Tension of three ropes?

Question

There are two weights, each of which are 37.73N and 90.944N. They are arranged like so:

http://i91.photobucket.com/albums/k285/happygolucky2442/aaaaphs.jpg

By using triangle trig, it is possible to find the tension in the three ropes (lines touching the weights). Can anyone help me find the answers (the three tensions in newtons)?

Answer 1

Simply keep in mind that summation of forces equal to zero.

Sum (Fx)=0
Sum (Fy)=0
let th
F1- force o string supporting W1 (weight on the left)
F2- force o string supporting W2 (weight on the right)
F3- force o string between the weights

Fx=0=F1cos(54) - F2cos(68) +F3cos (13)
Fy=0= W1+W2-F1sin(54) - F2sin(68) -F3sin (13)

we have 2 equations and three unknowns.
Since the dimensions or length of the ropes ar enot given we can't employ moments.

let's use superposition and solve one side at a time

W1=F1sin(54) then F1=W1/sin(54)=37.73/sin(54)=46.6N
similarly
F2=W2/sin(68)=90.944/sin(68)=98.08

since
F1cos(54) - F2cos(68) +F3cos (13)

then F3=(F1cos(54) - F2cos(68))/cos (13)

F3=(46.6cos (54)-98.08cos(68))/cos(13)= - 9.6 N

(Negative sign means that the originally assumed 'positive' direction to the left was wrong. F3 is pointing to the right.)

Answer 2

T1 is tension in Left hand rope, T2 in the middle rope and T3 is the tension in the Right hand rope. By resolving the acting weights and tensions, you get for the Left hand system at the point at which the 37.73N force acts, the following:
37.73N = T1sin54 - T2sin13 & T2cos13 = T1cos54, i.e. 2 equations. From these, we find, T1 = 56.04N and T2 = 33.8N.
Now we work on the Right hand system, this simplified if we only take the horizintal balanced forces acting and the value for T2 just derived, i.e. T2cos13 = T3cos60 => T3 = 65.86N......Hope that helped!!

Answer 3

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