A 75.0 kg ladder that is 3.00 m in length is placed against a wall at an angle theta. The center of gravity of the ladder is at a point 1.20 m from the base of the ladder. The coefficient of static friction at the base of the ladder is 0.800. There is no friction between the wall and the ladder. What is the vertical force of the ground on the ladder?
To find the vertical force of the ground on the ladder, we need to analyze the forces acting on the ladder and use Newton's second law of motion.
First, let's identify the forces acting on the ladder:
1. Weight (mg): This is the force due to gravity acting straight downward. The weight is given by the mass of the ladder (m = 75.0 kg) multiplied by the acceleration due to gravity (g ≈ 9.8 m/s^2).
2. Normal force (N): This is the force exerted by the ground on the ladder perpendicular to the surface. It counteracts the component of the weight that is perpendicular to the ground.
3. Friction force (F_friction): This is the force of static friction at the base of the ladder. It acts parallel to the ground and opposes the tendency of the ladder to slide or tip over.
Since the ladder is in equilibrium (neither sliding nor tipping), the sum of the forces in the horizontal and vertical directions is zero. We are interested in the vertical force, so let's focus on the vertical equilibrium equation:
Sum of vertical forces = 0
In the vertical direction, we have two forces: the weight (mg) acting downward and the normal force (N) acting upward. The friction force does not contribute to the vertical equilibrium equation since it acts parallel to the ground.
mg - N = 0
Now, let's solve for N:
N = mg
Substituting the given values:
N = (75.0 kg)(9.8 m/s^2)
N = 735 N
Therefore, the vertical force of the ground on the ladder is 735 N.