# Leaning sticks on a horizontal surface

1. Mar 4, 2013

### zeralda21

1. The problem statement, all variables and given/known data

A body is composed of two straight pins that are joined at a right angle. They have lengths α and β and the mass per unit length is ρ. 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.

Picture: http://i.imgur.com/Hr1RBRF.jpg

2. Relevant equations

I think it is:
Moment: M=F*s
Equilibrium equation

3. The attempt at a solution

I begin by drawing a free body diagram and mark out all forces. The following four so far:

Two upward normal forces from the ground located at each endpoint. Call them N_1 and N_2.
Two downward gravitational forces, m_1g and m_2g, located at a/2 and b/2.

Clearly it is not moving so the equations equilibrium and moment must therefore be zero.

x-direction: No acting force
y-direction:N_1+N_2=m_1g+m_2g

I have computed the moment around several points but I end up nowhere. Especially not the desired result D. The fact that the sticks are joined at a right angle means that the Pythagorean Theorem can be used, hence sqrt(a^2+b^2).

Last edited: Mar 4, 2013
2. Mar 4, 2013

### haruspex

Pls post your attempt that uses the left-hand contact point.

3. Mar 6, 2013

### zeralda21

This is how far I got:

Equilibrium:
x-direction: No acting forces in this direction.
y-direction:$$N_1+N_2-m_1g-m_1g=0$$ and therefore $$N_1+N_2=m_1g+m_2g$$
We know that the density is constant along the sticks and is ρ. So:
$$m_1g=(ρag)/2$$ and $$m_2g=(pbg)/2$$.

So we can conclude that $$N_1+N_2=\frac{1}{2}ρg(a+b)$$.

Now we proceed by calculating the moment around the left-hand contact point:

$$\frac{pag}{2}\frac{a}{2}\frac{\sqrt{2}}{2}+\frac{pbg}{2}\frac{b}{2}\frac{\sqrt{2}}{2}-N_2\sqrt{a^2+b^2}=0$$ and therefore

$$N_2=\frac{ρg}{4\sqrt{2}}\sqrt{a^2+b^2}$$. Clearly false.

4. Mar 6, 2013

### zeralda21

I should maybe motivate why I use sin(45). Drop a bisection h from the right angle and call the base of the new triangle, c. Since we bisect the triangle, we split the 90 degrees. We know also have two uniform triangle.

5. Mar 6, 2013

### zeralda21

I have recently made some progress so I update. still wrong though, really confusing.

As I said previously, we can bisect the angle with a downward altitude, call it h. Recall that the area of the triangle is:

$$\frac{ab}{2}=\frac{\sqrt{a^2+b^2}h}{2}$$ and we have $$h=\frac{ab}{\sqrt{a^2+b^2}}$$. It follows now that $$sin\alpha=\frac{h}{a}=\frac{b}{\sqrt{a^2+b^2}}$$.

We proceed now by calculating the moment around the left contact-point and recall that $$M=Frsin\alpha$$

$$(\frac{ρag}{2})(\frac{a}{2})(\frac{b}{\sqrt{a^2+b^2}})+(\frac{ρbg}{2})(\frac{b}{2})(\frac{a}{\sqrt{a^2+b^2}})-N_2\sqrt{a^2+b^2}=0$$

Simplifying and express in terms of N_2 gives:

$$N_2=\frac{ρabg}{4(a^2+b^2)}(a+b)$$ which is still WRONG....

6. Mar 6, 2013

### haruspex

$(\frac{ρag}{2})$ ... $(\frac{ρbg}{2})$
why /2 ?
...$(\frac{b}{\sqrt{a^2+b^2}})$
Don't you want cos α, not sin α?
...$(\frac{a}{\sqrt{a^2+b^2}})$
Now you want sin α, but even then (b/2) sin α is only the horizontal distance from the right hand contact (or from the apex) to the centre of mass of the B rod. You want the entire distance from the left-hand contact.

7. Mar 7, 2013

### zeralda21

No wait, I cannot be true that $$cos\alpha=sin\alpha$$. If $$sin\alpha=\frac{b}{\sqrt{a^2+b^2}}$$ then it follows from the pythagorean theorem that $$cos\alpha=\sqrt{1-\frac{b^2}{a^2+b^2}}=\frac{a}{\sqrt{a^2+b^2}}$$

Ok, I have now solved it. I'll post my solution soon.

Last edited: Mar 7, 2013
8. Mar 7, 2013

### zeralda21

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 the false answer$$N_2=ρg$$. 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:

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

9. Mar 7, 2013

### haruspex

If you go back to my previous post you'll find two errors in the above:
1. The first two terms should be a3 and b3 respectively in the numerator. (What you have is dimensionally wrong.)
2. You're still not allowing for the fact that the centre of mass of the B rod is displaced by an extra distance a cos α from the left-hand contact.

10. Mar 7, 2013

### zeralda21

Thanks haruspex, I'll have a serious check on this before I post anymore stupid posts.