Objects in equilibrium problem

[BrainMan]

The problem is a uniform plank of length 6 m and mass 30 kg rests horizontally on a scaffold, with 1.5 m of the plank hanging over one end of the scaffold. How far can a 70 kg person walk on the overhanging part of the plank before it tips?

Relevant equations
Sum of all the forces = 0
Sum of the torque must= 0Attempt-
(1) I make the pivot point at the point of overhang so 1.5 of of the board is over the pivot point and -4.5 is to the left of the pivot point.
(2) 1.5 is .25 of the length so it is .25 of the mass so the mass over the pivot point is 7.5. 4.5 is .75 of the length so it is .75 of the mass so the mass to the left of the pivot point is 22.5 kg.
(3) since the sum of all the torque is 0 then 22(-4.5)+7.5(1.5)+70(x)= 0
(4) solving for x I got 1.29 the correct answer is 0.643.

 

[Patolord]

First you calculate the torque you need to lift the bar
Torque=F*d then T=300 * 1.5 = 450 N*m
now you do sum of the torque must be = 0
-450 + 700 *x = 0
solve for x = 0,643

 

[BrainMan]

What do you mean by "lift" the bar?

 

[Patolord]

Tip the plank down. my bad xD

 

[BrainMan]

Ok I understand. Where did you get the 300 in calculating the torque needed to lift the bar?

 

[Patolord]

its 30kg * 10 m/s²

 

[Patolord]

http://gyazo.com/7f481406e5d2058a94ef5c48023f1d18

 

[BrainMan]

Ok thanks a lot!