# Equilibrium Ladder Problem

**1. The problem statement, all variables and given/known data**

A uniform ladder is 10 m long and weighs 180 N. In the figure below, the ladder leans against a vertical, frictionless wall at height h = 8.0 m above the ground. A horizontal force is applied to the ladder at a distance 1.0 m from its base (measured along the ladder).

(a) If F = 50 N, what is the force of the ground on the ladder, in unit-vector notation?

(b) If F = 150 N, what is the force of the ground on the ladder, in unit-vector notation?

(c) Suppose the coefficient of static friction between the ladder and the ground is 0.38; for what minimum value of F will the base of the ladder just start to move toward the wall?

**2. Relevant equations**

∑[itex]\tau[/itex] = r x F and ∑F = ma

**3. The attempt at a solution**

I got both parts (a) and (b) right so I don’t need those.

For part (c), I summed the horizontal forces: ∑F[itex]_{x}[/itex] = 0 = -F[itex]_{f}[/itex] + F[itex]_{A}[/itex] – N[itex]_{W}[/itex]

where F[itex]_{A}[/itex] is the applied force, F[itex]_{f}[/itex] is the frictional force, and N[itex]_{W}[/itex] is the normal force from the wall.

Then I summed the vertical forces: ∑F[itex]_{y}[/itex] = 0 = N – W

Then I summed the torques using the base of the ladder as the axis of rotation: ∑[itex]\tau[/itex] = 0 = Fsin(53) + 5(180)sin(37) – 10N[itex]_{W}[/itex]sin(53)

Then I used N[itex]_{W}[/itex] = μmg – F[itex]_{f}[/itex] and substituted it into the torque equation and then plugged in all my numbers to solve for F.

I got F = 133 N, which is wrong and I’m not sure what I’m doing wrong.

http://ift.tt/1drLaGR

## Leave a comment