Finding mass with given stress, strain, and original area

[Taco John]

Homework Statement

figure whatever represents an insect caught at the midpoint of a spider-web thread. The thread breaks under a stress of 8.2 x 10^8 N/m^2 and a strain of 2. Initially, it was horizontal and had a length of 2cm and a cross-sectional area of 8 x 10^-12 m^2. 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 weight?

Homework Equations

I would imagine the different variations of stress and strain

The Attempt at a Solution

I know the answer will be small (as the insect is supposed to be a bumble bee assuming the picture in the book isn't lying to me.

I don't know where to begin exactly.

I did stress = F/A to get F = .00656N, which means the bumble bee would have to apply just under that amount of force, as that is what the web will break at. But I'm unsure on what to do next.

 

[Shooting Star]

Draw a diagram. Suppose A and B are points where the thread is attached. The midpoint C is now vertically displaced to D. At the point D, the forces acting are the tensions T in the thread, and the weight W.

Write down the eqn for the equilibrium of the vertical forces.
Find the relation between BD and BC, from geometry.
The common thing between these eqns is the angle CBD.

That should be enough to get you started. If you show some work based on this, perhaps we could help you more, if you need it. You can do this.

 

[Moetasim]

As a scientist, your first step would be to identify the relevant equations and concepts related to this problem. In this case, you would need to use the equations for stress (σ = F/A) and strain (ε = ΔL/L) to solve for the unknown variables, which in this case is the force and the weight of the insect.

Using the given values for stress (8.2 x 10^8 N/m^2) and strain (2), we can rearrange the equation for stress to solve for force: F = σA. Plugging in the original area (8 x 10^-12 m^2), we get a force of 6.56 x 10^-3 N.

Next, we can use the equation for weight (W = mg) to find the weight of the insect. We know the mass of the insect is negligible, so we can assume that the weight is equal to the force exerted by the insect. Therefore, the weight of the insect is also 6.56 x 10^-3 N.

In conclusion, the weight of the insect caught in the spider-web thread is 6.56 x 10^-3 N. This is a very small weight, which is consistent with the assumption that the insect is a bumble bee.