Question

A firefighter with a weight of 707 N slides down a vertical pole with an acceleration of 2.79 m/s2, directed downward. (a) What is the magnitude of the vertical force on the firefighter from the pole?

Answer 1

Answer:

The vertical force acting on the firefighter = 908.27 N

Explanation:

Force: Force of a body is defined as the product of mass and its acceleration. The S.I unit of force is Newton (N)

The vertical force acting on the firefighter = Force due to the weight of the firefighter + force due to acceleration.

Ft = Fw - Fa

Where Ft = The vertical force acting on the firefighter, Fw = Force due to the weight of the firefighter, Fa = force due to acceleration.

Fw = mg

Making m the subject of formula in the equation above

m = Fw/g................... Equation 1

Where m = mass of the firefighter, g = acceleration due to gravity,

Given: Fw = 707 N,

Constant: g = 9.8 m/s²

Substituting these values into eqaution 1

m = 707/9.8

m = 72.14 kg.

But, Fa = ma

Where a = acceleration of the firefighter.

Given: a = 2.79 m/s²,

And m = 72.14 kg

Fa = 72.14 × 2.79

Fa = 201.27 N

Therefore, Ft = 707 + 201.27  = 908.27 N

Ft = 908.27 N

The vertical force acting on the firefighter = 908.27 N