325 - 12-07 - elasticity - 43-51: in which we solve problems 43-51. d
(1.1)
Chapter 12, Problem 44: The figure shows the stress-strain curve for a material. material. The scale of the 6 2 stress axis is set by s = 300, in units of 10 N/m . What are (a) the Young's modulus and (b) the approximate yield strength for this material?
(a) The Young’s modulus is given by E
stress strain
slope of the stress-strain curve
150 106 N/m 2 0.002 8
7.5 1010N/m 2.
2
(b) Since the linear range of the curve extends to about 2.9 × 10 N/m , this is approximately the yield strength for the material. Chapter 12, problem 45: In Fig. 12-54, a lead brick brick rests horizontally on cylinders cylinders A and B. The areas of the top top faces of of the cylinders are related by A A= 2A B; the Young's moduli of the cylinders are related by E A A= 2 E B. The cylinders had identical lengths before the brick was placed on them. What fraction of the brick's mass is supported (a) by cylinder A and (b) by cylinder B? The horizontal distances between the center of mass of the brick and the centerlines of the cylinders are d A for cylinder A and d B B for cylinder B. (c) What is the ratio d A A /d B?
(a) Since the brick is now horizontal and the cylinders were initially the same length , then both have been compressed an equal amount . Thus,
E
F A
F A A A EA
FB AB EB
FA FB
AA EA AB EB
(2 AB )(2 EB ) AB EB
4; F A FB W
F A W
4 5
0.80; (1.2)
(b) This also leads to the result F result F B/W = 1/5 = 0.20. (c) Computing torques about the center of mass, we find F Ad A = F Bd B, which leads to, A
B FA d A sin 90 FB d B si s in 90
d A d B
F B F A
1 4
0.25
(1.3)
Chapter 12, Problem 46 (Flying Circus of Physics): The figure shows an approximate plot p lot of stress versus strain for a spider-web thread, out to the point of breaking at a strain of 2.00. The vertical axis scale is set set by a 2 2 2 = 0.12 GN/m ,b = 0.30 GN/m , and c = 0.80 GN/m . Assume that the thread has an initial length of 0.80 cm, an
-12
2
initial cross-sectional area of 8.0 × 10 m , and (during stretching) a constant volume. The strain on the thread is the ratio of the change in the thread's length to that initial length, and the stress on the thread is the ratio of the collision force to that initial cross-sectional area. Assume also that w hen the single thread snares a flying insect, the insect's kinetic energy is transferred to the stretching of the thread. (a) How much kinetic energy (in μJ) would put the thread on the verge of breaking? Would (b) the fruit fly of mass 6.00 mg and speed 1.70 m/s and (c) the bumble bee of mass 0.388 g and speed 0.420 m/s break the thread? (If break then type 1,
else type 0). Since the force is (stress × area) and the displacement is (strain × length), we can write the work integral (eq. 7-
32) as W = Fdx =
(stress) A (differential strain) L
= AL (stress) (differential strain)
which means the work is (thread cross-sectional area) × (thread length) × (graph area under curve). The area under the curve is 1 1 1 1 graph area as1 (a b)( s2 s1 ) (b c )( s3 s2 ) as2 b( s3 s1 ) c( s3 s 2 ) 2 2 2 2 1 (0.12 109 N/m 2 )(1.4) (0.30 109 N/m 2 )(1.0) (0.80 10 9 N/m 2)(0.60) 2
4.74 108 N/m 2. (a) The kinetic energy that would put the thread on the verge of breaking is simply equal to W :
K W AL(graph area) (8.0 1012 m2 )(8.0 103 m)(4.74 108 N/m2 )
3.03 105 J. (b) The kinetic energy of the fruit fly of mass 6.00 mg and speed 1.70 m/s is 1 1 K f m f v2f (6.00 10 6 kg)(1.70 m/s) 2 8.67 10 6 J. 2 2 (c) Since K f W , the fruit fly will not be able to break the thread. (d) The kinetic energy of a bumble bee of mass 0.388 g and speed 0.420 m/s is 1 1 Kb mb vb2 (3.99 104 kg)0.420 m/s) 2 3.42 10 5 J. 2 2 (e) On the other hand, since Kb W , the bumble bee will be able to break the thread.
Chapter 12, problem 48: Figure 12-57 shows the stress versus strain plot for an aluminum wire that is stretched by a machine pulling in opposite directions at the two ends of the wire. The scale of the stress axis is set by s = 7.00, in units of 7 2 -6 2 10 N/m . The wire has an initial length of 0.800 m and an initial cross-sectional area of 2.00 × 10 m . How much work -3 does the force from the machine do on the wire to produce a strain of 1.30 × 10 ?
