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A body is composed of two straight pins that are joined at a right angle. They have lengths $a$ and $b$ and the mass per unit length is $\rho$. When the body is balanced on a flat surface, as shown, how large is the normal force against the ground in the right point of contact? 4 options as can be seen in the picture.


Let me point out that this is a conceptual question. When I first tried to solve this problem I decided to choose to calculate the moment around the left contact-point in order to reduce one term(left normal force). This seems like a natural way but gives a false answer($N_2= \rho g$. If I instead calculate about the vertex of the triangle and use Newton's second law I get the correct solution.(answer is D).

So how should I choose the "correct" point?

Around left contact point:

By dropping an altitude h at the right angle we get that:

Notice that: $\cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$, $\cos\beta=\frac{b}{\sqrt{a^2+b^2}}$ and $m_ag=\rho ag$ and $m_b=pbg$

$-\frac{\rho a^2g}{\sqrt{a^2+b^2}}-\frac{\rho b^2g}{\sqrt{a^2+b^2}}+N_2\sqrt{a^2+b^2}=0$. Simplifying gives $N_2=\rho g$

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Related: (Which you should probably have linked just so that it didn't look like you were trying to sneak it past...). I don't believe we have a policy on problem solving strategy questions as yet. I'm inclined (heh!) to be against, but won't take unilateral action on the matter. In any case, answerers should stick to the strategy question and not to answering the particulars of this problem. – dmckee Mar 7 '13 at 21:59
Assuming that the last line there is meant to be the torques (moments in engineer-speak) about the point where the short piece hits the ground you've written both contributions from the masses incorrectly. They both act through their own CoG... – dmckee Mar 7 '13 at 22:27
dmckee: Yes,that point. I am not really following: Both masses act from the midpoint on each stick, that is $\frac{a}{2}$ and $\frac{b}{2}$. Altitude h can easily be computed since $\frac{ab}{2}=\frac{h\sqrt{a^2+b^2}}{2}$. It follows that $h=\frac{ab}{\sqrt{a^2+b^2}}$ Hence the values for $\cos\alpha$ and $\cos\beta$ – EricAm Mar 7 '13 at 22:36
Personally, I wouldn't chose a method that required computing $h$. I mean, why bother? You problem when working from one contact point is that while the effective distance for the torque on that leg is easy the effective distance for the toque on the other leg has a messy form involving a subtraction from $\sqrt{a^2 + b^2}. Yuck. Working from the joint means using all four forces, but they have more tractable distances. – dmckee Mar 7 '13 at 22:43
Sure, I see your point. Meanwhile, I see no way of finding $\cos\alpha$ and $\cos\beta$ without introducing h. Right? – EricAm Mar 7 '13 at 22:50
up vote 4 down vote accepted

There is no "correct" point. Using any point will give you the same answer. If you didn't get the same answer using both points then you've done something wrong.

Hint: you can treat the gravitational force as though it were acting at the shape's center of mass.

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If there is no "correct" point then one would naturally choose the left hand since it will 1)reduce one term and 2)reduce N_1 from the equation. However I end up with two different answers. For the left contact-point I get [tex]N_2=\rho g[/tex] and when choosing about the vertex of the triangle I get answer D. I can edit and include it in my post above. – EricAm Mar 7 '13 at 22:02
While there are no correct point there are some for which the computation is simpler. In this case doing it from the joint means that you don't have to compute one of the distances as a difference with $\sqrt{a^2 + b^2}$ and simplifies the algebra. Alas, unless you great intuition sometimes the only way to know that is to try it a couple of ways. – dmckee Mar 7 '13 at 22:36

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