0150 m)
0 p − p =116 Pa
Figure 14.31a
(b) The pressure at the interface is interface a oil p = p + ρ g(0.100 m). The lower face of the block is 1.50 cm below the
interface, so the pressure there is interface water p = p + ρ g(0.0150 m). Combining these two equations gives
a oil water p − p = ρ g(0.100 m) + ρg(0.0150 m)
3 3 2
a p − p = [(790 kg/m )(0.100 m) + (1000 kg/m )(0.0150 m)](9.80 m/s )
a p − p = 921 Pa
(c) IDENTIFY and SET UP: Consider the forces on the block. The area of each face of the block is
A = (0.100 m)2 = 0.0100 m2. Let the absolute pressure at the top face be t p and the pressure at the bottom face be
bp . In Eq.(14.3) use these pressures to calculate the force exerted by thefluids at the top and bottom of the block.
The free-body diagram for the block is given in Figure 14.31b.

EXECUTE: y y ΣF = ma
b t p A − p A − mg = 0
b t ( p − p )A = mg
Figure 14.31b
Note that b t b a t a ( p − p ) = ( p − p ) − ( p − p ) = 921 Pa −116 Pa = 805 Pa; the difference in absolute pressures equals
the difference in gauge pressures.
2
b t
2
( ) (805 Pa)(0.0100 m ) 0.821kg.
9.80 m/s
m p p A
g

= = =
And then ρ = m/V = 0.821 kg/(0.100 m)3 = 821 kg/m3.
EVALUATE: We can calculate the buoyant force as oil oil water water B = (ρ V + ρ V )g where 2
oil V = (0.0100 m )(0.850 m) =
8.50×10−4 m3 is the volume of oil displaced by the block and 2 4 3
water V = (0.0100 m )(0.0150 m) =1.50×10− m is the volume
of water displaced by the block. This gives B = (0.821kg)g. The mass of water displaced equals the mass of the block.
14.32. IDENTIFY: The sum of the vertical forces on the ingot is zero. ρ = m/V . The buoyant force is water obj B = ρ V g .
SET UP: The density of aluminum is 2.7×103 kg/m3 . The density of water is 1.00×103 kg/m3 .
EXECUTE: (a) T = mg = 89 N so m = 9.08 kg . 3 3
3 3
9.08 kg 3.36 10 m 3.4 L
2.7 10 kg/m
V m
ρ
= = = × − =...