Answer
No. The climber only got $90\%$ of the way across the log
Work Step by Step
Please see the attached image first.
Let's consider the static equilibrium of the board.
$\uparrow \sum \vec F=0$
$n_{1}-m_{1}g-m_{2}g=0$
$n_{1}=(m_{1}+m_{2})g-(1)$
$\rightarrow \sum\vec F=0$
$F_{S}-n_{1}=0$
$F_{S}=n_{2}-(2)$
$\sum \tau_{A}\curvearrowright=0$
$m_{1}g\frac{L}{3}cos26^{\circ} + m_{2}gxcos26^{\circ}\space -n_{2}Lsin26^{\circ}=0$
$(2)=\gt(3)$
$m_{1}g\frac{L}{3}cos26^{\circ} + m_{2}gxcos26^{\circ}\space -F_{S}Lsin26^{\circ}=0$
$\frac{m_{1}gcot26^{\circ}}{3}+\frac{m_{2}gxcot26^{\circ}}{L}=F_{S}-(4)$
Let's apply x = L
$\frac{m_{1}gcot26^{\circ}}{3}+\frac{m_{2}gcot26^{\circ}}{L}=F_{S}$
$\frac{(224g+77.3g)}{3}\times2.05=F_{S}$
$311.53g=F_{S}-(A)$
Now let's find the maximum static friction of the horizontal wall.
$F_{S}=\mu n_{1}= 0.982(m_{1}+m_{2})g$
$F_{S}= 295.88g-(B)$
$(A)\gt (B)$ , So the climber can't get to the righthand end of the log.
$(B), (4)=\gt$
$(\frac{m_{1}}{3}+\frac{m_{2}x}{L})gcot26^{\circ}=295.88g$
$77.3\frac{x}{L}=144.31-\frac{224}{3}=0.9L$
The climber can only get $90\%$ of the way across the log.
