A person is stretching by putting their arms straight over their heads and bending forward at their ankles until their hands are pushing against the wall in front of them. As they stretch, they keep stepping back from the wall, leaning further forward. The coefficient of static friction between their hands and the wall and their feet and the ground is 0.4. Each hand experiences the same normal force from the wall, and each foot experiences the same normal force from the floor, though the hand normal force and the foot normal force may not be equal. Their center of mass is at the middle of their total length (feet to hands). At what angle between the floor and their body do they slip?
To determine the angle at which the person slips, we need to analyze the forces acting on the person at that moment. Let's break it down step by step.
1. When the person is not slipping, the static friction force between their hands and the wall and their feet and the ground provides the necessary force to prevent slipping. Thus, the maximum static friction force on each hand is equal to the normal force on the hand multiplied by the coefficient of static friction, i.e., F_hand = μ_s * N_hand. Similarly, the maximum static friction force on each foot is F_foot = μ_s * N_foot.
2. The normal force on each hand is equal to half the person's weight, as the center of mass is in the middle of their total length. Thus, N_hand = weight / 2.
3. The normal force on each foot is also half the person's weight, as both feet hold the same weight. Hence, N_foot = weight / 2.
4. At the slipping point, the person's weight will exert a torque around the point of contact between their feet and the ground. This torque must be equal to the torque exerted by the static friction force between their hands and the wall to keep the person from rotating.
The torque due to weight is given by the weight multiplied by the distance between the center of mass and the point of contact with the ground. In this case, the point of contact is the feet, and since the center of mass is at the middle of the person's total length, the distance is half the length.
Let's denote the length of the person from feet to hands as L. Thus, the torque due to weight is Torque_weight = (weight / 2) * (L / 2) = weight * L / 4.
The torque due to static friction on the hands is given by the static friction force multiplied by its perpendicular distance from the person's center of mass. The perpendicular distance when the person is leaning forward at an angle θ can be approximated as (L / 2) * sin(θ), as the person's body forms a triangle.
Hence, the torque due to static friction, Torque_static_friction = F_hand * (L / 2) * sin(θ) = (μ_s * N_hand) * (L / 2) * sin(θ) = (0.4 * weight / 2) * (L / 2) * sin(θ) = weight * L * sin(θ) / 16.
To prevent slipping, the torque due to static friction must equal the torque due to weight. Therefore, weight * L / 4 = weight * L * sin(θ) / 16.
Simplifying the equation, we get 4 = L * sin(θ) / 16.
To find θ, we isolate sin(θ) and solve for it: sin(θ) = 4 * 16 / L.
Finally, we can determine θ by taking the inverse sine (sin⁻¹) of the right-hand side: θ = sin⁻¹(4 * 16 / L).
Please note that the exact value of L (length from feet to hands) is not provided in the question, so you would need to substitute the appropriate value to find the specific angle.