L=4mL=4mL=4m
mass of plank, m=20kgm=20kgm=20kg
take counterclockwise torques are positive
Apply torque equation about one end
F1(0)−mg(L/2)+F2(L−1m)=0F_{1}(0) - mg(L/2)+F_{2}(L-1m)=0F1(0)−mg(L/2)+F2(L−1m)=0
F2(L−1m)=mg(L/2)F_{2}(L-1m)=mg(L/2)F2(L−1m)=mg(L/2)
F2=mg(L2(L−1))=20kg×9.8m/s2×4m2(4m−1m)=130.66NF_{2}=mg(\frac{L}{2(L-1)})=20kg\times9.8m/s^{2}\times \frac{4m}{2(4m-1m)} = 130.66NF2=mg(2(L−1)L)=20kg×9.8m/s2×2(4m−1m)4m=130.66N
=131N=131N=131N
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