A 75kg firefighter slides down a 2.5m fire pole from rest. It takes the firefighter 1.8s to reach the ground. If the firefighter bends his knees when he reaches the ground such that his centre of mass mives 41cm while coming to a stop, what must the normal force of the ground acting on him have been during the stopping phase of his motion?
(I thought if I just found the Force of gravity i could solve for the normal force but I think that's too simple. Basically, not sure how to approach this question.)
To solve this problem, you need to consider the forces acting on the firefighter during various phases of their motion. Let's break it down step-by-step:
1. Initial phase (sliding down the fire pole):
During this phase, only the force of gravity is acting on the firefighter, pulling them downwards. The force of gravity is given by the formula F_gravity = m * g, where m is the mass of the firefighter and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Calculating the force of gravity: F_gravity = 75 kg * 9.8 m/s^2 = 735 N
2. Final phase (coming to a stop with knees bent):
At the moment when the firefighter's knees bend and stop their downward motion, they experience an upward force known as the normal force. This is the force exerted by the ground on the firefighter, opposing the force of gravity.
To determine the normal force, you need to consider the change in momentum of the firefighter as their center of mass moves 41 cm (converted to meters) while coming to a stop. The momentum change is given by the formula ΔP = m * Δv, where ΔP is the change in momentum, m is the mass of the firefighter, and Δv is the change in velocity.
Calculating the change in velocity: Δv = 0 m/s (since the firefighter comes to a stop)
Calculating the change in momentum: ΔP = m * Δv = 75 kg * 0 m/s = 0 kg⋅m/s
According to the impulse-momentum principle, the change in momentum is equal to the net force acting on the firefighter multiplied by the time taken to come to a stop. In this case, the net force is the normal force and the time is 1.8 seconds.
Using the formula for impulse: ΔP = F_net * Δt
Substituting the values: 0 kg⋅m/s = F_net * 1.8 s
Solving for the net force (normal force): F_net = 0 kg⋅m/s / 1.8 s = 0 N
Therefore, the normal force of the ground acting on the firefighter during the stopping phase of their motion is 0 N. This implies that the firefighter's knees must have absorbed the entire force required to bring them to a stop.