Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 10, Problem 91P

(a)

To determine

The maximum stress on a single strand and also the angle between the web and the horizontal.

(a)

Expert Solution
Check Mark

Answer to Problem 91P

The angle between the web and the horizontal is 42.2°.

Explanation of Solution

The Young’s modulus of the spider silk is 4.0GPa and the maximum stress is 1.4 GPa.

Write the expression for maximum stress

FA=YΔLL                                                                                       (I)

Here, FA is the maximum stress, Y is the Young’s modulus, ΔL is the change in length of the strand and L is the original length of the strand.

Rearranging the above expression

ΔLL=F/AY                                                                                    (II)

Substitute 4.0GPa for Y and 1.4 GPa for F/A  in (II) to find ΔLL

ΔLL=1.4GPa4.0GPa=0.35                                                                            (III)

The original length of the strand is the radius of the strand. Thus, ΔLr=0.35

Here, r is the radius of the web

Write the expression for angle between the web and the horizontal (Refer figure 1)

Physics, Chapter 10, Problem 91P , additional homework tip  1

cosθ=rr+ΔL                                                                           (IV)

Here, θ is the angle between the web and the horizontal.

Multiplying both numerator and denominator by r and taking cos-1 on both sides

θ=cos1(11+ΔLr)                                                                          (V)

Substitute 0.35 for ΔLr in the above equation (V) to find θ

θ=cos1(11+0.35)=42.2°

Thus, the angle between the web and the horizontal is 42.2°.

(b)

To determine

The mass of the bug given the maximum stress.

(b)

Expert Solution
Check Mark

Answer to Problem 91P

The mass of the bug is 48g.

Explanation of Solution

The cross sectional area of each web strand is 1.0×1011 m2

Physics, Chapter 10, Problem 91P , additional homework tip  2

Write the expression for total force in vertical y-direction at equilibrium position. (Refer figure 2)

Fy=0Tsinθmg=0     

Here, T is the tension on the strand, m is the mass of the bug and g is the acceleration due to gravity.

Rearranging the above equation for T

T=mgsinθ                                                                            (VII)

Dividing (VII) by A

TA=mgAsinθ                                                                     (VIII)

Here, A is the cross sectional area of all the strands.

The tensile force per unit area is the tensile stress i.e. TA=FA

Rearranging (VIII) for m

m=FA(Asinθ)g                                                   (IX)

Substitute 1.4 GPa for F/A, 50×1.0×1011 m2 for A, 9.8ms2 for g and 42.2° for θ in (IX) to find m

m=1.4 GPa(50×1.0×1011 m2)sin(42.2°)9.8ms2=1.4×109 Nm2(50×1.0×1011 m2)sin(42.2°)9.8ms2=48g

Thus, the mass of the bug is 48g.

(c)

To determine

The extension of the web given the radius of the web.

(c)

Expert Solution
Check Mark

Answer to Problem 91P

The depth of the web is 9.1cm.

Explanation of Solution

The radius of the web is 0.10m.

Refer figure 1 and write the expression for the depth of the web

d=(r+ΔL)sinθ                                                   (X)

Here, d is the depth of the web.

Rearrange (X)

d=r(1+ΔLr)sinθ                                                   (XI)

Substitute 0.10m for r, 42.2° for θ and 0.35 for ΔLr in (XI) to find d

d=(0.10m)(1+0.35)×sin(42.2°)=9.1×102m=9.1cm

Thus, the depth of the web is 9.1cm.

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