Question:
A uniform plank $AB$ of $4m$ length and $15kg-wt$ weight rests horizontally on two supports/pillars: one at $A$ and the other at a point $0.5m$ from $B$. Find the weight of a boy if he can stand on top of the plank at point B without upsetting the stability of the plank.
My book's attempt:
Let $O$ be the middle point of $AB$ so that the weight $15 kg-wt$ acts at $O$. Let the other support be at $C$ and $Wkg-wt$ be the weight of the boy.
$$\text{So, } AB=4m,\ AO=OB=2m,\ BC=0.5m$$
According to the question, the force acting at $C$ and the resultant of $15kg-wt$ acting at $O$ and $Wkg-wt$ acting at $B$ must be equal and act in opposite directions along the same line of action.
$$\frac{15}{BC}=\frac{W}{OC}=\frac{15+W}{OB}$$
$$\text{Each force is proportional to the distance between the points of application of the other two.}$$
From the 1st and 2nd ratios,
$$\frac{15}{BC}=\frac{W}{OC}$$
$$15\cdot OC=W\cdot BC$$
$$15\cdot (2-0.5)=W\cdot 0.5$$
$$W=45$$
Desired weight of the boy is $45 kg-wt\ \text{(Ans.)}$
My comments:
Notice that in the book's attempt, it didn't even acknowledge the existence of the pillar/support at $A$. The book didn't consider any force coming out of $A$. It's as if the pillar/support at $A$ didn't even exist. Why then does the question give us the pillar at $A$ anyways? From common sense I think that the plank would become unstable if the support from $A$ was removed; however, I don't know if I can trust my common sense. Is the support at $A$ necessary for the stability of the plank? If yes, then why didn't the book consider any force coming out of $A$?
My question:
- Is the support at $A$ necessary for the stability of the plank? If yes, then why didn't the book consider any force coming out of $A$?
Best Answer
Let $R_A$ be the reaction force at point $A.$ Then, by the principle of moments, $$R_A\times AC+W\times BC=15\times OC.$$ Therefore, minimising $R_A$ maximises $W$ (noting that all the other quantities are constant); in other words, $W$ is maximum when $R_A$ is zero.