Question

A fireman clings to a vertical ladder and directs the nozzle of a hose horizontally toward a burning building. The rate of water flow is 6.31kg/s, and the nozzle speed is 12.5 m/s. The hose passes vertically between the fireman’s feet, which are 1.30 m below the nozzle. Choose the origin to be inside the hose between the fireman’s feet. What torque must the fireman exert on the hose? (This could also be stated, what is the rate of change of the angular momentum of the water?)

Answer 1

Answer:

A torque of 102.5375 Nm must be exerted by the fireman

Explanation:

Given:

The rate of water flow = 6.31 kg/s

The speed of nozzle  = 12.5 m/s

Now, from the Newton's second law we have  

The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction

thus we have,

F = 6.31 kg/s x 12.5m/s

or

F = 78.875 N  

Now,

The torque (T) exerted by water force about the fireman's will be

T = (F x d)

or

T = 78.875 N x 1.30 m

T = 102.5375 Nm

hence,

A torque of 102.5375 Nm must be exerted by the fireman