Solve the reactions for the beam shown below. The applied moment acts as show in the clockwise direction.
Hello CeliaSuppose that the reactions at A and B are $\displaystyle R_A$ and $\displaystyle R_B$ respectively. Then, take moments about A:
$\displaystyle M(A):14 \times 3 + 50 = R_B \times 10$
$\displaystyle \Rightarrow R_B = \frac{92}{10} = 9.2$
$\displaystyle M(B): R_A \times 10 + 50 = 14\times 7$
$\displaystyle \Rightarrow R_A = \frac{48}{10} = 4.8$
You can check by resolving vertically that $\displaystyle R_A+R_B = 14$
Grandad