Well you've got no picture but lets see what we can do description-wise.
a) Sum of torques must be zero (I'm assuming B is between center of board and man). Mg of the board times distance to middle of board to B = weight of man times distance to B
b) Sum of forces must be zero. Your answer looks ok as A should be zero.
c) Now sum torques around A
Clockwise:
weight of board times distance from a to center of board + weight of man times distance from A to man
Counter clockwise:
Normal force from B times distance from A to B. To find normal at A, sum of forces = zero
d)Same thing. sum torques, find A; sum forces, find B
A 20.0 m long uniform beam weighing 800 N rests on walls A and B, as shown in the figure below.(a) Find the maximum weight of a person who can walk to the extreme end D without tipping the beam.(b) Find the forces that the walls A and B exert on the beam when the same person is standing at point D.
(c) Find the forces that the walls A and B exert on the beam when the same person is standing at a point 2.0 m to the right of B.
(d) Find the forces that the walls A and B exert on the beam when the same person is standing 2.0 m to the right of A.
I have the first two parts done but I don't really know what equations I should be using for the last two parts. my answer for part A was 800N and my answers for part B were 1600N for wall B and 0N for wall A. like I said I just don't know what equations to use, I don't necessarily need someone to go through the entire process of solving the equation.
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