Question

A firefighter who weighs 712 N slides down a vertical pole with an acceleration of 3.00 m/s², directed downward. (a) What is the magnitude of the vertical force on the firefighter from the pole and (b) What is the direction (up or down) of the vertical force on the firefighter from the pole and (c) What is the magnitude of the vertical force on the pole from the firefighter? and (d) What is the direction of the vertical force on the pole from the firefighter?

Short answer

1. The magnitude of vertical force 494 N.
2. The direction of vertical force on the firefighter from the pole is vertically upwards.
3. Magnitude is 494 N.
4. The direction of vertical force on the pole from the firefighter is vertically downwards.

Step-by-step solution

Given information

1. The weight of a firefighter is W = 712 N,
2. Mass of a firefighter is, $\left(m=\frac{W}{9.8{\text{m/s}}^2}\right)=72.5\text{kg}$
3. The upward direction is positive.

Understanding the concept of force

As per Newton’s second law, the force on the object is equal to the product of mass and its acceleration. The net force on the object is the vector sum of all the forces acting on it. Newton’s third law states that for every action there is an equal and opposite reaction.

The weight of the firefighter, which is sliding down with a certain acceleration, is given. Therefore, we can calculate the desired result by using Newton’s law of motion and condition for the net force.

Formulae:

${\mathit{v}}_{\mathbf{f}}^{\mathbf{2}}\mathbf{=}{\mathit{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{\Delta }\mathit{x}\mathbf{}\mathbf{}\mathbf{}\mathbf{}$ (1)

${\mathit{v}}_{\mathbf{f}}\mathbf{=}{\mathit{v}}_{\mathbf{0}}\mathbf{+}\mathit{a}\mathit{t}$ (2)

F = ma (3)

(a) Calculate the force exerted by the pole on the firefighter

The net force on the object is equal to the difference between the force exerted by the pole (F_p) and the weight of the firefighter (F_g) . The net force is equal to the product of the mass of the firefighter and the acceleration. Therefore,

$\begin{array}{rcl}{F}_p-F_g& =& ma\\ F_p& =& ma+F_g\end{array}$

Substitute the given values in the above equation,

$\begin{array}{rcl}{F}_p& =& \left(72.7\text{kg}\right)\left(-3.00{\text{m/s}}^2\right)+712\text{N}\\ & =& 494\text{N}\end{array}$

Hence, the force exerted by the pole on the firefighter is 494 N.

(b) Calculate the direction of the force exerted by the pole on the firefighter

From part (a), the direction of vertical force is along upwards since it has a positive value.

(c) Calculate the magnitude of vertical force on the pole from the firefighter

Let’s assume that the vertical force by the firefighter on the pole is F_f. From Newton’s third law, we can write,

$F_p=-F_f$

That means,

$F_f=-494\text{N}$

(d) Calculate the direction of vertical force on the pole from the firefighter

From part (c), the direction of vertical force on the pole by the firefighter is downward as it has a negative value.