Physics 2001 Problem Set 9 Solutions

Transcription

1 Physics 2001 Problem Set 9 Solutions Jeff Kissel December 4, A cube of concrete has a side of length l = m. Within the volume of the cube, there are two spherical cavities, each with radius r = m. What is the weight of the block? The density ρ of concrete is 2200 kg m 3. To find the weight, we ll use the typical definition, W mg (1) However, we re given the density instead of the mass. We therefore need to manipulate the equation of density for mass, ρ m V m = ρv (2) So Eq. 1 becomes W = ρv g (3) Finally, we need the volume. Because of the two holes inside the cube, the volume will not just be the length of a side cubed, but the volumes of the two spheres subtracted from it, V = V cube 2V sphere 1

2 ( ) 4 = l πr3 V = l πr3 (4) Plugging this in, we get the weight of the Swiss concrete cube, W = ρg (l 3 83 ) πr3 (5) = (2200 kg m 3 )(9.8 m s 2 ) ((0.15 m) 3 83 ) π(0.025 m)3 W = N (6) 2. Suppose you pour water into a container until it reaches a depth of d w = 12 cm. Then, you carefully pour in d o = 7.2 cm of olive oil on top of the water. What is the pressure at the bottom of the container? The densities of water ρ w and oil ρ o are 1000 kg m 3 and 920 kg m 3, respectively. The definition of pressure is P F A (7) which have the many equivalent units [P] [Pa] ( Pascals ) = [N m 2 ] = [J m 3 ] = [kg m 1 s 2 ] If we rearrange the above equation for F, then we can solve this problem via a process we already know how do: summation of forces. The forces in 2

3 this problem (as shown in the figure above), are the weight of the oil, the weight of the water, the force due to atmospheric pressure, and the force from the pressure on the bottom of the beaker for which we ll solve. Note, the question does not explicitly state that one must include the effects of atmospheric pressure (P A = 1 atm Pa), so if you leave it out, it s OK. Again, we re given densities instead of masses, so we ll use Eq. 2. However, we re also not given the volume, so we must extrapolate that each liquid will fill some cross-sectional area, A, which we can multiply by the heights given. This will give us the cylindrical volume filled by each liquid. (It s not given, so as we ve found doing many physics problems before this, it ll cancel.) H ok, let s go. The system is in equilibrium, so the summation of forces in the ŷ direction is 0 = P B A P A A m o g m w g (m = ρv = ρah) = P B A P A A ρ o Ah o g ρ w Ah w g P B A = P A A + ρ o Ah o g + ρ w Ah w g ( told ya!) P B = P A + ρ o h o g + ρ w h w g (8) = ( Pa) + (920 kg m 3 )(0.072 m)(9.8 m s 2 ) + (1000 kg m 3 )(0.12 m)(9.8 m s 2 ) P B = Pa = atm (9) If you didn t include atmospheric pressure, the result is considerably less, P B = ρ o h o g + ρ w h w g (10) = (920 kg m 3 )(0.072 m)(9.8 m s 2 ) + (1000 kg m 3 )(0.12 m)(9.8 m s 2 ) P B = Pa = atm (11) 3. A sphere of cork (less dense than water) is tied to one end of a string and pulled halfway under the surface of fresh water in a beaker. The other end of the string is tied to the bottom of the beaker (see figure below). The sphere of cork has a radius r = 0.15 m. What is the tension T in the string? (ρ cork = 240 kg m 3 ). As with the last problem, if we sum the forces on the cork, we can obtain the tension because the system is in equilibrium. So, 0 = F B T W c T = F B W c (12) 3

4 But wait! We ve got a new force here: the buoyancy force, F B. Archimedes principle defines this force as F Buoyancy = W fluid (13) in which W fluid is the would-be weight of the amount of fluid displaced by the volume of a given object placed in the fluid. Thus, for the water displaced in this problem (by exactly half the volume the sphere of cork), this weight is F B = m w g = ρ w V w g = ρ w g ( 1 2 [ ]) 4 3 πr3 c F B = 2 3 πρ wgr 3 c (14) Also we need the weight of the cork, W c in terms of the information given; W c Plugging Eq. 14 and 15 into Eq. 12, T = F B W c = 2 3 πρ wgr 3 c 4 3 πρ cgr 3 c = m c g = ρ c V c g W c = 4 3 πρ cgr 3 c (15) T = 2 3 πgr3 c(ρ w 2ρ c ) (16) = 2 3 π(9.8 m s 2 )(0.15 m) 3 ( (1000 kg m 3 ) 2(240 kg m 3 ) ) T = N (17) 4

5 4. Quiz: A solid hemispherical piece of silver with a 1.0 m rests on a horizontal surface. What is the pressure (in Pa) beneath the piece? (Ignore the pressure due to the atmosphere.) ρ silver = 10, 500 kg m 3. As with the other problems in this set, we ll use the summation of forces on the sphere to get the pressure. PA 0 = PA W s = W s P = W s A (18) For the weight of the sphere, we ll again use m = ρ s V, where here the volume will be that of a half-sphere. The area, A in this case is that touching the surface. This will be the area of a circle, which is the shape of any cross-section of a sphere. This sphere happens to be split exactly in half, so the radius r will be the same as that of the sphere. So, P = ρ1 2 V sphere A circle πr 3 = ρ π r 2 P = 2 3 ρrg = 2 3 (10, 500 kg m3 )(1 m)(9.8m s 2 ) P = Pa (19) 5