Is this correct? (spider web tension question)

[phy_]

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 tensions in the other three strands are as follows:
strand 1: 21 mN (20 degrees East of North)
strand 2: 16 mN (60 degrees East of South)
strand 3: 18 mN (40 degrees West of South)

if i solve y

T1 cos20(0.021N)
=0.019 N
T2 cos60(0.016N)
=0.008 N
T3 cos 40(0.018N)
=0.013 N

T4 y total 0.014

solve for x
sin20(0.021N)
=0.00718 N
sin60(0.016)
=0.013 N
sin(40)(0.018N)
=0.011N

T4 x total 0.0162

c squared = 0.0162 squared + 0.014 squared
c=0.0214

tan theta = 0.014/0.0162 = 0.0214 N40.8W degrees

is this correct?

 

[Carid]

You don't see to deal with the directions correctly.

0.019 - 0.008 - 0.0013 = ?

 

[phy_]

could you explain why you would subtract?

 

[physics girl phd]

I find your work hard to follow. I think (like Carid notes) you aren't quite dealing with directions well. In a problem like this, I would first take all individual strings and break them into their components along perpendicular directions: in your case, east and north... and maybe you'd even like to note west and south instead of using negative east and negative north (these negatives would explain the subtraction that Carid is noting).

for instance I would say:
string 1: 0.019 N North + 0.00718 N East
string 2: ...
string 3: ...

 

[Stovebolt]

From the looks of the numbers, I'm guessing you averaged the magnitudes of each of the forces? If so, that is not the correct approach. Otherwise it looks like you took the right steps.

I highly recommend drawing a diagram of the web, keeping in mind that you are looking for the sum of the forces about the center point where the strands intersect.

Once that is done, take physics girl's advice. Choose one direction as positive for each component (x and y), then add any positive force and subtract the negative forces.

 

[phy_]

Thank-you very much. Question solved.