
Static Equilibrium help
I can't figure this one out...
A man carries a 3 m, 32N board over his shoulder. If the board extends 1.8m behind the shoulder, and 1.2m in front of the shoulder, how much force must the man apply vertically downwards with his that rests on the board at .2m infront of the shoulder. (assume weight of board is evenly distributed)
Answer: 48N
helllpppp!!!!!

This is basically a torque problem. You have to draw a diagram!
The board is pivoted at the shoulder. The force of gravity is considered to be acting at the geometric center of the board, a little behind the shoulder. His arm exerts a force 0.2 m in front of the shoulder. Our objective here is to maintain the resting state of the board.
$\displaystyle \implies \Sigma \tau = 0$
Therefore:
$\displaystyle \tau_{\text{counterclockwise}} = \tau_{\text{clockwise}}$
Where the torque produced by the gravity force is counterclockwise and the torque produced by the force exerted by the arm is clockwise. Note here that the direction of the forces will be assumed to be perpendicular.
$\displaystyle F_g\cdot d_1 = F \cdot d_2$
Where d is the distance from pivot. Note that distance of the gravity force from pivot is:
$\displaystyle d_1=1.81.5$
Rearrange the equation:
$\displaystyle F = \frac{F_g \cdot d_1}{d_2}$
$\displaystyle F = \dots$

I tried that but I still get a different answer to the one originally provided.
Any chance someone can provide a worked example?

I did work it out for you, and it delivered the right answer (48). Post your work here so I can see where you went wrong and then correct you.