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.

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Video Transcript

This is College Physics Answers with Shaun Dychko. This spider hanging from a vertical thread in part A has a mass of eight times ten to the minus five kilograms and our job is to find the tension in this spider silk thread. So the net force vertically is mass times acceleration and the net force is the tension upwards, minus the gravity downwards. We know that acceleration is zero because the spider is motionless we're told, and the force of gravity is the mass of the spider times the gravitational field strength

*g*. So we can substitute for*mg*in place of*Fg*here and then well, since acceleration is zero, that means this term becomes zero. So add*mg*to both sides and you solve for force of tension and it is the weight in other words. So that's eight times ten to the minus five kilograms times 9.8 newtons per kilogram, giving 7.84 times ten to the minus four newtons. Now in part B, the spider is sitting in the middle of the horizontal section of the spider thread. We're told that the angle of depression of this thread is 12 degrees and that'll also be this angle here because these are -- these dotted lines are parallel lines and these are interior opposite angles, if you remember your parallel line rules for angles, and 12 degrees there as well. So this tension force is going to be exerted along the thread up into the right as well as along the thread up into the left and combined, these two tension forces are keeping the spider stationary. Now speaking about the vertical direction, the y direction, the net force will be two times the vertical component of each of these tension forces. Here is a vertical component of one of the tension forces here and it also exists on the other side so we have two of them supporting the spider, and that is the opposite leg of this triangle and so we use the sine function to find the opposite leg, multiplied by the hypotenuse which is*Ft*. So*Ft y*is sine*theta*times*Ft*. So, two times the vertical component of the tension upwards and then minus the weight downwards is the net force which is mass times acceleration. But acceleration is zero and so we can replace*ma*with zero.*Fg*is*mg*and so we replace that and then*Ft y*component is tension force times sine*theta*. So I replace that as well in this formula re-written here with a whole bunch of substitutions. Then we add*mg*to both sides, and then we have*2 Ft**sine**theta**on the left and**mg*on the right. Then we'll solve for*Ft*by dividing both sides by two sine*theta*. So the tension force is the weight of the spider divided by two times the sine of the angle of depression of the thread. So that's 8 times ten to the minus five kilograms times 9.8, divided by two times sine of 12 which is 1.89 time ten to the minus three newtons.
## Comments

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

Submitted by jgrimes14 on Tue, 08/18/2020 - 16:16

In reply to This instructor is going way… by drewboffeli

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

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

notshow 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

In reply to Why do we only consider the… by charlotte3452

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

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

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

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

In reply to Why the answer only… by pansapinsa