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%3Cp%20class%3D%22c1%22%3EThe%20figure%20represents%20an%20insect%20caught%20at%20the%0Amidpoint%20of%20a%20spider-web%20thread.%20The%20thread%20breaks%20under%20a%20stress%0Aof%209.8%26nbsp%3B%C3%97%26nbsp%3B10%3Csup%3E8%3C%2Fsup%3E%26nbsp%3BN%2Fm%3Csup%3E2%3C%2Fsup%3E%26nbsp%3Band%0Aa%20strain%20of%202.5.%20Initially%2C%20it%20was%20horizontal%20and%20had%20a%20length%20of%0A3.0%20cm%20and%20a%20cross-sectional%20area%0Aof%26nbsp%3B7.0%26nbsp%3B%C3%97%26nbsp%3B10%3Csup%3E-12%3C%2Fsup%3E%26nbsp%3Bm%3Csup%3E2%3C%2Fsup%3E.%26nbsp%3BAs%0Athe%20thread%20was%20stretched%20under%20the%20weight%20of%20the%20insect%2C%20its%20volume%0Aremained%20constant.%20If%20the%20weight%20of%20the%20insect%20puts%20the%20thread%20on%0Athe%20verge%20of%20breaking%2C%20what%20is%20the%20insect's%20mass%3F%20(A%20spider's%20web%0Ais%20built%20to%20break%20if%20a%20potentially%20harmful%20insect%2C%20such%20as%20a%20bumble%0Abee%2C%20becomes%20snared%20in%20the%20web.)%3C%2Fp%3E%0A%3Cp%20class%3D%22c2%22%3E%3Cimg%20src%3D%0A%22http%3A%2F%2Fedugen.wileyplus.com%2Fedugen%2Fcourses%2Fcrs4957%2Fart%2Fqb%2Fqu%2Fc12%2Fq06f.jpg%22%20%2F%3E%3C%2Fp%3E%0AExplanation / Answer
Breaking stress:
Sut = 9.8*10^8 Pa
Breaking axial strain:
epsilon = 2.5
Use the strain to find the new length of each half of the thread
L0 = 1/2*Lnet0
L = L0*(1 + epsilon)
L = Lnet0*(1 + epsilon)/2
Cross sectional area @ breaking. Use the concept of constant volume:
Aut*L = A0*L0
Aut*L0*(1+epsilon) = A0*L0
Aut = A0/(1+epsilon)
Stress at breaking related to tension:
Sut = F/Aut
F = Sut*Aut = Sut*A0/(1+epsilon)
Result in breaking tension
F = Sut*A0/(1+epsilon)
Now, consider the FBD on the insect: the thread is divided in two when the insect is caught. Half of the thread pulls up and to the left, the other half pulls up and to the right. Since it is cut in half, both angles (theta, from horizontal) are the same.
Add up vertical forces:
W = 2*F*sin(theta)
Express in terms of known tension force expression:
W = 2*Sut*A0/(1+epsilon)sin(theta)
To find theta, draw a triangle, containing L0 as its base and L as its hypotenuse. Use cosine to relate the two.
cos(theta) = L0/L
cos(theta) = L0/(L0*(1+epsilon))
cos(theta) = 1/(1+epsilon)
Substitute to master expression:
W = 2*Sut*A0/(1+epsilon)*sin(acos(1/(1+epsil...
Simplify:
W = 2*Sut*A0/(1+epsilon)*
sqrt(epsilon*(2+epsilon)/
(1+epsilon)^2)
Use data: Sut = 9.8e8 Pa; A0 = 7e-12 m^2; epsilon = 2.5;
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