L=4mmass ofplank ,m=20kgtake counterclockwise torques arepositiveApply torque equation about one endF1(0)−mg(L2)+F2(L−1m)=0F2(L−1m)=mg(L2)F2=mg(L2(L−1))=20kg×9.8m/s2×4m2(4m−1m))=130.66N=131NL=4m \\mass \space of plank\space, m=20kg \\ take \space counterclockwise \space torques\space are positive\\ Apply\space torque \space equation\space about\space one\space end\\ F_{1}(0)-mg(\frac{L}{2})+F_{2}(L-1m)=0 \\F_{2}(L-1m)=mg(\frac{L}{2}) \\F_{2}=mg(\frac{L}{2(L-1)})\\ \quad=20kg\times 9.8m/s^{2}\times\frac{4m}{2(4m-1m)})\\ \quad=130.66N\\ \quad=131NL=4mmass ofplank ,m=20kgtake counterclockwise torques arepositiveApply torque equation about one endF1(0)−mg(2L)+F2(L−1m)=0F2(L−1m)=mg(2L)F2=mg(2(L−1)L)=20kg×9.8m/s2×2(4m−1m)4m)=130.66N=131N
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