A firefighter of mass 120 kg slides down a vertical pole with an acceleration of 4 m/s per second. What is the friction force that acts on the firefighters?
In this case, the friction force is the force that is acting against gravity.
Use F=M*a
If g=9.8m/s^2 and the firefighter's acceleration is 4m/s^2 then (9.8-4)m/s^2 is the deceleration suppied by the firefighter's hands (or legs, whatever part of the body he/she is using to decelerate). So the force is
F=M*(9.8-4)m/s^2 where M is the firefighter's mass.
480kgm/s^2
Thank you
First, calculate the firefighter's weight:
Weight = mass x gravity
Weight = 120 kg x 9.8 m/s^2
Weight = 1176 N
Next, calculate the net force acting on the firefighter:
Net force = mass x acceleration
Net force = 120 kg x 4 m/s^2
Net force = 480 N
Since the firefighter is sliding down the pole, the net force is equal to the friction force plus the weight:
Friction force + Weight = Net force
Friction force = Net force - Weight
Friction force = 480 N - 1176 N
Friction force = -696 N (Negative because it acts in the opposite direction of the weight)
Therefore, the friction force that acts on the firefighter is 696 N.
To calculate the friction force acting on the firefighter, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's determine the magnitude of the acceleration caused by gravity. Near the surface of the Earth, the acceleration due to gravity is approximately 9.8 m/s^2.
Therefore, the firefighter's acceleration is 4 m/s^2 downwards.
To find the friction force, we need to subtract the acceleration caused by gravity from the firefighter's acceleration. So, the deceleration caused by the firefighter's hands or legs (used to control the descent) is (9.8 - 4) m/s^2.
Now, we can calculate the friction force using the formula F = M * a, where F is the force, M is the mass of the firefighter, and a is the deceleration.
Given that the firefighter's mass is 120 kg and the deceleration is (9.8 - 4) m/s^2, we can substitute these values into the formula:
F = 120 kg * (9.8 - 4) m/s^2
F = 120 kg * 5.8 m/s^2
Calculating this expression results in a friction force of 696 N (Newtons) acting on the firefighter during the slide down the vertical pole.