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Question
(a) Calculate the tension in a vertical strand of spider web if a spider of mass $8.00 \times 10^{-5} \textrm{ kg}$ hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in Figure 4.17. The strand sags at an angle of $12^\circ$ below the horizontal. Compare this with the tension in the vertical strand (find their ratio).

a) $7.84 \times 10^{-4} \textrm{ N}$

b) $1.89 \times 10^{-3} \textrm{ N}$, which is 2.4 times the tension in the vertical strand.

Solution Video

# OpenStax College Physics Solution, Chapter 4, Problem 19 (Problems & Exercises) (3:42)

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Video Transcript

Submitted by drewboffeli on Tue, 10/15/2019 - 16:58

This instructor is going way to fast, will be not be a subscriber anymore.

Submitted by charlotte3452 on Wed, 04/01/2020 - 17:44

Why do we only consider the y component for part b)?

Submitted by ShaunDychko on Wed, 04/01/2020 - 21:24

Thank you for the question. The y-component of the tension in each segment of the wire is the only force exerted by the wire on the spider. The horizontal components of the tension are not exerted on the spider, so they have no place in the net Force calculations. Strictly speaking, part (b) does not show a free body diagram of the spider, which perhaps makes it confusing. A free body diagram of the spider would have one arrow down representing gravity, and one arrow of equal length pointing up representing the total of the y-components of the tension in each segment of wire. This means only the y-components appear in our calculations.
Hope this helps,
Shaun

Submitted by ckw666 on Thu, 09/24/2020 - 10:11

Can you explain what the x-components of the wire?

Submitted by pansapinsa on Thu, 08/19/2021 - 22:47

Why the answer only considers tension in 1/2 of the string, not the string as a whole?

Submitted by ShaunDychko on Mon, 08/23/2021 - 14:06

Hi pansapinsa,
Thank you for the question. I think what you mean is why do we consider each half of the string separately? Since we add the y-component of the force due to each half together, we do get the total y-component of force on the spider. The approach of considering each half separately is necessary since we want to know the total force on the point where the spider is positioned, which is the middle. Tension in a stationary string is pulling equally in both directions. Since the strand isn't moving we know the net force is zero. The tension at that point is pulling both up-to-the-right and up-to-the-left, so we need to find the y-component of each of those forces.
Hope this helps a bit,
Shaun