A light horizontal beam $AB$, of length $9m$, supported at its ends by a force $S$ acting vertically and a force $R$ acting at an angle α to the line of the beam. A force of $30N$ is applied to the beam, at an angle of $30^o$, $3m$ from $B$. The beam is in equilibrium and $S=10N$, $α=arctan(\frac{1}{3√3})$ and $R=10√7N$.
Calculate the magnitude and sense of the necessary moment that would have to be applied at $A$ to reduce the reaction at $B$ to zero.
(Answer given as $90Nm$ anticlockwise)
What is the best approach to the last part of the question? Summing the forces at each point or taking moments?
Assuming the conditions are the same as in the first part of the question, where $R$, $\alpha$ and $S$ had to be worked out, we now add an extra force at $A$ to create an anticlockwise moment, which is pointing directly downwards at $A$.
We then take moments about $B$, and with $Rsinα = 5N$:
$$M(B): -9Rsinα - 3*30sin30^o + 9F = 0$$
So $$-45 -45 = -9F$$
So $$ -90 = -9F$$
Therefore $$ 9F = 90 Nm$$
So the necessary moment about $A$ to reduce the reaction at $B$ to zero is $90Nm$ anticlockwise.
(It turns out that $F=10N$, which is the same as $S$, so $9F$ is the necessary moment, in $Nm$.)