Spider silk has a Young's modulus of $4.0 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}\( and can withstand stresses up to \)1.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} . \mathrm{A}$ single webstrand has across-sectional area of \(1.0 \times 10^{-11} \mathrm{m}^{2}\) and a web is made up of 50 radial strands. A bug lands in the center of a horizontal web so that the web stretches downward. (a) If the maximum stress is exerted on each strand, what angle \(\theta\) does the web make with the horizontal? (b) What does the mass of a bug have to be in order to exert this maximum stress on the web? (c) If the web is \(0.10 \mathrm{m}\) in radius, how far down does the web extend? (IMAGE NOT COPY)

To calculate the force exerted on each strand, we can use the formula for stress. Stress = Force / Area Given the stress is \(1.4 \times 10^9 \mathrm{N} / \mathrm{m}^2\) and the cross-sectional area is \(1.0 \times 10^{-11} \mathrm{m}^2\), we can find the force exerted on a single strand. Force = Stress × Area Force = \(1.4 \times 10^9 \mathrm{N} / \mathrm{m}^2\) × \(1.0 \times 10^{-11} \mathrm{m}^2\) Force = \(1.4 \times 10^{-2} \mathrm{N}\)

Since there are 50 radial strands in the web, we can find the total force on the web. Total Force = Force on each strand × Number of strands Total Force = \(1.4 \times 10^{-2} \mathrm{N}\) × 50 Total Force = 0.7 N

Considering a right triangle formed by one horizontal radial strand, one vertical strand, and the diagonal strand connecting them, we can use the property of tension forces in the web to solve for θ. The total tension force (0.7 N) forms an angle θ with the radial tension force, whereas the vertical tension force is perpendicular to the radial strands. Thus, radial tension = total tension × cos(θ), and vertical tension = total tension × sin(θ). The number of radial strands is 50, so the total radial force is 50 times the radial tension. Then, we have: Total radial force = total vertical force 50 × total tension × cos(θ) = total tension × sin(θ) 50 × cos(θ) = sin(θ) Divide by cos(θ) on both sides and use the identity sin(θ)/cos(θ) = tan(θ) 50 = tan(θ) Now, we find the angle θ: θ = arctan(50) θ ≈ 89°

To find the mass of the bug that exerts the maximum stress on the web, we will use the vertical tension force in the web and Newton's second law of motion. We obtained the total force on the web as 0.7 N and it can be used as the vertical tension force. Force = mass × acceleration due to gravity mass = Force / acceleration due to gravity Considering the acceleration due to gravity (g) as 9.81 \(m/s^2\): mass = 0.7 N / 9.81 \(m/s^2\) mass ≈ 0.071 kg

To find the distance the web extends downward with a 0.1 m radius, we can use the sine function: sin(θ) = opposite side (vertical extension) / hypotenuse (radius) vertical extension = radius × sin(θ) As θ ≈ 89° and the radius is 0.1 m: vertical extension = 0.1 m × sin(89°) vertical extension ≈ 0.1 m The web extends downward by approximately 0.1 m.