Figure 12-55 shows an approximate plot of stress versus strain for a spider-web

Question

Figure 12-55 shows an approximate plot 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 by values a = 0.12 GN/m2, b = 0.30 GN/m2, and c = 0.80 GN/m2. Assume that the thread has an initial length of 0.80 cm, an initial cross-sectional area of 8.0 × 10-12 m2, and (during stretching) a constant volume. Assume also that when 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 would put the thread on the verge of breaking? What is the kinetic energy of (b) a fruit fly of mass 6.00 mg and speed 1.70 m/s and (c) a bumble bee of mass 0.388 g and speed 0.420 m/s? Would (d) the fruit fly and (e) the bumble bee break the thread?

Explanation / Answer

a)

K = integration{F d}

The force is equal to stress multipled by area or "F = P A"

displacement is equal to strain multipled by length of the thread: "d = S L"

S: strain

P: stress

L: length of the thread

A: cross-sectional area of thread

Therefore:

K = integration{F d} = integration{(P A) (S L)} = A * L * integration{P S}

integration{P S} can be found from the area under the stress-strain diagram:

integration{P S} = (1*a)/2 + (1.4-1)*(b+a)/2 + (2-1.4)*(c+b)/2

integration{P S} = ((1*0.12e9)/2 + (1.4-1)*(0.30e9+0.12e9)/2 + (2-1.4)*(0.80e9+0.30e9)/2)

integration{P S} = 474000000

==> K = 474000000* A * L = 474000000 *(8e-12)*(0.80e-2) = 0.000030336 J

==> K = 3.03 * 10^-5 J

b)

K = 0.5 m v^2 = 0.5 * 6e-6 * 1.7*1.7 = 8.67 * 10^-6 J

(c) a bumble bee of mass 0.388 g and speed 0.420 m/s? Would

K = 0.5 m v^2 = 0.5 * 0.388 * 0.420*0.420 = 0.0342 J

(d) the fruit fly and

No

(e) the bumble bee break the thread

Yes

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