# Tensions in Cable

1. Nov 10, 2014

### SilverAu

1. The problem statement, all variables and given/known data
A vendor hangs an 8.6kg sign in front of his shop with a cable held away from the building by a lightweight pole. The pole is free to pivot about the end where it touches the wall,

2. Relevant equations
T*L*sinΘ = mgd
where d is the distance from the sign to the wall
and L is the distance from the cable to the wall
and T the cable tension.

3. The attempt at a solution
T = mgd / sinΘ = 8.6kg * 9.8m/s² * d / Lsin60 = 97N * L/d
T = 97 N
Is this right? because I am confused on what the distance is.

2. Nov 10, 2014

### SteamKing

Staff Emeritus
Instead of trying to decipher some formula of unknown source, why don't you draw a free body diagram of the pole supporting the sign and write equations of static equilibrium? If you do, you might find that the equation you were using is not entirely correct.

3. Nov 10, 2014

### Simon Bridge

... lets see your reasoning for each step of calculation you did?
2nd what Steamking says above.