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  Chapter 13. ElasticityPhysics, 6 th  Edition Chapter 13. Elasticity   Elastic Properties of Matter 13-1. When a mass of 500 g is hung from a spring, the spring stretches 3 cm. What is the spring constant ! m = 0.500   g#  x = 0.03 m,  F = W = mg $  F = -kx; % &0.50 g'&(.) m*s'0.03 m  F k  x = = ; + 163 *m13-%. What ill e the increase in stretch for the spring of Prolem 13-1 if an additional 500-g mass is hung lo the first !  F = W = mg   $ % &0.500 g'&(.) m*s'163 *m  F  xk  ∆∆ = = # ∆ / + 3.00 cm13-3. he spring constant for a certain spring is found to e 3000 *m. What force is reuired tocompress the spring for a distance of 5 cm  F = kx = &3000 *m'&0.05 m'#  F = 150 13-2.  6-in. spring has a 2-l eight hung from one end, causing the ne length to e 6.5 in. What is the spring constant What is the strain ! ∆ / + 6.5 in. 4 6.0 in. + 0.50 in. $&2 l'#0.5 in.  F k  x = =  k = ).00 l*in.0.50 in.6.00 in.  LStrain L ∆= = # Strain = 0.0)3313-5.  coil spring 1% cm long is used to support a 1.)- g mass producing a strain of 0.10. o far did the spring stretch What is the spring constant # &'&1%.0 cm'&0.10' o  LStrain L L strain L ∆= ∆ = = # ∆  + 1.%0 cm % &1.) g'&(.) m*s'#0.01%0 m  F k  L ∆= =∆  k = 1270 *m172  Chapter 13. ElasticityPhysics, 6 th  Edition13-6. 8or the coil of Prolem 13-5, hat total mass should e hung if an elongation of 2 cm is desired  F = mg = kx; % &1270 *m'&0.02 m'(.)0 m*s kxm g  = = ; m = 6.00 g Young’s Modulus 13-7.  60- g eight is suspended y means of a cale ha9ing a diameter of ( mm. What is the stress ! 8 + mg + &60 g'&(.) m*s % '# 8 + 5)) # : + 0.00( m $ %%-5% &0.00( m'6.36 / 10 m22  D A  π π  = = = # % 5)) 0.00707 m  F Stress A = = # Stress + (.%2 / 10 6  Pa13-).  50-cm length of ire is stretched to a ne length of 50.01 cm. What is the strain ∆  + 50.01 4 50 cm# 0.01 cm50 cm  LStrain L ∆= = # Strain = %.00 / 10 -2 13-(.  1%-m rod recei9es a compressional strain of -0.0002. What is the ne length of the rod# &'&1%.0 m'&-0.0002' o  LStrainLLstrain L ∆= ∆ = = # ∆  + -0.002)0 m  L = L o  + ∆  L = 1%.000 m 4 0.002)0 m ; L + 11.((5 m13-10. ;oung<s modulus for a certain rod is 2 / 10 11  Pa. What strain ill e produced y a tensile stress of 2%0 =pa 611 2%0 / 10Pa# 2 / 10Pa StressStressYStrainStrainY  = = = Strain = 1.05 / 10 -3 175  Chapter 13. ElasticityPhysics, 6 th  Edition13-11.  500- g mass is hung from the end of a %-m length of metal ire 1 mm in diameter. >f the ire stretches y 1.20 cm, hat are the stress and strain What is ;oung<s modulus for this metal !  F = mg   + &500 g'&(.) m*s % '#  F   + 2(00 #  D  + 0.001 m# ∆  L  + 0.012 m $ %%-7% &0.001 m'7.)5 / 10 m22  D A  π π  = = =   -7% 2(00 7.)5 / 10 m  F Stress A = = # Stress + 6.%2 / 10 (  Pa 0 0.012 m%.00 m  LStrain L ∆= = # Strain = 7.00 / 10 -3(-3 6.%2 / 10Pa7 / 10 StressY Strain = = # ; + ).(1 / 10 11  Pa 13-1%.  16 ft steel girder ith a cross-sectional area of 10 in. %  supports a compressional load of %0 tons. What is the decrease in length of the girder ! ; + 30 / 10 6  Pa# 1 ton + %000 l $ 6%% &20,000 l'&16 ft'&1% in*ft'# &30 / 10l*in.'&10 in.'  FLFLYL ALYA −= ∆ = =∆ # ∆  L = - 0.0%56 in.13-13. o much ill a 60 cm length of rass ire, 1.% mm in diameter, elongate hen a 3- g mass is hung from an end ! Y   + )(.6 / 10 (  Pa#  D  + 0.001% m ; L o  + 0.60 m# m  + 3 g $ %%-6% &0.001% m'1.13 / 10 m22  D A  π π  = = =  # 8 + &3 g'&(.) m*s % ' + %(.2 # #  FLY  AL =∆ (-6% &%(.2 '&0.60 m'&)(.6 / 10'&1.13 / 10m'  FL LYA ∆ = = # ∆  L = 1.72 / 10 -2  m176  Chapter 13. ElasticityPhysics, 6 th  Edition?13-12.  ire of cross-section 2 mm %  is stretched 0.1 mm y a certain eight. o far ill a ire of the same material and length stretch if is cross-sectional area is ) mm %  and the same eight is attached 11%%  FLFLY  ALAL = =∆ ∆ #  A 1 ∆  L 1  = A 2 ∆  L 2 %11%%% &2 mm'&0.10 mm'&) mm'  AL L A ∆∆ = = ; ∆  %  + 0.0500 mm13-15.  ire 15 ft long and 0.1 in. %  in cross-section is found to increase its length y 0.01 ft under a tension of %000 l. What is ;oung<s modulus for this ire Can you identify the material % &%000 l'&15 ft'#&0.10 in.'&0.01 ft'  FLY  AL = =∆   Y + 30 / 10 6  l*in. %    , stee  Shear Modulus 13-16.  shearing force of 20,000  is applied to the top of a cue that is 30 cm on a side. Whatis the shearing stress !  A = &0.30 m'&0.30 m' + 0.0( m %  $ % 20,000 0.0( m  F Stress A = = # Stress = 2.22 / 10 5  Pa?13-17. >f the cue in Prolem 13-16 is made of copper, hat ill e the lateral displacement of the upper surface of the cue 5( **2.22 / 10Pa# 2%.3 / 10Pa  FAFAS S  φ φ  = = = # φ  + 1.05 / 10 -5  rad# &0.30 m'& ! !   φ φ  = = = 1.05 / 10 -5  rad'# !   + 3.15 µ m177
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