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Need Help- Finding the tension and Force wind using newton's laws

  • #1
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0

Homework Statement


A 1.20 g spider hangs on its thread from the branch of a tree. A horizontal win blows the spider and the tread to an angle of 35 deg from the vertical. Find the force of win on the sider and the tension in the tread.

Diagram:
http://s1176.beta.photobucket.com/user/LolaGoesLala/media/g.jpg.html


Homework Equations


Fnet= Fg + Fw + Tcosθ + Tsinθ


The Attempt at a Solution



Fg = mg
Fg= (0.0012 kg)(9.8 m/s^2)
Fg= 0.01176 N

Fnet = 0.01176 N + Fw + Tcos55° + Tsin55°
I got the 55° by subtracting the 35° from 90°

Vertical equilibrium:
0.01176 N = Tsin55°
0.014356 N = T

Horizontal equilibrium:
Tcos55°=Fw
(0.014356 N)(cos55°)=Fw
0.00823 N = Fw
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi Lolagoeslala! :smile:

Yes, that looks ok.

Two points …

i] why did you convert from 35° to 55°? it's correct, but it's unnecessary, and it gives you an extra chance to make a mistake!

ii] your line "Fnet= Fg + Fw + Tcosθ + Tsinθ" is simply wrong
(but it didn't matter, because your component equations were correct, so you didn't need that line)
… you can't add components in different directions!

correct would be "Fnet = Fg + Fw + T"

or "Fnet = Fg + Fw + (Tcosθ,Tsinθ)" :wink:
 
  • #3
217
0
Hi Lolagoeslala! :smile:

Yes, that looks ok.

Two points …

i] why did you convert from 35° to 55°? it's correct, but it's unnecessary, and it gives you an extra chance to make a mistake!

ii] your line "Fnet= Fg + Fw + Tcosθ + Tsinθ" is simply wrong
(but it didn't matter, because your component equations were correct, so you didn't need that line)
… you can't add components in different directions!

correct would be "Fnet = Fg + Fw + T"

or "Fnet = Fg + Fw + (Tcosθ,Tsinθ)" :wink:
Oh akay by the way for your first point how you are like why did you convert to 55° from 35°
is because when i was making the FBD.... i needed the angle of the horizontal. But how would you take the 35° and work with that.... i have not learned how to work with angles that are like on the opposite sides.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
But how would you take the 35° and work with that.... i have not learned how to work with angles that are like on the opposite sides.
ah, this is elementary Euclidean geometry …

if you have a letter "N", the two angles are equal (i think they're called "alternate angles") …

so the "downward" angle (35°) between the thread and the vertical wall is the same as the "upward" angle between the thread and the imaginary vertical through the spider :wink:

(the thread corresponds to the diagonal of the "N")
 
  • #5
217
0
ah, this is elementary Euclidean geometry …

if you have a letter "N", the two angles are equal (i think they're called "alternate angles") …

so the "downward" angle (35°) between the thread and the vertical wall is the same as the "upward" angle between the thread and the imaginary vertical through the spider :wink:

(the thread corresponds to the diagonal of the "N")
Oh i know that LOLLL :D
Thanks for reviewing my memory :D thanks for checking by the way :D
 

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