# Homework Help: Tension in a spiderweb

1. Jan 26, 2012

### steven1495

1. The problem statement, all variables and given/known data

A spider builds its web in a window frame that is lying on the ground. It is supported by four main strands. Calculate the force of tension in strand 4 assuming the web is stable. The tension s in the other three strands are as follows:
strand 1: 21.00 mN [22.00°E of N]
strand 2: 16.00 mN [66.00°E of S]
strand 3: 18.00 mN [44°W of S]

2. Relevant equations

I honestly have no idea how to do this, because our teacher never taught it to us, and I need it for my physics exam tomorow

3. The attempt at a solution

2. Jan 26, 2012

### steven1495

it would also be great if you could show all the steps for me

Thanks

3. Jan 26, 2012

### 256bits

You have 4 strands with the tension and direction of each strand.
Your teacher never taught you about tension, so you think you cannot solve the problem. Or he didn't teach you about spider webs?

The tension in each strand can be broken down into x and y directions.

solve for forces in x direction to be zero.
Same for the y-direction.

4. Jan 26, 2012

### steven1495

He didn't mention anything about tension, he even admitted it when someone asked him in front of the class. My friends are saying that they got ‎9.987mN [0.06884° W of S], but I originally got 17.45mN [48.06°W of N], but then redid the question and got 23mN [56°W of N].

If someone could varify what one is right, preferably before 10pm (eastern standard time) because I have an exam tomorow :/

And what does mN stand for?

Thanks

5. Jan 27, 2012

### 256bits

I should mention that after finding the resultant for strands #1, #2, #3, the tension of force in strand #4 would have to be in the opposite direction so as to balance the spider web, but that you already knew.

m stands for milli, as in miilimeter, milliNewton, millilitre, milligram

I get 9.94 mN acting 89.98 West of South

My spider web forces (tension in the strands ) looks like this:

#### Attached Files:

• ###### spider.bmp
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Last edited: Jan 27, 2012
6. Jan 28, 2012

### physicsisgrea

My attempt:
Consider the forces acting on the web.
Since the question has not provided the weight of web, it may be negligible.

It is in equilibrium. Net force and torque = 0.
Along x-direction, we have:
S4 sin theta = S1 sin 22 + S2 sin 66 - S3 sin 44 = 9.979615116 mN
Along y-direction, we have
S4 cos theta = S1 cos 22 - (S2 cos 66 + S3 cos 44) = 0.01495825 mN

theta here stands for "(N theta W).

Solving for simple geometry:
S4 = 9.9796 mN (approx.)
theta = 89.9141 (approx.), so direction = N89.9141W.

All angles are in degree measure.