Answer
3m
Work Step by Step
Please see the attached image first.
Let's consider the static equilibrium of the board.
$\uparrow \sum \vec F=0$
$n_{1}-m_{1}g-m_{2}g=0$
$n_{1}=(m_{1}+m_{2})g-(1)$
$\sum \tau_{B}\curvearrowright=0$
$n_{1}\times Lcos70^{\circ}-m_{1}g\frac{L}{2}cos70^{\circ}-m_{2}gxcos70^{\circ}-F_{S}Lsin70^{\circ}=0$
$n_{1}L-F_{S}Ltan70^{\circ}-\frac{m_{1}gL}{2}=m_{2}gx-(2)$
According to equation (2), we'll get the maximum climbing length when static friction is greatest. At that time,
$F_{S}=\mu n_{1}=\mu (m_{1}+m_{2})g-(3)$
$(1),(3)=\gt (2)$
$(m_{1}+m_{2})gL-\mu (m_{1}+m_{2})gLtan70^{\circ}-m_{1}g\frac{L}{2}=m_{2}gx$
Let's plug known values into this equation.
$(75.27)4-0.265(75.27)4\times2.75-6.47\times\frac{4}{2}=68.8x$
$1=x$
Maximum climbing length = 4 - x = 3m
