3 stings are attached on a ring. 2 of the strings make 70 degrees angle and each is pulled with 7.0 N . What force must be applied to keep the 3rd string to keep stationary?

Answer 1

You are looking for the equilbrant.

Add the three forces, and set to zero.

let one be at zero, one at 70, and the third unknown.

F=0=7 (cos000N+sin000E+cos070N+sin070E) + Fx(cosYYY N+ sinXXXE)

now, sum North forces=0
7N+7cos70+Fxcosxxx=0
Fxcosxxx=-9.39

and, sum East forces=0
7sin70+FxsinXXX=0
Fx*sinXXX=-6.58

divide one equation by the other

tan XXX=6.58/9.39

angxxx=180+35=215

Fx= 6.58/sin215=11.4

Remember, I have used N as 000, then angles measured clockwise.

check my work.