Question

A gardener waters the plants by a pipe of diameter 1 mm. The water comes out from the pipe at
the rate of 10 cm3/sec. The reactionary force exerted on the hand of the gardener is :

Answer 1

A gardener waters the plants by a pipe of diameter 1 mm. The water comes out from the pipe at

the rate of 10 cm³/sec. The reactionary force exerted on the hand of the gardener is:

**Solution:**

$$V = 10^{-2} \mathrm{m}^3 - \text{volume of water for 1 second};$$ $$\Delta t = 1 \mathrm{s};$$ $$D = 10^{-3} \mathrm{m} - \text{diameter of the pipe};$$

F – the force with which the water acts on the hand of the gardener ;

N – reactionary force exerted on the hand of the gardener;

$$\rho = 1000 \frac{\mathrm{kg}}{\mathrm{m}^3} - \text{water density};$$

First we need to find the velocity of the water flow:

$$\theta = \frac{V}{S} = \frac{4V}{\pi D^2}$$

Change of momentum can be found from the formula

$$F \Delta t = p_1 - p_2; \quad p_1 = m \cdot \theta, p_2 = 0$$ $$F \Delta t = m \cdot \theta$$ $$m = \rho \cdot V$$

(1)in(2):

$$F \Delta t = \rho V \cdot \frac{4V}{\pi D^2} = \rho \frac{4V^2}{\pi D^2}$$ $$F = \rho \frac{4V^2}{\pi D^2 \Delta t}$$

When water exerts a force on a hand, the hand simultaneously exerts a force equal in magnitude and opposite in direction to that of the water.

$$\vec{N} = -\vec{F}; \quad |\vec{N}| = |\vec{F}|$$ $$x: N = F = \rho \frac{4V^2}{\pi D^2 \Delta t} = 1000 \frac{\mathrm{kg}}{\mathrm{m}^3} \cdot \frac{4 \cdot 10^{-2} \mathrm{m}^3}{\pi \cdot 10^{-3} \mathrm{m} \cdot 1 \mathrm{s}^2} = 12.7 \times 10^3 \mathrm{N}$$

Answer: the reactionary force exerted on the hand of the gardener is $12.7 \times 10^3 \mathrm{N}$