Answer
86.67 kilograms
Work Step by Step
We know that the net torque must be 0. Thus, calling the bottom of the log the axis of rotation, we find:
$2.1Mgsin63^{\circ}+6.3m_cgsin63^{\circ}=(6.3)(.92)(M+m_c)(g)sin27^{\circ}$
$2.1(340)(9.81)sin63^{\circ}+6.3m_c(9.81)sin63^{\circ}=(6.3)(.92)(340+m_c)(9.81)sin27^{\circ}$
$m_c=\fbox{86.67 kilograms }$
