Question

A firefighter with a weight of 756 N slides down a vertical pole with an acceleration of 2.96 m/s2, directed downward. What are the magnitudes and directions (choose the positive direction up) of the vertical forces

(a) on the firefighter from the pole and
(b) on the pole from the firefighter?

Answer 1

a) F = 527.65 N, Force applied is upwards.

b)F = - 527.65 N, where, negative sign depicts Force is applied downwards.

Step-by-step explanation:

Data provided:

Weight of the firefighter = 756 N

Mass of the firefighter = 756/9.8 = 77.14 Kg

Acceleration, a = 2.96 m/s²

a) In the absence of the pole the firefighter would have been moving down with an acceleration of 9.8 m/s (i.e the acceleration due to the gravity), but due to the presence of the pole the acceleration of the firefighter has been reduced. thus, a force is applied by the pole on the firefighter to reduce the acceleration.

therefore, we have

F = ma(net) = 77.14 × (9.8-2.96) = 527.65 N, Force applied is upwards.

B) According to the Newton's third law, the force will be equal and opposite to the force in the part a)

thus, we have

F = - 527.65 N