ch12-mechanical properties of matter

7/30/2019 Ch12-Mechanical Properties of Matter

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CHAPTER 12 Mechanical Properties of Matter

Question 12.1

The Youngs modulus of a wire is 1.2 1011 N m2

(a) Calculate the stress needed to increase the length of the wire by 0.1%.(b) What is the force required to produce this extension, if the cross-section area of the

wire is 2.0 mm2.

Solution 12.1

stress(a) Youngs modulus E =

strain

Stress = E strain

but strain =

if 100% = 0.1%

then = 0.001

Hence stress = (1.2 1011)(0.001)= 1.2 108 N m2

F(b) Stress =

A

F = (stress)(cross-section area)= (1.2 108)(2.0 106)= 2.4 102 N

Question 12.2

A 8.0 m steel wire is used to support a 10 kg load. If the Youngs modulus of steel is2.0 1011 N m2 and the extension in the wire is 5.0 mm, find the cross-section area of thewire.

Solution 12.2

F/A F E = or E =

e / Ae

F 10 9.81 8.0 A = =

Ee 2.0 1011 5.0 103

= 7.85 106 m2


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Question 12.3

A cylindrical steel bar of length 1.50 m and diameter 4.0 mm is supporting a 36 kN load.Given that the Youngs modulus of steel is 2.0 1011 N m2 find

(a) the tensile stress in the bar, and(b) the original length of the bar.Assume that the limit of proportionality is not exceeded.

Solution 12.3

force(a) Stress =

cross-section aread2

Cross-section area of bar = 4

(4.0 103)2=

4

= 1.26 105 m2

36

103

Tensile stress = N m21.26 105

= 2.86 109 N m2

(b) Let the original length of the bar = L Compression of the bar = L 1.50

F LUsing Youngs modulus =

A e

and e = L 1.5

L 2.0 1011 = (2.86 109)()

L 1.50

2.0 1011 (L 1.50) = (2.86 109)L L = 1.52 m

Question 12.4

A steel wire of length 1.5 m and radius 1.0 mm is joined end-to-end with another copperwire of the same length and cross-section area. A force of 200 N is applied to the wire.Calculate(a) the stress on each wire,

(b) the extension of each wire and(c) the total extension of the wires, and the total force required to produce totalextension of 4.0 mm.

(Youngs modulus of steel Es

= 2.0 1011 N m2 and Youngs modulus of copperE

c= 1.2 1011 N m2)

Solution 12.4

(a) Force on each wire is the same as the total force applied to the wires.

200 Stress = N m2

(1.0 103)2

= 6.4 107 N m2


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(1.5)(b) For the steel wire 2.0 1011 = 6.4 107

es

e = 4.8 104 m

1.5For the copper wire 1.2 1011 = 6.4 107 ()

ec

ec

= 8.0 104 m

(c) The total extension = es

+ ec

= (4.8 104 + 8.0 104) m

= 12.8 104 m

= 1.28 103 m

4.8(d) Fraction of e

sto total extension = = 0.375

12.8

Thus if the total extension is 3.00 mm = 3.0 103 me

s, the extension in the steel wire = 0.375 4.0 103 m

es

= 1.5 103 m

F (1.50) 2.0 1011 =

(1.0 103)2 1.5 103

F = 628 N

Question 12.5

A 40 kg load is hung at the lower end of a steel wire of length 0.80 m and radius 1.0 mm.Assuming that the proportionality limit is not exceeded, calculate

(a) the stress,(b) the extension and(c) the energy stored per unit volume in the wire.

Youngs modulus of steel is 2.0 1011 N m2.

Solution 12.5

F mg(a) Stress = =

A r2

(40)(9.81)= N m2

(1.0 103)

= 1.25 108 N m2

F L(b) E =

A eF L 0.80

e = = 1.25 108 () mA E 2.0 1011

= 5.0 103 m

1(c) Energy stored per unit volume =

21 5.0 103

= (1.25 108)()2 0.8

= 3.9 105 J m3


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0

0.2

0.4

0.6

2 4 6

F/ N

X/cm

Question 12.7

The graph below shows the variation of the forces as a function of compression X for aspring used to launch an object on a smooth horizontal surface.

The mass, m of object is 5.0 g and the spring is compressed 6.0 cm before it is released.Calculate(a) the force constant Kof the spring,(b) the elastic energy stored in the spring before it is released and(c) the velocity of the object after it leaves this spring.

Solution 12.7

(a) From Hookes law, F = KX

FK = = gradient of the graph

X0.6 N

= 6.0 102 m

= 10 N m1

1(b) Elastic energy stored = KX2

2

1= (10)(6.0 102)22

= 1.8 102 J

Question 12.6

In an experiment to measure the force constant of a spring, a 20.0 g marble is placed on acompressed spring. When the marble is released, the marble is launched vertically

upwards. It is observed that when the spring is compressed 9.0 cm, the maximum heightreached by the marble is 15.0 m. Calculate(a) the change in the gravitational potential energy of the marble during the ascend and

(b) the force constant of the spring.

Solution 12.6

(a) Change in gravitational P.E. = mgh= (20.0 103)(9.81)(15.0)= 2.94 J

(b) From initial elastic P.E. stored in spring = change in gravitational P.E. of marble

1Ke2 = mgh2

1(K)(9.0 102)2 = 2.942

K = 726 N m2


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Question 12.8

A light steel wire 1.50 m long and of cross-section area of 2.5 mm 2 is attachedhorizontally to two points 1.50 m apart. A load attached to the mid-point of the wirecauses it to depress by 6.0 102 m. Calculate(a) the extension of the wire,(b) the tension in the wire,(c) the elastic potential energy stored in the wire when the load is attached and(d) the weight of the load.

(Youngs modulus of steel = 2.0 1011 N m2)

Solution 12.8

(a) Original length of wire = 1.50 mExtended length of wire = AM+ MBUsing Pythagoras Theorem

MB = 0.752 + (6 102)2

= 0.7524 m extended length of wire = 2(0.7524) m

= 1.5048 mbecause AM= MB = 0.7524 m

extension of wire = (1.5048 1.50) m= 0.0048 m= 4.8 103 m

F (b) From E =

A e

EAeF =

(2.0 1011)(2.5 106)(4.8 103)=

1.50

= 1.6 103 N

1(c) Elastic pot. energy stored in the wire = Fe

21

= (1.6 103)(4.8 103) J2

= 3.84 J

(d) mg = 2Fcos

6.0 102= 2(1.6 103)

0.7524

= 2.6 102 N

mg

M

A B1.50 m

(c) Kinetic energy gained by object = elastic P.E. in the spring

1 mv2 = 1.8 1022

1 (5.0 103)v2 = 1.8 1022

v = 7.2 m s1

mg

Fsin Fsin

2Fcos

ch12-mechanical properties of matter

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