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This is a question from an old O level additional mathematics book by T A Humphreys (1967). It is given as follows:

Strings of length 4 ft and 3 ft are attached to a fixed point O and to the ends A, B respectively of a rod 5 ft long and weighing 15 lb. wt. The rod rests in equilibrium in a horizontal position supported by a smooth peg at C. If AC = 1 ft find the tensions in the strings and the reaction at the peg.

Solutions are:

$T_A$=$6$$\frac{3}{22}$$lb. wt.$
$T_B$=$8$$\frac{2}{11}$$lb. wt.$
$R$=$4$$\frac{17}{22}$$lb. wt$

I have, after considerable effort, managed to obtain the answers given. However, I am unable to obtain them in the order given in the question and I was wondering if this was normal.

My solution involved finding R first by taking moments about O, therefore eliminating $T_A$ and $T_B$. It was only at this point that I could solve the two equations I had formed for $T_A$ and $T_B$. I should add that this chapters is devoted to moments of forces so I would imagine this is the way to solve - there may be other more novel solutions but I'm not sure.

Apologies if I'm either being pedantic here or have asked a question with an obvious answer. I would just like to know if I'm missing something.

Many thanks in advance for any advice given.

GR L
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1 Answers1

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I will show how to obtain $R$, considering in competent units

$$ \cases{ OA=4\\ OB=3\\ AB=5\\ AC=1\\ w = 15 } $$

Solving for $AX,XB,OX$

$$ \cases{ AX+XB=AB\\ \frac{XB}{OX} =\frac{OB}{OA}\\ AX\cdot XB = OX^2 } $$

where $X$ is the vertical projection of $O$ onto $AB$ we have

$$ \cases{ AX = \frac{AB\cdot OA^2}{OA^2+OB^2}\\ XB = \frac{AB\cdot OB^2}{OA^2+OB^2}\\ OX = \frac{AB\cdot OA\cdot OB}{OA^2+OB^2}\\ } $$

now computing the momenta equilibrium regarding $O$ we have

$$ \left(\frac 12 AB-XB\right)w - (AB-AC-XB)R=0 $$

and solving for $R$ we obtain

$$ R = \frac 12\frac{AB(OA^2-OB^2)}{AB\cdot OA^2-AC(OA^2+OB^2)}w = \frac{105}{22}=4\frac{17}{22} $$

To find $T_A, T_B$ we need to solve

$$ T_A \vec{OA}+T_B\vec{OB}+(R+w)\vec{e}_y = 0 $$

where

$$ \cases{ \vec{OA} = \frac{1}{OA}\left(AX,OX\right)\\ \vec{OB} = \frac{1}{OB}\left(-XB,OX\right)\\ \vec e_y = (0,-1) } $$

Cesareo
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