Static Equilibrium: Find Normal Force and x-Coord of CM

[biomajor009]

Homework Statement

A 6.3-m long plank is supported by two concrete blocks, one that is placed 2 m from the left end of the plank and another that is placed 1.2 m from the right end. The mass of the plank is 44.5 kg.

(a) Find the normal force from each of the concrete blocks on the plank. Call the normal force from the left block NL and the normal force from the right block NR.

(b) Find the normal force from each of the concrete blocks if a 62.4 kg person stands on the plank, 2.7 m from the left end.

(c) Find the x-coordinate of the center of mass of the plank-person system described in the previous part, using the left end of the plank as x = 0.

(d) How close to the left end of the plank can the person get before the plank starts to tip over?

Homework Equations

$$\Sigma$$Fx = 0
$$\Sigma$$Fy = 0

The Attempt at a Solution

Found NL = 566.9 and NR = 1003 but I'm not 100% confident

 

[PhanthomJay]

No, that's not correct; all forces must sum to zero and you must also consider rotational equilibrium $$\Sigma M = 0$$ about any point. Hint: the 44Kg mass acts at the cm of the plank. Then sum moments of all forces about one of the concrete block support points.