Asked by Joseph
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?
Answers
Answered by
bobpursley
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.
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.
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