Answer :
Answer:
The answer is "[tex]\bold{MB= \frac{L^2 \ WB}{12}}[/tex]"
Explanation:
please find the complete question in the attached file:
From A to B
[tex]reqAB = \int^{\frac{1}{2}}_{0} \frac{WB}{\frac{L}{2}} x \ dx = \frac{L\ WB}{4}[/tex]
From symmetry [tex]Ay = Cy = FeqAB[/tex]
The FeqAB position is found on a table of common centres:
[tex]\frac{2}{3}( \frac{L}{2})[/tex]
[tex]\to \Sigma_{NB=0} FX = NB=0\\\\\to \Sigma_{VB=0} Fy = \frac{LWB}{4} - \frac{LWB}{4} -VB =0 \\\\\to \Sigma M_{.CCW \ A} = \frac{L}{3} - \frac{LWB}{4} - \frac{L}{2}VB+MB =0 \\\\MB= \frac{L^2 \ WB}{12}[/tex]
