Question

A 96.97 -kg person is halfway up a uniform ladder of length \(3.433 \mathrm{~m}\) and mass 24.91 kg that is leaning against a wall. The angle between the ladder and the wall is \(\theta=27.30^{\circ}\). Assume that the friction force between the ladder and the wall is zero. What is the minimum value of the coefficient of static friction between the ladder and the floor that will keep the ladder from slipping?

Short answer

Answer: The minimum value of the coefficient of static friction is approximately 0.365.

Step-by-step solution

Identify the forces acting on the ladder and the person

The forces acting on the ladder are: - The gravitational force of the ladder (downward) at its center of mass: \(F_{grav,ladder} = m_{ladder} g\) - The gravitational force of the person (downward) at the person's position on the ladder: \(F_{grav,person} = m_{person} g\) - The normal force exerted by the wall on the ladder (horizontal) at the top: \(F_{normal,wall}\) - The normal force exerted by the floor on the ladder (vertical) at the bottom: \(F_{normal,floor}\) - The friction force exerted by the floor on the ladder (horizontal) at the bottom, opposite to the direction of the normal force exerted by the wall: \(F_{friction} = \mu F_{normal,floor}\)

Set up the static equilibrium conditions

The ladder is in static equilibrium, which means the net force and the net torque acting on it must be zero. Let's set up the equations for the net force in the x-direction, the net force in the y-direction, and the net torque about the bottom point of contact with the floor. 1. Net force in the x-direction: \(F_{friction} - F_{normal,wall} = 0\) 2. Net force in the y-direction: \(F_{normal,floor} - F_{grav,ladder} - F_{grav,person} = 0\) 3. Net torque about the bottom point of contact with the floor: \(F_{normal,wall}L\sin(\theta) - F_{grav,ladder}(\frac{1}{2}L\cos(\theta)) - F_{grav,person}(L\cos(\theta)) = 0\)

Solve for the normal forces

From the net force in the x-direction equation, we find the normal force exerted by the wall: \(F_{normal,wall} = F_{friction}\) From the net force in the y-direction equation, we find the normal force exerted by the floor: \(F_{normal,floor} = F_{grav,ladder} + F_{grav,person}\)

Solve for the minimum coefficient of static friction

To find the minimum value of the coefficient of static friction, plug the normal forces in the net torque equation: \(F_{friction}L\sin(\theta) - F_{grav,ladder}(\frac{1}{2}L\cos(\theta)) - F_{grav,person}(L\cos(\theta)) = 0\) Divide both sides by \(L\cos(\theta)\): \(\frac{F_{friction}}{\cos(\theta)} - \frac{F_{grav,ladder}}{2} - F_{grav,person} = 0\) Now, substitute the expression for \(F_{friction}\) and solve for the minimum coefficient of static friction: \(\frac{\mu F_{normal,floor}}{\cos(\theta)} = \frac{F_{grav,ladder}}{2} + F_{grav,person}\) Rearrange to solve for \(\mu\): \(\mu = \frac{\frac{F_{grav,ladder}}{2} + F_{grav,person}}{F_{normal,floor}\cos(\theta)}\)

Calculate the minimum coefficient of static friction

Plug in the given values and calculate the minimum coefficient of static friction: \(\mu = \frac{\frac{(24.91 kg)(9.81 m/s^2)}{2} + (96.97 kg)(9.81 m/s^2)}{(24.91 kg + 96.97 kg)(9.81 m/s^2)\cos(27.30^{\circ})}\) After calculating the expression, we get: \(\mu \approx 0.365\) The minimum value of the coefficient of static friction between the ladder and the floor that will keep the ladder from slipping is approximately 0.365.