P0632 Mechanics › Rotational Motion

Static Equilibrium of a Leaning Ladder

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Rotational Motion Beginner rigid body balance

Source: High school physics (Chinese)

Problem Sets:

rigid body discussion set

Problem

A ladder AB of length $L = 4.0$ m and weight $G = 100$ N leans against a smooth vertical wall. Its center of mass C is located a distance $d = 1.5$ m from its lower end B. The ladder is in static equilibrium, making an angle $\theta = 20^\circ$ with the wall.

  1. What is the normal force exerted by the wall on the ladder?
  2. What are the components of the force exerted by the ground on the ladder?

N_A = 13.6 N N_B = 100 N f_B = 13.6 N

For the ladder to be in static equilibrium, the net force and net torque acting on it must be zero. Let $N_A$ be the normal force from the smooth wall, $N_B$ be the normal force from the ground, and $f_B$ be the static friction force from the ground.

  1. Sum of vertical forces: $\Sigma F_y = 0$
$$N_B - G = 0 \implies N_B = G$$
  1. Sum of horizontal forces: $\Sigma F_x = 0$
$$N_A - f_B = 0 \implies f_B = N_A$$
  1. Sum of torques about point B (the base of the ladder): $\Sigma \tau_B = 0$. Let counter-clockwise be positive. The lever arm for the weight $G$ is $d \sin\theta$, and for the wall force $N_A$ it is $L \cos\theta$.
$$N_A (L \cos\theta) - G (d \sin\theta) = 0$$ $$N_A = \frac{G d \sin\theta}{L \cos\theta} = \frac{G d}{L} \tan\theta$$

Then, we can find $f_B$ using the horizontal force balance.