Since the force is (stress × area) and the displacement is (strain × length), we can write,
W dW F dx
A
dx
f
A
( ) d
0
2.00 10 6 m2 0.800 m
1 A max 2
1.00 10 7.00 7 10 2
1
3
7
N
(1.4)
0.0560 J ;
m2
chapter 12, problem 49: In Fig. 12-58, a 103 kg uniform log hangs by two steel wires, A and B, both of radius 1.20 mm. Initially, wire A was 2.50 m long and 2.00 mm shorter than wire B,
writing these dimensions,
r A rB r 1.20 103 m; LA 2.50 m L f , A LA ; LB LA L f , B LB ;
2.00 103 m; (1.5)
The log is now horizontal. This means the final lengths are the same, L f , A L f , B L;
(1.6)
What are the magnitudes of the forces on it from (a) wire A and (b) wire B? (c) What is the ratio d A /d B?
(a) Let F A and F B be the forces exerted by the wires on the log and let m be the mass of the log. Since the log is in equilibrium,
F
y
0 FA FB mg ;
(1.7)
Noting that wire A originally had a length L A and stretches by L A , in which we use (1.6) and A
If
is
E A
F A A A
E
LA Lf ,A
LA
FA L A A E
;
B
E B LB
FB L B A E
E ,
; A r 2 ;
(1.8)
the amount by which B was originally longer than A then, since they have the same length after the log
is attached, L A LB . Using (1.8) in this statement, and subsequently solving for F B ,
L A LB
F A LA AE
FB LB AE
solve for F F B B
FAL A L B
AE
;
LB
(1.9)
Putting (1.9) into the force-balance (1.7), and subsequently solving for F A ,
F A LA AE L LB B
F A
mgLB AE solve for F ; mg 0 F A L L A B
(1.10)
A
Putting numbers into (1.10), in which we use A r , 2
F A
(103kg)(9.8m/s2 )(2.50m+2.00 103 m) (4.52 106 m2 )(200 109 N/m2 )(2.0 103 m) 2.50m 2.50m+2.00 103 m
866 N ; (1.11)
(b) From the condition F A + F B – mg = 0, we obtain F B mg F A (103kg) (9.8m/s2 ) 866 N 143 N.
(c) The net torque must also vanish. We place the origin on the surface of the log at a point directly above the center of mass. The force of gravity does not exert a torque about this point. Then, the torque equation becomes, d F 143N solve for d A / d B 0 F A d A – FB d B A B 0.165 ; (1.12) d B F A 866N
Chapter 12, problem 50: An insect is caught at the midpoint of a spider-web thread. The thread breaks under a stress of 8.20×108 N/m2 and a strain of 2.00. Initially, it was horizontal and had a length of 2.00 cm and a cross-sectional area of 8.00 × 10 -12 m2. As the thread was stretched under the weight of the insect, its volume remained constant. If the weight of the insect puts the thread on the verge of breaking, what is the insect's mass? (A spider's web is built to break if a potentially harmful insect, such as a bumble bee, becomes snared in the web.)
On the verge of breaking, the length of the thread is L L0 L L0 (1 L / L0 ) L0 (1 2)
3 L0 , where
L0 0.020 m is the original length, and strain L / L0 2 , as given in the problem. The free-body diagram of the system is shown below. The condition for equilibrium is mg 2T sin , where m is the mass of the insect and
T A(stress) . Since the volume of the thread remains constant as it is being stretched, we have V A0 L0 AL , or
A A0 ( L0 / L) A0 / 3
. The vertical distance y is
y ( L / 2) ( L0 / 2) 2
2
9 L20 4
L20 4
2 L0 .
Thus, the mass of the insect is
m
2T sin g
2( A0 / 3)(stress) sin g
2 A0 (stress)
y
3g
3L0 / 2
4 2 A0 (stress) 9 g
4 2(8.00 10 12 m 2 )(8.20 10 8 N/m 2) 9(9.8 m/s 2 )
4.2110 4 kg or 0.421 g.
Chapter 12, Problem 51 SN: The figure is an overhead view of a rigid rod that turns about a vertical axle until the identical rubber stoppers A and B are forced against rigid walls at distances 7.0 cm = rA and 4.0 cm = rB from the axle. Initially the stoppers touch the walls without being compressed. Then a force of magnitude 220 N = F is applied perpendicular to the rod at a distance 5.0 cm = R from the axle. Find the magnitude of the force compressing (a) stopper A and (b) stopper B, in terms of the variables given.
Let the forces that compress stoppers A and B be F A and F B, respectively. Then equilibrium of torques about the axle requires FR = r A F A + r B F B. If the stoppers are compressed by amounts | y A| and | y B|, respectively, when the rod rotates a (presumably small) angle (in radians), then | y A | rA
and | yB | r B . Furthermore, if
their “spring constants” k are identical, then k = |F / y| leads to the condition F A/r A = F B/r B, which provides us with enough information to solve. (a) Simultaneous solution of the two conditions leads to Rr (5.0 cm)(7.0 cm) (220 N) 118 N F A 2 A 2 F r A r B (7.0 cm)2 +(4.0 cm) 2 (b) It also yields F B
Wileyplus answer
Rr B r A2 r B2
F
(5.0 cm)(4.0 cm) (7.0 cm) 2 +(4.0 cm) 2
(220 N) 68 N.
1.2 10
2
N.
