A woman who weighs 476 N is leaning against a smooth vertical wall at an agnle 60 degrees above the horizontal. Find (a) the force FN (directed perpendicular to the wall) exerted on her shoulders by the wall and the (b) horizontal and (c) vertical components of the force exerted on her shoes by the ground. I know that the the vertical component is 476 N, the other one's I have no clue of how to do. I made the sum of the forces in the x direction zero. So I know that the force of static friction will equall the normal force exerted by the wall. I also made the sum of the torques to equal zero. So, FnLn = FwLw. I get my answer to be 137.41 N, but this is not right. Ln represents the lever arm's lengh. I used trig functions to get Lw=cos60 (0.75), and Ln=sin60 (1.5) This is the drawing they provide the distance (hypotenuse) from the Fn to the ground as 1.5. This number is broken into 1.1 m for her feet to the waist and 0.4 m from waist to shoulders. The head isn't included since there is no force for it.