Answer to Question #130127 in Mechanics | Relativity for Ariyo Emmanuel

Question

Water is accelerated by a nozzle to an average speed of 20 m/s, and strikes
a stationary vertical plate at a rate of 10 kg/s with a normal velocity of 20
m/s (Fig. below). After the strike, the water stream splatters off in all
directions in the plane of the plate. Determine the force needed to prevent
the plate from moving horizontally due to the water stream.

Answer 1

Solution:

The momentum equation for steady one-dimensional flow is given as:

$\Sigma \vec{F} = \Sigma_{(out)} \beta \dot{m}\vec{V}-\Sigma_{(in)} \beta \dot{m}\vec{V}$

Writing it for this problem along the x-direction (without forgetting the negative sign for forces and velocities in the negative x-direction) and noting that $V_{(1,x)} = V_{1}$ and $V_{(2,x)}=0$ gives

$$-F_{R}=0-\beta \dot{m}\vec{V}_{1}$$

Substituting the given values,

$F_{R}=\beta \dot{m}\vec{V}_{1}=(1)(10 \tfrac{kg}{s} )(20 \tfrac{m}{s} )( \tfrac{1N}{1kg* \tfrac{m}{s^{2}} } ) = 200 N$

Therefore, the support must apply a 200-N horizontal force (equivalent to the weight of about a 20-kg mass) in the negative x-direction (the opposite direction of the water jet) to hold the plate in place.