Cliff and Will are carrying a uniform 2.0m board of mass 75kg. Will is supporting the board at the end while cliff is 0.6m from the other end as shown in the following figure. Cliff has attached his lunch to his end of the board, and the tension in the string supporting the lunch is 190N. Find the normal forces exerted by Cliff and Will.

Question

Can someone help with the formulas for these? Thank you.

Answer 1

190 down at left
C up at x = .6
75*9.81 down at x = 1
W up at x = 2

forces up = forces down
C + W = 190 + 75*9.81 = 926 Newtons

moments about x = 0 sum to 0
-.6 C + 75*9.81*1 - 2 W = 0
or 2W +.6C = 736 Newton meters

Answer 2

Thanks for your time! I don't understand this answer. Maybe someone else can help me understand the formulas to find the force for Cliff and the force for Will.
Thank you

Answer 3

C + W = 926 so
2 W = 1852 - 2 C

then
(1852 - 2 C) + .6 C = 736

1852 - 736 = 1.4 C
etc

Answer 4

Thanks again, Damon. Still not getting this at all. I appreciate your time, but the steps still don't make sense to me.

Answer 5

add up the forces. That is one equation
add up the torques (moments) about x = 0. That is the second equation