Question Number 73255 by TawaTawa last updated on 09/Nov/19
Answered by mr W last updated on 09/Nov/19
$${there}\:{is}\:{no}\:{support}\:{on}\:{the}\:{right}\:{end} \\ $$$${of}\:{the}\:{beam}! \\ $$$$ \\ $$$$\mathrm{0}.\mathrm{4}{L}×{R}_{{Mid}} −\mathrm{0}.\mathrm{2}{L}×\mathrm{0}.\mathrm{6}{W}−\mathrm{0}.\mathrm{8}{L}×\mathrm{0}.\mathrm{15}{W}=\mathrm{0} \\ $$$$\Rightarrow{R}_{{Mid}} =\frac{\mathrm{0}.\mathrm{2}×\mathrm{0}.\mathrm{6}+\mathrm{0}.\mathrm{8}×\mathrm{0}.\mathrm{15}}{\mathrm{0}.\mathrm{4}}{W}=\mathrm{0}.\mathrm{6}{W} \\ $$$$\mathrm{0}.\mathrm{4}{L}×{R}_{{Left}} −\mathrm{0}.\mathrm{2}{L}×\mathrm{0}.\mathrm{6}{W}+\mathrm{0}.\mathrm{4}{L}×\mathrm{0}.\mathrm{15}{W}=\mathrm{0} \\ $$$$\Rightarrow{R}_{{Left}} =\frac{\mathrm{0}.\mathrm{2}×\mathrm{0}.\mathrm{6}−\mathrm{0}.\mathrm{4}×\mathrm{0}.\mathrm{15}}{\mathrm{0}.\mathrm{4}}{W}=\mathrm{0}.\mathrm{15}{W} \\ $$$$ \\ $$$${check}:\:{R}_{{Mid}} +{R}_{{Left}} =\mathrm{0}.\mathrm{6}{W}+\mathrm{0}.\mathrm{16}{W}=\mathrm{0}.\mathrm{75}{W} \\ $$
Commented by TawaTawa last updated on 09/Nov/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$