Question #87218
Q1.A square plate of uniform thickness and length of side 30 cm hangs vertically from hinges at its top edge. When a horizontal jet strikes the plate at its centre, the plate is deflected and comes to rest at an angle of 30 to the vertical. The jet is 25 mm in diameter and has velocity of 6 m/s. Calculate the mass of the plate and give the distance along the plate from the hinge, of the point at which the jet strikes the plate in its deflected position.
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Expert's answer
2019-04-03T09:10:38-0400


Let find the force which water presses on plate:

FwΔt=m(v0)F_w \Delta t = m(v-0) (second Newnon's law)

FwΔt=ρVv=ρSvΔtvF_w \Delta t = \rho V v= \rho S v \Delta t v then

Fw=ρSv2=ρπ(d2)2v2=1000kgm3π(2,5m1022)236m2s2=17.67NF_w = \rho S v^2 = \rho \pi (\frac{d}{2})^2 v^2 = 1000 \frac{kg}{m^3} \cdot \pi (\frac{2,5m \cdot 10^{-2}}{2})^2 \cdot 36 \frac{m^2}{s^2} = 17.67N


AC - plate start position.

AC' - plate deflected position.

α=30\alpha = 30^\circ





AB=ABcosα=l2cosα=17.32cmAB' = \frac{AB}{\cos \alpha} = \frac{l}{2\cos \alpha} = 17.32cm - second Question






h2=ABcosα=l2h_2 = AB' \cdot \cos \alpha = \frac{l}{2}

h1=l2sinα=l2sin30=l4h_1 = \frac{l}{2} \cdot \sin \alpha = \frac{l}{2} \cdot \sin 30^\circ = \frac{l}{4}


moment of mg = moment of water force:

mgh1=Fwh2mgh_1 = F_wh_2

m=Fwh2gh1=2Fwgm = \frac{F_w h_2}{g h_1} = \frac{2F_w}{g}

m=217.67N10Nkg=3.534kgm = \frac{2\cdot 17.67N}{10 \frac{N}{kg}} = 3.534kg


So the answer:

1) 3.534kg

2) 17.32cm


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