# Physics 107 HOMEWORK ASSIGNMENT #9

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1 Physics 07 HOMEORK ASSIGNMENT #9 Cutnell & Johnson, 7 th edition Chapter : Problems 6, 8, 33, 40, 44 *6 A 58-kg skier is going down a slope oriented 35 above the horizontal. The area of each ski in contact with the snow is 0.3 m. Determine the pressure that each ski exerts on the snow. *8 Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. ater and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished. 33 Interactive Solution.33 presents a model for solving this problem. Multiple-Concept Example 8 also presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of 830 kg/m 3. The weight of the input piston is negligible. The radii of the input piston and output plunger are and 0.5 m, respectively. hat input force F is needed to support the N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is.30 m above that of the input piston? 40 The density of ice is 97 kg/m 3, and the density of sea water is 05 kg/m 3. A swimming polar bear climbs onto a piece of floating ice that has a volume of 5. m 3. hat is the weight of the heaviest bear that the ice can support without sinking completely beneath the water? 44 An object is solid throughout. hen the object is completely submerged in ethyl alcohol, its apparent weight is 5. N. hen completely submerged in water, its apparent weight is 3.7 N. hat is the volume of the object?

2 6. REASONING Pressure is defined in Equation.3 as the magnitude of the force acting perpendicular to a surface divided by the area over which the force acts. The force acting perpendicular to the slope is due to the component of the skier s weight that is directed perpendicular to the slope. In the drawing at the right, this component is labeled perpendicular. Note that the fact that the skier is moving is of no importance. perpendicular SOLUTION perpendicular e assume that each ski bears the same amount of force, namely (see the drawing). According to Equation.3, the pressure that each ski applies to the snow is P perpendicular A where A is the area of each ski in contact with the snow. From the drawing, we see that perpendicular cos 35, so that the pressure exerted by each ski on the snow is 58 kg 9.80 m/s cos 35 perpendicular cos35 3 P.8 0 Pa A A 0.3 m where we have used the fact that mg (Equation 4.5). 8. REASONING AND SOLUTION The mercury, being more dense, will flow from the right container into the left container until the pressure is equalized. Then the pressure at the bottom of the left container will be P ρ w gh w + ρ m gh ml and the pressure at the bottom of the right container will be P ρ m gh mr. Equating gives Both liquids are incompressible and immiscible so ρ w gh w + ρ m g(h ml h mr ) 0 () h w.00 m and h ml + h mr.00 m Using these in () and solving for h ml gives, h ml (/)(.00 ρ w /ρ m ) 0.46 m. So the fluid level in the left container is.00 m m.46 m from the bottom.

3 33. REASONING e label the input piston as and the output plunger as. hen the bottom surfaces of the input piston and output plunger are at the same level, Equation.5, F F A / A, applies. However, this equation is not applicable when the bottom surface of the output plunger is h.50 m above the input piston. In this case we must use Equation.4, P P + ρgh, to account for the difference in heights. In either case, we will see that the input force is less than the combined weight of the output plunger and car. SOLUTION a. Using A π r for the circular areas of the piston and plunger, the input force required to support the N weight is 3 ( m) A π F F ( N) 93.0 N A π ( 0.5 m) (.5) b. The pressure P at the input piston is related to the pressure P at the bottom of the output plunger by Equation.4, P P + ρgh, where h is the difference in heights. Setting ( P ) F / A F / π r, ( P ) F / π r, and solving for F, we have F F gh r π r + ρ π r ( N) π π ( π ) 3 ( m) ( 0.5 m) 3 3 π kg/m 9.80 m/s.30 m m 94.9 N (.4) 40. REASONING The ice with the bear on it is floating, so that the upward-acting buoyant force balances the downward-acting weight ice of the ice and weight bear of the bear. The magnitude F B of the buoyant force is the weight of the displaced water, according to Archimedes principle. Thus, we have F +, the expression with which B ice bear we will obtain bear. e can express each of the weights and ice as mass times the magnitude of the acceleration due to gravity (Equation 4.5) and then relate the mass to the density and the displaced volume by using Equation..

4 SOLUTION Since the ice with the bear on it is floating, the upward-acting buoyant force F B balances the downward-acting weight ice of the ice and the weight bear of the bear. The buoyant force has a magnitude that equals the weight stated by Archimedes principle. Thus, we have B ice bear bear ice of the displaced water, as F + or () In Equation (), we can use Equation 4.5 to express the weights and ice as mass m times the magnitude g of the acceleration due to gravity. Then, the each mass can be expressed as m ρv (Equation.). ith these substitutions, Equation () becomes ( ρ ) m g m g ( ρ V ) g V g () bear ice ice ice hen the heaviest possible bear is on the ice, the ice is just below the water surface and displaces a volume of water that is V V. Substituting this result into Equation (), we find that ( ρ V ) g ρ V g ( ρ ρ ) V g bear ice ice ice ice ice kg/m 97 kg/m 5. m 9.80 m/s 5500 N ice 44. REASONING hen an object is completely submerged within a fluid, its apparent weight in the fluid is equal to its true weight mg minus the upward-acting buoyant force. According to Archimedes principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object. The weight of the displaced fluid depends on the volume of the object. e will apply this principle twice, once for the object submerged in each fluid, to find the volume of the object. SOLUTION The apparent weights of the object in ethyl alcohol and in water are: Ethyl alcohol 5. N mg ρ gv alcohol eight in True Magnitude of alcohol weight buoyant force () ater 3.7 N mg ρ gv water eight True Magnitude of in water weight buoyant force () These equations contain two unknowns, the volume V of the object and its mass m. By subtracting Equation () from Equation (), we can eliminate the mass algebraically. The result is

5 5. N 3.7 N gv ρ ρ water alcohol Solving this equation for the volume, and using the densities from Table., we have V 5. N 3.7 N.5 N m g ρ ( water ρalcohol ) ( 9.80 m/s )(.00 0 kg/m 806 kg/m ) 4 3

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