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376_PartUniversity Physics Solution

# 376_PartUniversity Physics Solution - Equilibrium and...

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Equilibrium and Elasticity 11-35 E VALUATE : Our estimate is based solely on compressive stress; other injuries are likely at a much lower height. 11.91. I DENTIFY and S ET U P : 0 / Y F l A l = Δ (Eq.11.10 holds since the problem states that the stress is proportional to the strain.) Thus 0 / . l F l AY Δ = Use proportionality to see how changing the wire properties affects . l Δ E XECUTE : (a) Change 0 l but F (same floodlamp), A (same diameter wire), and Y (same material) all stay the same. 0 constant, l F l AY Δ = = so 1 2 01 02 l l l l Δ Δ = Δ Δ 2 1 02 01 1 ( / ) 2 2(0.18 mm) 0.36 mm l l l l l Δ = Δ = Δ = = (b) 2 2 1 4 ( / 2) , A d d π π = = so 0 2 1 4 F l l d Y π Δ = , F 0 , l Y all stay the same, so ( ) 2 1 0 4 ( ) / constant l d F l Y π Δ = = 2 2 1 1 2 2 ( ) ( ) l d l d Δ = Δ 2 2 2 1 1 2 ( / ) (0.18 mm)(1/2) 0.045 mm l l d d Δ = Δ = = (c) , F 0 , l A all stay the same so 0 / constant lY F l A Δ = = 1 1 2 2 l Y l Y Δ = Δ 10 10 2 1 1 2 ( / ) (0.18 mm)(20 10 Pa/11 10 Pa) 0.33 mm l l Y Y Δ = Δ = × × = E VALUATE : Greater l means greater , l Δ greater diameter means less , l Δ and smaller Y means greater . l Δ 11.92. I DENTIFY : Apply Eq.(11.13) and calculate V Δ . S ET U P : The pressure increase is / w A , where w is the weight of the bricks and A is the area 2 r π of the piston. E XECUTE : 2 5 2 (1420 kg)(9.80 m/s ) 1.97 10 Pa (0.150 m) p π Δ = = × 0 V p B V Δ Δ = − gives 5 0 8 ( ) (1.97 10 Pa)(250 L) 0.0542 L 9.09 10 Pa p V V B Δ × Δ = − = − = − × E VALUATE : The fractional change in volume is only 0.022%, so this attempt is not worth the effort. 11.93. I DENTIFY and S ET U P : Apply Eqs.(11.8) and (11.15). The tensile stress depends on the component of F ! perpendicular to the plane and the shear stress depends on the component of F ! parallel to the plane. The forces are shown in Figure 11.93a Figure 11.93a (a) E XECUTE : The components of F are shown in Figure 11.93b. The area of the diagonal face is /cos . A θ Figure 11.93b 2 cos tensile stress cos /( /cos ) . ( /cos ) F F F A A A θ θ θ θ = = = (b) sin cos sin2 shear stress sin /( /cos ) ( /cos ) 2 F F F F A A A A θ θ θ θ θ θ = = = = " (using a trig identity). E VALUATE : (c) From the result of part (a) the tensile stress is a maximum for cos 1, θ = so 0 . θ = ° (d) From the result of part (b) the shear stress is a maximum for sin2 1, θ = so for 2 90 θ = ° and thus 45 θ = °

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11-36 Chapter 11 11.94. I DENTIFY : Apply the first and second conditions of equilibrium to the rod. Then apply Eq.(11.10) to relate the compressive force on the rod to its change in length. S ET U P : For copper, 11 1.1 10 Pa Y = × . E XECUTE : (a) Taking torques about the pivot, the tension T in the cable is related to the weight by 0 0 (sin ) 2, so . 2sin mg T l mgl T θ θ = = The horizontal component of the force that the cable exerts on the rod, and hence the horizontal component of the force that the pivot exerts on the rod, is cot 2 mg θ and the stress is cot . 2 mg A θ (b) 0 0 cot . 2 l F mgl l AY AY θ Δ = = l Δ corresponds to a decrease in length. (c) In terms of the density and length, 0 ( ) , m A l ρ = so the stress is 0 ( 2)cot l g ρ θ and the change in length is 2 0 ( 2 )cot l g Y ρ θ .
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376_PartUniversity Physics Solution - Equilibrium and...

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