The figure represents an insect caught at the midpoint of a spider-web thread. T
ID: 2263784 • Letter: T
Question
The figure represents an insect caught at the midpoint of a spider-web thread. The thread breaks under a stress of
9.2
The figure represents an insect caught at the midpoint of a spider-web thread. The thread breaks under a stress of 9.2 times 108 N/m2 and a strain of 1.7. Initially, it was horizontal and had a length of 2.0 cm and a cross-sectional area of 7.0 times 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.)Explanation / Answer
Breaking stress:
Sut = 9.2*10^8 Pa
Breaking axial strain:
epsilon = 1.7
Use the strain to find the new length of each half of the thread
Lo = 1/2*Lnet0
L = L0*(1 + epsilon)
L = Lnet0*(1 + epsilon)/2
Cross sectional area @ breaking
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)
Using Sut = 9.2e8 Pa; A0 = 7e-12 m^2; epsilon = 1.7;
W = 1.6411 milliNewtons
Which corresponds to a mass of:
m = 164.1 micrograms