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(I've tried to ask about the specific concept that's giving me trouble so as to not make this an illegal question)

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:

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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:

I was told that as the book is considering $W$ as the maximum weight possible to be put at $B$, which doesn't upset the plank, there will be no normal force present at $A$. I'm having some trouble figuring out the addition/subtraction of forces that will result in our scenario. I think my question is about how/why the normal force at $A$ is zero.

My question:

  1. How/why is the normal force at $A$ zero?
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    $\begingroup$ Question: what's the upward force on "A" when the beam is barely balanced? $\endgroup$
    – David White
    Commented Mar 14, 2022 at 4:21
  • $\begingroup$ @DavidWhite Sir, do you mean when the beam/plank is balanced without the weight of the boy? $\endgroup$
    – tryingtobeastoic
    Commented Mar 14, 2022 at 4:23
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    $\begingroup$ No. When the boy is standing on the end of the beam, and the beam is just about to rotate due to the weight of the boy, what is the upward force on the beam from pillar "A"? $\endgroup$
    – David White
    Commented Mar 14, 2022 at 4:26
  • $\begingroup$ @DavidWhite There is no force. In other words, there is an upwards force of $0N$ from pillar $A$. $\endgroup$
    – tryingtobeastoic
    Commented Mar 14, 2022 at 4:27
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    $\begingroup$ A normal force arises out of a reaction to something. When the boy is standing on the right, and the beam just about to lose contact at point A, the force exerted by beam on pillar at A is zero. Hence normal force is also zero. $\endgroup$
    – sku
    Commented Mar 14, 2022 at 6:07

1 Answer 1

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When the beam is barely balanced, the plank nearly looses contact with the pillar labelled "A", which is why the normal force at "A" is zero. Any more weight at "B" will result in a torque imbalance, causing the plank to tip over.

To mathematically prove that a mass of $45\ \text{kg}$ at "B" is the maximum mass that results in a normal force of zero (at "A"), model the plank as a rigid body in static equilibrium.

These are the two equations that form the basis of the rigid object in static equilibrium model: $$ \sum \vec{F} = 0 \ \ \text{and} \sum \vec{\tau_z} = 0$$ The sum of torques can be made about any axis.

Call the normal forces at "A" and "C" as $\vec{n_A}$ and $\vec{n_C}$ respectively. Let the plank's mass be $M = 15 \ \text{kg}$, and the boy's weight be $\vec{F_g}$ = $(45 \text{kg}).\vec{g}$

The net torque about an axis perpendicular to point "A" is zero, so: $$\sum \tau_A = 0 \rightarrow \ -Mg(\overline{\text{AO}}) \ + n_C(\overline{\text{AC}}) -F_g(\overline{\text{AB}}) = 0$$

Solving this equation should yield : $n_C = 588N$

The net force on the plank is also zero, so: $$\sum F = 0 \rightarrow Mg \ + F_g - n_C \ -n_A = 0$$ which yields $\boxed{n_A = 0}$

As an exercise, try using this method to caluclate $n_A$ if the mass of the boy was $46 \ \text{kg}$. You will get a negative normal force at "A" which is unphysical.

Hope this helps.

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  • $\begingroup$ Thanks for your answer sir! However, I had a question: what would happen if the boy's weight was $44kg$ instead of being $45kg$? $\endgroup$
    – tryingtobeastoic
    Commented Mar 14, 2022 at 6:41
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    $\begingroup$ Put it in the same equations (Try it, I mean it!). You'll get a positive normal force at "A", signifying that the plank is at contact with the pillar. Make it exactly 45 kg and the plank barely has contact with the pillar (a limiting case). Make it greater than 45 kg and the plank is no longer in contact with the pillar. See the pattern? $\endgroup$
    – Cross
    Commented Mar 14, 2022 at 6:48
  • $\begingroup$ I understood your answer and I accepted it. Sir, I had another question. Is it the convention to consider downwards forces negative and upwards forces positive? If I break the convention, can my teacher deduct marks from me? $\endgroup$
    – tryingtobeastoic
    Commented Mar 14, 2022 at 11:57
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    $\begingroup$ Not at all. In my example I considered downward forces to be positive. It is generally convenient to use the Cartesian sign convention, but it's certainly not a fixed rule. As for whether or not you'll lose marks, you'd have to ask your teacher that question. I personally wouldn't. $\endgroup$
    – Cross
    Commented Mar 14, 2022 at 12:09
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    $\begingroup$ It indicates that we are finding the net external torque through an axis parallel to the z - axis. I used this because most problems involve forces that lie on the xy - plane ( such as this one ). It is not a necessity though, and the general condition is just $\sum \vec{\tau_{ext}} = 0$ $\endgroup$
    – Cross
    Commented Mar 15, 2022 at 4:15

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