How do I find the forces acting on a ladder?

[Eclair_de_XII]

Homework Statement

"To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be friction-less. The center of mass of the ladder is 2.00 m from the bottom. The person is standing 3.00 m from the bottom. Find the normal reaction and friction forces on the ladder at its base."

Homework Equations

$\theta=cos^{-1}(\frac{1}{3})=tan^{-1}(2\sqrt{2})$
$m_0=10kg$
$s_0=\frac{2 m}{tan\theta}$
$m_1=70kg$
$s_1=\frac{3 m}{tan\theta}$
$||s||=6m$

The Attempt at a Solution

I'm choosing my lever arm to be the point at which the ladder touches the ground.

$∑|τ|=-g(m_0s_0+m_1s_1)+||s||cos\theta⋅F_r=0$
$∑F_x=-gcos\theta(m_0+m_1)-F_r+f_s=0$

$F_r=f_s-gcos\theta(m_0+m_1)$
$F_r⋅||s||⋅cos\theta=g(m_0s_0+m_1s_1)$
$F_r=\frac{g}{||s||}sec\theta(m_0s_0+m_1s_1)$
$f_s-gcos\theta(m_0+m_1)=\frac{g}{||s||}sec\theta(m_0s_0+m_1s_1)$
$f_s=gcos\theta(m_0+m_1)+\frac{g}{||s||}sec\theta(m_0s_0+m_1s_1)$

Can anyone tell me if I'm going in the right direction? Thanks.

 

[TSny]

You have the correct approach. But I noticed a few things:

(1) It is always helpful to define the symbols that you are using. For example, I had to try to decipher your equations to decide what angle you are calling $\theta$. Likewise for the symbols $s_1$, $s$, and $s_0$.

(2) The problem states that "the center of mass of the ladder is 2.00 m from the bottom". I think this probably means that the center of mass is 2.00 m along the ladder starting from the bottom of the ladder. If they meant it the way you interpreted it, I think they would have said that the center of mass is 2.00 m above the ground. Similarly for the position of the person.

(2) You wrote the equation $∑|τ|=-g(m_0s_0+m_1s_1)+||s||cos\theta⋅F_r=0$. Check to make sure that the expression for the lever arm for $F_r$ is correct.

 

[TSny]

On more thing, it looks like you have assumed that the forces of gravity have a horizontal component. Is this right?

 

[Eclair_de_XII]

Check to make sure that the expression for the lever arm for $F_r$ is correct.

Oh, it's $||s||sin\theta⋅F_r$, right? $F_r$ acts horizontally, so the lever arm is the vertical component of $||s||$.

Is this right?

Forgive me. I think I should treat the ladder as a point mass for the summation of forces expression.

$∑F_x=f_s-F_r=0$
$∑F_y=-(m_0+m_1)g+F_N=0$

 

[TSny]

Oh, it's $||s||sin\theta⋅F_r$, right? $F_r$ acts horizontally, so the lever arm is the vertical component of $||s||$.

$∑F_x=f_s-F_r=0$
$∑F_y=-(m_0+m_1)g+F_N=0$

That all looks good.

 

[Eclair_de_XII]

$\theta=cos^{-1}(\frac{1}{3})$
$m_0=70 kg$
$s_0=3 m$
$m_1=10kg$
$s_1=32 m$

$∑|τ|=-gcos\theta(m_0s_0+m_1s_1)+F_r(||s||sin\theta)=0$
$F_r=(\frac{g}{||s||})(cot\theta)(m_0s_0+m_1s_1)=(\frac{9.8\frac{m}{s^2}}{6m})(\frac{1}{2\sqrt{2}})((70kg)(3m)+(10kg)(2m))=133 N$
$∑F_x=f_s-F_r=0$
$f_s=F_r=133N$

$∑F_y=F_N-g(m_0+m_1)=0$
$F_N=g(m_0+m_1)=784N$

 

[TomHart]

I tried to do a quick check and it looks like your results for F_r and F_N are correct.

 

[Eclair_de_XII]

Is that right? I probably need a new physics problem book because it says that $F_N=784N$ but $f_s=376N$.

 

[TomHart]

I just repeated the sum of the moment equation and once again calculated 132.8 N for F_r, which is equal to -f_s, since those are the only 2 horizontal forces. So I think that is the correct answer unless you and I are both doing something fundamentally wrong.

 

[TomHart]

It looks like the book incorrectly used (F_r)(||s||)(cosθ), instead of (F_r)(||s||)(sinθ) when calculating the moment from force F_r.