Question

A ladder of length L = 2.8 m and mass m = 16 kg rests on a floor with coefficient of static friction ?_s = 0.51. Assume the wall is frictionless.

1)

What is the normal force the floor exerts on the ladder?

2) What is the minimum angle the ladder must make with the floor to not slip?

3) A person with mass M = 69 kg now stands at the very top of the ladder. What is the normal force the floor exerts on the ladder?

4) What is the minimum angle to keep the ladder from sliding?

Answer 1

Fw is the force from the wall on the ladder. It is horizontal.
N is the Normal Force. It is vertical.
Ff is the friction force, it points to the right.
u is the coefficient of friction.
Wp is the weight of the person equals Mass times gravity (M g).
Wl is the weight of the ladder, it is mass times gravity. (m g)
L is the length of the ladder.

The diagram has no angle, lets call it theta.
All of our forces are horizontal or vertical. Lets try the old "sum of the forces = zero".
In the y direction we get:
0 = N - Wp - Wl
N = Wp + Wl
and at the minimum angle,
Ff = u * N
Ff = u * (Wp + Wl)
but for my next trick I will need Fw. Well the sum of the forces in the x direction equals zero
Fw = Ff = u * (Wp + Wl)




Again, ignore the T parts.

We have to break the other forces (Wp, Wl and Fw) into components that are either perpendicular to the ladder or tangential, along the ladder.

Now sum of the moments about the base.....yields this

0 = (L/2) * Wl * cos theta + L * Wp * cos theta - L *Fw * sin theta



L * Fw * sin theta = (L/2) * Wl * cos theta + L * Wp * cos theta

First pull out all the L's and get rid of them and pull out the cos
.
Fw * sin theta = cos theta ( (1/2) * Wl + Wp)

sin theta = cos theta * ( (1/2) * Wl + Wp) /Fw

sin theta /cos theta = ((Wl / 2) + Wp) /Fw

tan theta = ((Wl / 2) + Wp) /Fw

Now substitute in Fw = u * (Wp + Wl)
tan theta = ((Wl / 2) + Wp) /(u * (Wp + Wl))

Sorry if you don't like keeping it all in variables, feel free to stop and crunch out any numbers along the way, for N, for Fw and Fw, for Wp and Wl. Personally, I like to try to get one equation that has what the question asked for on the left side and only the variables given in the question on the left.

So now we can put in Wp = M g and Wl = m g for all weights, and then delete the gravity on the top and on the bottom.

tan theta = ((m / 2) + M) /(u * (M + m))

theta = arctan ((m / 2) + M) /(u * (M + m))

There now just plug in m, M and u

theta = arctan ((16 kg / 2) + 69 kg) /(0.56 * (69 kg + 16 kg))

theta = arctan [77 kg /(0.56 * 85 kg)[

theta = arctam (1.6176)

theta = 58.276 degrees