I like to solve physics problems.  Here is one for you (I just made it up).

A 4 meter ladder leans against a frictionless wall at a 30 degree angle.  The mass of the ladder is 10 kg.  A human stands 1 meter up the ladder and has a mass of 70 kg.  What is the minimum coefficient of static friction between the floor and the ladder so that the ladder doesn’t slip?

Here is the solution—in video form.

But wait! There’s more.  Let’s do another problem.

Suppose you have the same ladder and the same human.  However, in this case the coefficient of static friction between the ladder and the ground is 0.55.  How far up can the human move before the ladder slips?

I like this question because I don’t know how the answer will turn out.  That makes it fun.  So, let’s do it.

But what kind of problem is this?  I’ll make this a multiple choice question.  Here are your options.

1. A friction problem.
2. An equilibrium problem.
4. A work-energy problem.

Did you pick?  OK, I’ll tell you that there will be quite a few students that will say this is a friction problem because it has a coefficient of friction.  That’s not untrue—but it’s not a good way to classify the problem.  You could also say this is a “ladder problem”—again, not untrue but not helpful.

This is an equilibrium problem.  We are trying to find the point at which the ladder slips—the point it leaves equilibrium, but it’s still sort of in equilibrium.

For an object in equilibrium, there are two main ideas—especially for a rigid object.  First, it must have zero net force.  Second, it must have zero net torque about any point.  In two dimensions, I can write these two conditions as the following three equations.

$F_\text{net-x}=0$

$F_\text{net-y} = 0$

$\tau_\text{net-o} = 0$

Let’s talk about the torque stuff.  I don’t want to get into a whole thing about torque, so let me just say that torque is like a “rotational force” and it can be calculated as:

$\tau = Fr\sin\theta$

In this expression, F is the applied force, r is the distance from the point at which you are calculating torque, and $\theta$ is the angle between r and F.  This equation is equivalent to F-perpendicular times r or F times r-perpendicular.

Oh, torques that would make an object rotate clockwise are negative.

One last thing about torques—especially the sum of the torques.  If an object is in rotational equilibrium about some point (point “o”) then it is also in rotational equilibrium about any other point.  When you set up the torque equation, you can pick whatever point you like to sum the torques—it’s your choice (but choose wisely).

Now we can start setting up some equations for equilibrium.  I’ll start with a force diagram for the ladder.

I know that’s a little busy—but it will have to do.  Here are some comments.

• There are two normal forces.  One from the wall (labeled 1) and one from the floor.
• There are two gravitational forces—and this is a cheat.  There is a gravitational force on the ladder and it is as though it is one force acting at the center of mass for the ladder. If the ladder has a uniform density, the center of mass is the center of the ladder.
• The other gravitational force is fake.  This force $m_2\vec{g}$ is there to represent the weight of the human.  But that gravitational force acts on the human, not the ladder.  The ladder pushes UP on the human the same as the weight.  Since forces come in pairs, an upward pushing force from the ladder means there is a downward pushing force from the human on the ladder.  It’s equal to mg, but not mg.
• The friction force is parallel to the floor.  The maximum magnitude of this friction force would be: $F_f = \mu_s N_2$ where $\mu_s$ is the coefficient of static friction.

Now for some equations.  First, this is the sum of the forces in the y-direction.

$F_\text{net-y} = 0 = N_2 - m_1 g-m_2g$

Just a quick reminder.  These are not vectors.  These are components of vectors in the y-direction.  That’s why the two gravitational forces have a negative sign.  I guess I can simplify this a little bit.

$N_2 = (m_1 + m_2)g$

Next is the sum of forces in the x-direction.

$F_\text{net-x} =0=N_1 - F_f$

Since we are at the point of maximum friction, I can include the expression for the frictional force in terms of the coefficient.

$N_1 = \mu_s N_2$

Notice that I know the value of $N_2$ since it only depends on the mass of stuff—but I don’t know $N_1$.  I’m going to need another equation. That’s where the sum of the torques comes in.

In order to write down the sum of the torques, I need to pick a point about which I calculate the torque.  I’m going to go with the bottom of the ladder.  At this point, there are two forces applied.  Since they are applied at the point about which the torque is calculated, they contribute zero torque and they won’t appear in the equation.  Winning.

Here is the sum of the torques about point O (at the bottom of the ladder).

$\tau_\text{net-o} = 0 = m_2gs\cos\theta +m_1g\left(\frac{L}{2}\right)\cos\theta - N_1 L\sin\theta$

What do I want to solve for here?  I want the distance the human goes up the ladder.  This is the value “s” in the expression above.  Really, I have all the values I need in that expression to solve for s.  I just need to set $N_1$ to the maximum frictional force.  But that would be boring.

Instead, let me make a plot of the frictional force as a function of human distance up the ladder.  That will be more fun, right?

Here’s what I get.

From this plot, the human can go up 1.16 meters before the required frictional force exceeds the maximum.  Oh, here is the code for that plot.

Now that you have the code, you can change stuff—like the angle of the ladder or the mass of the human or whatever.

The end.