Leaning sticks on a horizontal surface

[zeralda21]

Homework Statement

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

Homework Equations

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

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).

 

[haruspex]

I have computed the moment around several points but I end up nowhere.

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

 

[zeralda21]

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

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.

 

[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.

 

[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...

 

[haruspex]

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

$(\frac{ρag}{2})$ ... $(\frac{ρbg}{2})$
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.

 

[zeralda21]

$(\frac{ρag}{2})$ ... $(\frac{ρbg}{2})$
why /2 ?

Since the force $$m_1g=\frac{ρag}{2}$$ is located in the center of the sticks.
...$(\frac{b}{\sqrt{a^2+b^2}})$
Don't you want cos α, not sin α?
I don't think it actually matters since if we bisect the right angle and create a 90-45-45 triangle which means that $$cos\alpha = sin\alpha$$
...$(\frac{a}{\sqrt{a^2+b^2}})$
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.

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.

 

[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.

 

[haruspex]

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

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.

 

[zeralda21]

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.

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