91Ó°ÊÓ

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.

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:

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

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

F = ma (3)

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.

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

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,

$F_f=-494\text{N}$

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