Question

Provided the diameter of the fire hose is 7.0 cm, and the water running through the hose flows at 420 L/min, how much force does she need to hold the nozzle, if its diameter is 0.75 cm?

Answer 1

the amount of force needed to hold the nozzle is 1100 N

Step-by-step explanation:

Given that;

The diameter of the fire horse is 7.0cm

The radius of the fire horse is d/2 = 7.0/2 = 3.5 cm = 0.035 m

the water running through the hose flows at 420 L/min = 0.42 m³/ 60 sec

The diameter of the nozzle = 0.75 cm

The radius of the nozzle = d/2 = 0.75/2 = 0.375 cm = 0.00375 m

Amount of force needed to hold the nozzle = ??

Using equation of continuity product

`A_1v_1 = A_2v_2`

where

`A_1` = cross sectional area of the hose

`v_1` = velocity of water flow at hose

`A_2` = cross sectional area of the nozzle

`v_2` = velocity of water flow at nozzle

Making
`v_1` the subject of the formula; we have

`v_1 = (A_2v_2)/(A_1)`

Also making
`v_2` the subject of the formula:

`v_2 = (A_1v_1)/(A_2)`

Also; from newton's law; we all know that F = ma

where ;

m = mass

a = acceleration

`a = (\Delta v)/(t) \\ \\ a = (v_2-v_1)/(t)`

So ;

`F = m (v_2-v_1)/(t)`

Also ; mass = density × volume

`m =\rho *V`

`F = \rho V (v_2-v_1)/(t)`

`F = (\rho V)/(t)[ (A_1v_1)/(A_2)- (A_2v_2)/(A_1)]`

Replacing
`A_2v_2` by
`A_1v_1` from above ; so

`F = (\rho V)/(t)[ (A_1v_1)/(A_2)- (A_1v_1)/(A_1)]`

`F = (\rho V)/(t) A_1 v_1 [ (1)/(A_2)- (1)/(A_1)]`

where the product of cross section area of velocity is equal to the volume of water and its time flow;

SO;

`A_1v_1 = (V)/(t)`

Replacing that into what we have above; we have:

`F = (\rho V)/(t) (V)/(t) [ (1)/(A_2)- (1)/(A_1)]`

`F ={\rho }( (V)/(t))^2 [ (1)/(A_2)- (1)/(A_1)]`

Where;

A = πr²

`F ={\rho }( (V)/(t))^2 [ (1)/(\pi r^2_2)- (1)/(\pi r^2_1)]`

`F ={(1000*10^3 \ kg/m^3) }( (0.42 m^3 )/(60 sec))^2 [ (1)/(\pi (0.00375)^2)- (1)/(\pi (0.035)^2)]`

F = 1100 N

Thus; the amount of force needed to hold the nozzle is 1100 N