376_PartUniversity Physics Solution

# 376_PartUniversity Physics Solution - Equilibrium and...

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Equilibrium and Elasticity 11-35 EVALUATE: Our estimate is based solely on compressive stress; other injuries are likely at a much lower height. 11.91. IDENTIFY and SET UP: 0 / YF lAl (Eq.11.10 holds since the problem states that the stress is proportional to the strain.) Thus 0 /. lF lA Y Δ= Use proportionality to see how changing the wire properties affects . l Δ EXECUTE: (a) Change 0 l but F (same floodlamp), A (same diameter wire), and Y (same material) all stay the same. 0 constant, Y Δ == so 12 01 02 ll ΔΔ = 2 1 02 01 1 ( / ) 2 2(0.18 mm) 0.36 mm l l l Δ =Δ= = (b) 22 1 4 (/ 2 ) , A dd ππ so 0 2 1 4 Fl l dY π , F 0 , l Y all stay the same, so ( ) 2 1 0 4 ( ) / constant ld Y = 11 2 2 () Δ 21 1 2 ( / ) (0.18 mm)(1/2) 0.045 mm d d Δ = = (c) , F 0 , l A all stay the same so 0 / constant lY F l A = lY Δ 10 10 1 2 ( / ) (0.18 mm)(20 10 Pa/11 10 Pa) 0.33 mm Y Y Δ = × × = EVALUATE: Greater l means greater , l Δ greater diameter means less , l Δ and smaller Y means greater . l Δ 11.92. IDENTIFY: Apply Eq.(11.13) and calculate V Δ . SET UP: The pressure increase is / wA , where w is the weight of the bricks and A is the area 2 r of the piston. EXECUTE: 2 5 2 (1420 kg)(9.80 m/s ) 1.97 10 Pa (0.150 m) p = × 0 V pB V Δ Δ=− gives 5 0 8 ( ) (1.97 10 Pa)(250 L) 0.0542 L 9.09 10 Pa pV V B Δ× = = × EVALUATE: The fractional change in volume is only 0.022%, so this attempt is not worth the effort. 11.93. IDENTIFY and SET UP: 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) EXECUTE: 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 ) . c o s) FF FA AA θθ = (b) sin cos sin2 shear stress sin /( /cos ) c o 2 F A = = " (using a trig identity). EVALUATE: (c) From the result of part (a) the tensile stress is a maximum for cos 1, = so 0 . (d) From the result of part (b) the shear stress is a maximum for sin2 = so for 2 90 and thus 45 = °

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11-36 Chapter 11 11.94. IDENTIFY: 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. SET UP: For copper, 11 1.1 10 Pa Y . EXECUTE: (a) Taking torques about the pivot, the tension T in the cable is related to the weight by 00 (sin ) 2, so . 2sin mg Tl m g l T 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 and the stress is cot . 2 mg A (b) cot . 2 lF mg l l AY AY Δ= = l Δ corresponds to a decrease in length. (c) In terms of the density and length, 0 () , mA l ρ = so the stress is 0 (2 ) c o t lg and the change in length is 2 0 ) c o t lg Y . (d) Using the numerical values, the stress is 5 1.4 10 × Pa and the change in length is 6 2.2 10 m. × (e) The stress is proportional to the length and the change in length is proportional to the square of the length, and so the quantities change by factors of 2 and 4.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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376_PartUniversity Physics Solution - Equilibrium and...

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