A person standing on a ladder : Static Equilibrium

[Chandasouk]

Homework Statement

A uniform ladder with mass m_2 and length L rests against a smooth wall. A do-it-yourself enthusiast of mass m_1 stands on the ladder a distance d from the bottom (measured along the ladder). The ladder makes an angle theta with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude f between the floor and the ladder. N_1 is the magnitude of the normal force exerted by the wall on the ladder, and N_2 is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive. None of your answers should involve pi (i.e., simplify your trig functions).

What is the minimum coeffecient of static friction mu_min required between the ladder and the ground so that the ladder does not slip?
Express mu_min in terms of m_1, m_2, d, L, and theta.

Suppose that the actual coefficient of friction is one and a half times as large as the value of mu_min. That is, \mu_{\rm s} = (3/2)\mu_{\rm min}. Under these circumstances, what is the magnitude of the force of friction f that the floor applies to the ladder?
Express your answer in terms of m_1, m_2, d, L, g, and theta. Remember to pay attention to the relation of force and mu_s.

Homework Equations

Net Torque = o

Net Force = 0

The Attempt at a Solution

I need a lot of help on this. First off, I found that

N_2 = (m_1+m_2)g

Then, upon Masteringphysic's suggestion

The best choice of origin in which to add up the torques would be the point at which the ladder touches the wall. By summing torques around this point we can ignore the normal force N_1 (because the moment arm has zero length) and find mu_s in terms of already given quantities.

So the pivot point is the point where the ladder touches the wall.

Here is where It went bad

I tried to "Sum the torques given in the problem about the point at which the ladder touches the wall."

Originally, I had

-(m_2*g)(L/2)-(m_1g)(L-d)-f+N_2

which was incorrect. I eventually gave up and found that the answer was

-m_1*g*(L-d)*cos(theta)-(m_2/2)*L*g*cos(theta)-f*L*sin(theta)+N_2*L*cos(theta) = 0

I do not understand how they obtained this answer. Mainly, I don't understand why it was necessary to include sin and cos theta.

Regardless, I tried to find \mu

I first wanted to find out what f equaled

so, algebraically solving for f from the net torque equation above

it yielded

f = \frac{-m_1*g(L-d)cos(theta)-m_2*g(L/2)cos(theta)+(m_1+m_2)gLcos(theta)}{Lsin(theta)}


I then divided f by N_2 which equated to (m_1+m_2)g and got

\frac{-m_1*g(L-d)cos(theta)-m_2*g(L/2)cos(theta)+(m_1+m_2)gLcos(theta)}{Lsin(theta)}

divided by

\frac{(m_1+m_2)g}{}

Which came out to be incorrect

 

[willem2]

Your answer can be much more simplified. You can get rid of g altogether.

 

[Chandasouk]

It'd be simplified to

f =\frac{-m_1(L-d)cos(theta)-m_2(L/2)cos(theta)+(m_1+m_2)Lcos(theta)}{Lsin(theta)}

divided by

{(m_1+m_2)}{} ?