A 1.6 m long ladder leans against a frictionless wall...?

Question

A 1.6 m long ladder leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.30. What is the minimum angle the ladder can make with the floor without slipping? Where do I start and how do I do this without knowing the mass?

Answer 1

Don't worry: the ladder's mass will cancel out of the final equation. :-) You do, however, need to know where the ladder's center of gravity is. If the ladder is top-heavy, it will slip at a different angle than if it's bottom-heavy. So we'll just assume that the center of gravity is right in the middle of the ladder.

You need to consider both the forces on the ladder and the _torques_ on the ladder.

Since the ladder is not accelerating (linearly), the sum of the forces must be zero. And since the ladder has no _angular_ acceleration, the sum of the _torques_ must also be zero.

For reference, let's say the ladder is leaning to the right.

First consider linear forces.

Vertical forces on ladder:
mg (down)
Fn (normal force from floor, pushing up)
They must cancel, so:
(1) Fn = mg

Horizontal forces on ladder:
Fw (force of wall, pushing to left)
Ff (force of friction from floor, pushing to right)

They must cancel, so:
(2) Fw = Ff

At the point of slippage, Ff = maximum friction = Fn(μ) = mg(μ), so let's substitute that value into (2):
(3) Fw = mg(μ)

Now consider the torques. Notice that, since the ladder is not actually rotating, the choice of a rotation axis is arbitrary. So, for convenience, let's choose the rotation axis to be at the point where the ladder touches the floor.

In the following, let "L" be the length of the ladder:

The wall exerts a counter-clockwise torque:
ccw_torque = Fw × (lever arm) = Fw(L)(sinθ)

Assume the weight mg acts act the center of the ladder (this is where it's important to know the center of gravity). The weight exerts a clockwise torque:
cw_torque = mg × (lever arm) = mg(L/2)(cosθ)

Note that the friction force Ff and normal force Fn (from the floor) don't contribute at all to the torque. That's because those lines of force pass directly through the axis. (And that's why this choice of an axis is "convenient"!)

So, the counter-clockwise torque must balance the clockwise torque:
(4) Fw(Lsinθ) = mg(L/2)(cosθ)

Combine this with (3) to get:
(5) mg(μ)(L)(sinθ) = mg(L/2)(cosθ)

Divide both sides by mgL (see, I told you "m" would cancel out) and rearrange to get:
(6) tanθ = 1/(2μ)