Problem: a movie stunt woman of weight 700 N walks out to the end of a uniform horizontal plank that projects perpendicularly over the edge of a roof. The plank is 6 m long and weighs 500 N. How far from the roof can the plank overhang?

Let x be the length of the overhang, from the side of the building to the end of the plank.

ΣF = 0 F = 500 N + 700 N = 1200 N (The upward force at the edge of the building.)

Στ = 0 when l = 0 at the person
(1200 N)(x) = (500 N)(3 m) x = 1.25 m

Στ = 0 when l = 0 at the edge of the building
(700 N)(x) = (500 N)(3 m - x) x = (1500 N-m)/(700 N + 500 N) = 1.25 m

Στ = 0 when l = 0 at the center of the plank.
(700 N)(3 m) = (1200 N)(3 m - x) x = (3600 N-m - 2100 N-m)/(1200 N) = 1.25 m

Στ = 0 when l = 0 at the end of the plank over the building.
(500 N)(3 m) + (700 N)(6 m) = (1200 N)(6 m - x) x = 1.25 m