Homework 9. Problems: 12.31, 12.32, 14.4, 14.21

Transcription

1 Homework 9 Problems: 1.31, 1.3, 14.4, 14.1

2 Problem 1.31 Assume that if the shear stress exceeds about 4 10 N/m steel ruptures. Determine the shearing force necessary (a) to shear a steel bolt 1.00 cm in diameter and (b) to punch a 1 cm diameter hole in a steel plate cm thick. a) -F F The above figure illustrates the formation of a shearing stress in the bolt. The connected pieces of metal exert opposite forces on the bolt. With good approximation we can assume that the shearing stress along the marked line is uniform. From the definition of shearing stress we can find the minimum shearing force that ruptures the bolt. F > df σda σ da σ Pa π ( 0.005m) 31.4kN A σ D π

3 b) D F h Shearing stress is created along the side of the punched disk. Note that the forces exerted on the disk are exerted along the circumference of the disk. The above figure illustrates the total force at its center only. With good approximation we can assume that the shearing stress along the side of the disk is uniform. This approximation simplifies the integration necessary to relate the shearing force with stress F > df σda σ da σ Pa π 0.005m 0.005m 6.kN A σ D π h Comment. Note for comparison that the weight of a car is about 15kN.

4 Problem 1.3 When water freezes, it expands by about 9 %. What would be the pressure increase inside your automobile engine block if the water in it froze? (The bulk modulus of ice is 10 9 N/m ) If water were free to expand it would change its volume by r 9% during the freezing process ( 1+ r) 0 1 where 0 is the original volume of water and 1 is the volume of the ice free to expand. With an assumption that the iced formed in the process cannot escape from the engine block the walls of the block compress the ice to the original volume 0 From the definition of bulk modulus, the change in pressure in a substance ΔP is related to the relative change in the volume (of the ice) Δ Δ P B The rest is algebra. Solving by substitution ( 1+ r) ( 1+ r) 1 0 ΔP B B 0 9 N Pa m r B 1+ r ( Note. It does not matter if the change is compared with the initial or the final volume of the substance r 1 Δ P B Br B Pa ) 1+ r 1 water water

5 Problem 14.4 Estimate the total mass of the Earth s atmosphere. (The radius of the Earth is m, and the atmospheric pressure at the Earth s is Pa.) The atmosphere is relatively thin (when compared with Earth s radius). One can make than a reasonable assumption that over the entire thickness of the atmosphere the (magnitude) of free fall acceleration has the same value. We can therefore estimate the atmosphere s weight first and then determine its mass. From the definition of pressure, the (magnitude of the) force exerted by the atmosphere on a differential of the Earth is related to the area of that (differential) df PdA The weight of the atmosphere (the force exerted on a (flat) with area equal to the area of the Earth) is therefore W PdA P 4πR Hence the mass of the atmosphere is m W g 4πPR g 4π Pa m 9.1 s 5 7 ( m) kg (Pretty heavy!)

6 Problem 14.1 Mercury is poured into a U-tube, as shown in Figure P14.1a. The left arm of the tube has a cross-sectional area A 1 of 10 cm, and the right arm has a cross-sectional area A of 5 cm. One hundred grams of water are then poured into the right arm, as shown in Figure P14.1. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm 3, what distance h does the mercury rise in the left arm? a) Water in the tube forms a cylinder of height H and the base area A. The density of water is uniform. From the definition of density ρ, the mass m of water is related to its volume, which for a cylinder can be expressed in terms of its height H and the area A of the base 1) ρ ρa H m In the above equation, only the height is not given in the problem. Solving we find m 100g ) H 0cm ρa 3 5cm 1g / cm Hg A 1 A H O h 1 h H b) At the level of the mercury-water interface in the right tube, the pressure has a value determined by the water above the considered level (and the atmospheric pressure P 0 ) 3) P P + ρ gh 0 H O

7 The mercury above the considered level also determines this pressure 4) P + ρ g(h h ) P 0 Hg 1 + The displacement of mercury by water has not significantly changed the pressure distribution and therefore we can assume that this process did not affect its volume. Therefore 5) A 1h1 Ah We obtained three equations with three unknowns (P-P 0, h 1, and h ). Solving simultaneously we find the answer h 1 ρ ρh O H A 1 + A Hg 1 1g / cm 13.6g / cm 0cm 10cm 1 + 5cm 0.49cm