Write an expression for the magnitude of the force, F, exerted on the firefighter by the pole.

Question

A firefighter, whose mass (including clothing and equipment) is m = 85 kg, hears the alarm and slides down the pole with a constant downward acceleration of magnitude a = 3.23 m/s^2.

Answer in terms of the variables from the problem statement as well as g for the acceleration due to gravity.

Answer 1

Hi Benson,

The firefighter descends vertically down with an acceleration a that is smaller than that of free fall g=9.8m/s^2

(the latter would be caused by the force of gravitation pull from Earth when no other vertical forces are present).

Now, when the firefighter slides down, he holes in the radio and the force of kinetic friction F_k between his hands/feets and the pole which is opposite to the (downward) velocity would be directed upwards, in other words it will be opposite to the force of gravitational pull. Therefore the net force acting on the firefighter of mass m would be equal to mg-F_k

and the net downward acceleration of the firefighter would be

a_net=(mg-F_k)/m (1)

From (1) we solve for F_k

F_k=m(g-a_net)=85kg(9.81m/s²-3.23m/s²)=560N

(retaining the lowest number (two) significant figures

Hope that would help

Dr Ariel B.

Answer 2

To solve this problem, let's first identify all of the forces acting on the firefighter as they slide down the pole. The force of gravity pulls the firefighter towards the ground while the friction force on the firefighter by the pole (F_fr) acts in the opposite direction of the force of gravity. The net force on the firefighter can therefore be expressed by the following equation:

F_net = F_g - F_fr = mg - F_fr

According to Newton's second law of motion, the net force on a body is equal to the mass of the body times its acceleration. Using this information, we get:

ma = mg - F_fr

F_fr = mg - ma

F_fr = m(g-a) = (85kg)(9.8m/s² - 3.23m/s²) = 560N