My Math Forum Magnitude and sense of the force applied at A to reduce the reaction at A to zero.

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February 7th, 2017, 02:30 PM   #1
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Magnitude and sense of the force applied at A to reduce the reaction at B to zero.

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 $\alpha$ 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 $S=10N$, $\alpha = arctan(\frac{1}{3\sqrt3})$ and $R= 10\sqrt7N$.

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, but should it be clockwise, as from$A$ the moment of the force $3m$ from $B$ on top of the beam is $$6*30sin30^o =90Nm$$ anticlockwise?
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Last edited by Statistics132; February 7th, 2017 at 02:58 PM.

 February 11th, 2017, 11:49 AM #2 Member   Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 No, it is anticlockwise. 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 $RsinĪ± = 5N$: $$M(B): -9RsinĪ± - 3*30sin30^o + 9F = 0$$ 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 $Nm$.) Last edited by Statistics132; February 11th, 2017 at 11:58 AM.

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