Answer
a) $F_L=310N$; up
$F_R=210N$; down
b) $0.65m$ from right hand
c) $1.2m$ from right hand
Work Step by Step
a) $\sum \tau_{right}=(0.32m)(F_L)-(1.00m)(F_g)=0$
$F_L=310N$; up
$\sum F_y=-F_R+F_L-F_g=0N$
$F_R=210N$; down
b) $F_L-F_R=98N$
$\sum \tau_{CoG}=(x)(F_L)-(1.00m)(F_R)=0$
$\sum \tau_{CoG}=(x)(150N)-(1.00m)(52N)=0$
$x=0.35m$
$0.65m$ from right hand
c) Assuming $F_L-F_R=98N$, the force on both hands cannot be smaller than or equal to 85N. Therefore, this means the distance from right hand is greater than 1m and $F_L+F_R=98N$
$\sum \tau_{CoG}=(x)(85N)-(1.00m)(13N)=0$
$x=0.15m$
$1.2m$ from right hand
