What is the Tension and Stretching of Spider Silk in Equilibrium?

[michael1291]

Homework Statement

The stretchy silk of a certain species of spider has a force constant of 1.50 . The spider, whose mass is 15mg, has attached herself to a branch as shown in the figure.
Figure:
http://session.masteringphysics.com/problemAsset/1348379/2/5-70.jpg
a)Calculate the tension in each of the three strands of silk.
b) Calculate the distance each strand is stretched beyond its normal length.

Homework Equations

Worked with the equations: Tc=Tacos(30) + Tbcos(40) for vertical components where Tc=mg=147 mN. And for the horizontal Tasin(30) = Tbsin(40). Found the ratio and plugged back into the equations to find Tb and then Ta.

The Attempt at a Solution

Using the relevant equations, I derived Ta = Tb x 1.286 and ultimately got answers:
Ta=100mN, Tb=78.2mN, and Tc=147mN.

 

[grzz]

re tensions: chech 147mN

re last part: use force = force constant x extension

 

[michael1291]

Does that say check* 147 mN, in that it is an incorrect magnitude for Tc?

 

[grzz]

A mass of 15mg = 15/(1000) = 0.015grams = 0.015/(1000) = 0.000015kg and now multiply by 9.81 to give weight in N to give 0.000147N = 0.147mN

 

[michael1291]

so it should come to .100mN, .0782mN, .147mN for Ta, Tb, and Tc respectively?

 

[michael1291]

the answer needs to be in 2 sig figs. I put in .100, .078, and .15. A message came up and said "Term 2: Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures."

I cannot find this error. Kept all numbers in my calculator.

Same response for the second part, in which I submitted .067, .052, and .098

 

[grzz]

I think that the angles were interpreted incorrectly.

The angle beteen tension A and tension B looks to me to be more than 90 degrees. So I think that the angles are 30deg and 70deg, together making up 100deg.

 

[michael1291]

worked out. Thanks a lot