Unformatted text preview: agnitude of
buoyant force (2) Magnitude of
buoyant force These equations contain two unknowns, the volume V of the object and its mass m. By
subtracting Equation (2) from Equation (1), we can eliminate the mass algebraically. The
result is
15.2 N − 13.7 N = gV ( ρ water − ρ alcohol )
Solving this equation for the volume, and using the densities from Table 11.1, we have
V= 48. 15.2 N − 13.7 N
1.5 N
=
= 7.9 × 10−4 m3
2
3
3
3
g ( ρ water − ρalcohol )
9.80 m/s 1.00 × 10 kg/m − 806 kg/m ( )( ) REASONING The freebody diagram shows the two forces
acting on the balloon, its weight W and the buoyant force FB.
Newton’s second law, Equation 4.2b, states that the net force
ΣFy in the y direction is equal to the mass m of the balloon
times its acceleration ay in that direction:
FB − W = ma y
1 24
43
ΣFy Our solution is based on this statement. (4.2b) FB W
Freebody diagram
for the balloon +y Chapter 11 Problems 591 SOLUTION Solving Equation 4.2b for the acceleration ay gives ay = FB − W
m (1) The weight W of an object is equal to its mass m times the magnitude g of the acceleration
due to gravity, or W = mg (Equation 4.5). The mass, in turn, is equal to the product of an
object’s density ρ and its volume V, so m = ρ V (Equation 11.1). Combining these two
relations, the weight can be expressed as W = ρ V g.
According to Archimedes’ principle, the magnitude FB of the buoyant force is equal to the weight of the cool air that the balloon displaces, so FB = mcool airg = (ρcool airV)g. Since we
are neglecting the weight of the balloon fabric and the basket, the weight of the balloon is
just that of the hot air inside the balloon. Thus, m = mhot air = ρhot airV and W = mhot airg = (ρhot airV)g . Substituting the expressions FB = (ρcool airV)g, m = ρhot airV, and W = (ρhot airV)g into
Equation (1) gives
ay = Vg − ρ hot airVg ( ρcool air − ρ hot air ) g
ρ
FB − W
= cool air
=
m
ρhot airV
ρhot air (1.29 kg/m3 − 0.93 kg/m3 )(9.80 m/s2 ) = 3.8 m/s2
=
0.93 kg/m3 49. SSM REASONING AND SOLUTION The weight of the coin in air is equal to the sum
of the weights of the silver and copper that make up the coin: e je j Wair = mcoin g = msilver + mcopper g = ρ silver Vsilver + ρ copper Vcopper g (1) The weight of the coin in water is equal to its weight in air minus the buoyant force exerted
on the coin by the water. Therefore,
Wwater = Wair − ρ water (Vsilver + Vcopper ) g
Solving Equation (2) for the sum of the volumes gives
V silver + V copper = Wair − Wwater ρ water g = mcoin g − Wwater ρ water g (2) 592 FLUIDS or (1.150 × 10 –2 kg) (9.80 m / s 2 ) − 0.1011 N
= 1.184 × 10 –6 m 3
3
3
2
(1.000 × 10 kg / m ) (9.80 m / s ) Vsilver + Vcopper = From Equation (1) we have
V silver = mcoin − ρ copper V copper ρ silver = 1.150 × 10 –2 kg − (8890 kg / m 3 )Vcopper
10 500 kg / m 3 or Vsilver = 1.095 × 10 –6 m 3 − ( 0.8467 )Vcopper (3) Substitution of Equation (3) into Equation (2) gives Wwater = Wair − ρ water 1.095 × 10 –6 m 3 − ( 0.8467 )Vcopper + Vcopper g
Solving for Vcopper gives Vcopper = 5.806 × 10 –7 m 3 . Substituting into Equation (3) gives
Vsilver = 1.095 × 10 –6 m 3 − ( 0.8467 )( 5.806 × 10 –7 m 3 ) = 6.034 × 10 –7 m 3
From Equation 11.1, the mass of the silver is
msilver = ρsilverVsilver = (10 500 kg/m3 )(6.034 × 10−7 m 3 ) = 6.3 × 10 −3 kg 50. REASONING AND SOLUTION The buoyant force exerted by the water must at least
equal the weight of the logs plus the weight of the people, FB = WL + WP ρwgV = ρLgV + WP
Now the volume of logs needed is
V= MP ρW − ρL = 4 ( 80.0 kg )
1.00 × 10 kg/m − 725 kg/m
3 3 3 = 1.16 m3 The volume of one log is VL = π(8.00 × 10 –2 2 m) (3.00 m) = 6.03 × 10 –2 m 3 Chapter 11 Problems 593 The number of logs needed is
–2 N = V/VL = (1.16)/(6.03 × 10 ) = 19.2
Therefore, at least 20 logs are needed . 51. SSM REASONING The height of the cylinder that is in the oil is given by
hoil = Voil / ( π r 2 ) , where V oil is the volume of oil displaced by the cylinder and r is the
radius of the cylinder. We must, therefore, find the volume of oil displaced by the cylinder.
After the oil is poured in, the buoyant force that acts on the cylinder is equal to the sum of
the weight of the water displaced by the cylinder and the weight of the oil displaced by the
cylinder. Therefore, the magnitude of the buoyant force is given by
F = ρ water gV water + ρ oil gV oil . Since the cylinder floats in the fluid, the net force that acts on
the cylinder must be zero. Therefore, the buoyant force that supports the cylinder must be
equal to the weight of the cylinder, or ρ water gV water + ρ oil gV oil = mg
where m is the mass of the cylinder. Substituting values into the expression above leads to
V water + ( 0.725)Voil = 7 .00 × 10 –3 m 3
From the figure in the text, Vcylinder = Vwater + (1) Voil . Substituting values into the expression for Vcylinder gives
V water + Voil = 8.48 × 10 –3 m 3 (2) Subtracting Equation (1) from E...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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