Question

A uniform 210 N ladder rests against a perfectly smooth wall, making a 30.0 degree angle with the wall.

Find the normal forces that the wall on the ladder. Express in N

Find the normal force that the floor exert on the ladder. Express in N

What is the friction force on the ladder at the floor? Express in N

Answer 1

if the ladder is not accelerating and not rotating, we know that:

1) sum of all horizontal forces =0
2) sum of all vertical forces =0
3) sum of all torques =0

the horizontal forces are the normal force from the wall (Nw) and friction; these must be equal and opposite, so we can write

Nw = friction

the vertical forces are the weight of the ladder (210N) and the upward normal force of the floor, Nf, these must be equal and opposite so we can write

Nf = 210N (this is the value of the normal force due to the floor)


finally, the sum of torques is zero; for simplicity, let's sum torques around the point where the ladder touches the floor, this way Nf and f will generate no torque around this point

the torque due to Nw is then Nw L cos 30 where L is the length of the ladder

the torque due to the weight of the ladder is

(210L)/2 sin 30 (the ladder is uniform so its weight acts at the center of gravity or at L/2)

these torques are equal so we have

Nw L cos30 = 210 (L/2) sin 30

or Nw=210/2 tan 30 = 60.62 N (notice that we do not need to know the length of the ladder)

since Nw = friction, this is the force of friction also

the normal force of the floor is 210N as determined earlier
(since Nw = friction = u Nf = 210u, we can find the coefficient of friction between the ladder and the floor:
210u=60.62 => u=0.2886)