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, we need to consider the forces acting on the firefighter during different phases of his motion. Let's break it down step-by-step:
Step 1: Find the acceleration of the firefighter during the sliding phase.
We can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration, and s is the distance traveled. Plugging in the values, we have:
v^2 = 0 + 2 * a * 2.5
v^2 = 5a
Step 2: Find the acceleration of the firefighter during the stopping phase.
Given that the firefighter bends his knees when he reaches the ground, we can assume that his deceleration is constant during this phase. We can use the equation:
v = u + at
where v is the final velocity (zero in this case), u is the initial velocity (the final velocity during the sliding phase), a is the acceleration, and t is the time taken. Plugging in the values, we have:
0 = v + a * 1.8
v = -1.8a
Step 3: Equate the accelerations during both phases.
The acceleration during the sliding phase and the stopping phase must be equal. Therefore, equating the acceleration values found in steps 1 and 2, we have:
5a = -1.8a
Solving for a, we find:
6.8a = 0
a = 0 m/s^2
Step 4: Calculate the force of gravity acting on the firefighter.
The force of gravity can be calculated as the mass of the firefighter multiplied by the acceleration due to gravity (9.8 m/s^2):
Force_gravity = mass * acceleration_due_to_gravity
Force_gravity = 75 kg * 9.8 m/s^2
Force_gravity = 735 N
Step 5: Calculate the normal force acting on the firefighter.
The normal force is the force exerted by the ground on the firefighter and acts perpendicular to the surface of contact. Since the acceleration is zero during the stopping phase, the net force acting on the firefighter is also zero.
Therefore:
Normal_force - Force_gravity = 0
Normal_force = Force_gravity
Normal_force = 735 N
So, the normal force of the ground acting on the firefighter during the stopping phase of his motion is 735 N.
To solve this problem, you need to consider the forces acting on the firefighter during different phases of his motion. Let's break it down step by step:
Step 1: Calculate the acceleration of the firefighter.
Using the formula for motion in one dimension, s = ut + (1/2)at^2, where:
s = distance (2.5 m),
u = initial velocity (0 m/s), and
t = time (1.8 s),
we can solve for the acceleration (a).
Rearranging the formula gives us: a = (2s - 2ut) / t^2.
Plugging in the values, we get: a = (2 * 2.5 - 2 * 0 * 1.8) / (1.8)^2.
Step 2: Calculate the net force acting on the firefighter while sliding.
The net force is equal to the product of mass (m) and acceleration (a), according to Newton's second law, F = ma. The force of gravity (weight) acts vertically downward, so the net force in the vertical direction is the difference between the upward normal force and the downward force of gravity.
Step 3: Calculate the force exerted by the firefighter's knees.
When the firefighter bends his knees and brings his center of mass to a stop, the force exerted by the knees acts vertically upward (opposite to the force of gravity).
Step 4: Calculate the normal force of the ground.
During the stopping phase, the normal force acting on the firefighter must equal the sum of the force of gravity and the force exerted by the knees to maintain equilibrium.
To summarize, the normal force is equal to the sum of the weight and the force exerted by the knees:
Normal force = weight + force exerted by the knees.
Now, you can calculate each of these forces and find the normal force.