Hey Help With Momentum Questions

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SUMMARY

The discussion focuses on calculating the supporting force required for a tank system against gravity, initially weighing 1000 N with 0.6 m³ of water. The calculated force to support the system is 6886 N, derived from converting water volume to kilograms and applying gravitational force. When a streamlined orifice with a diameter of 120 mm discharges water at a rate of 90.5 kg/s, the new supporting force required is 6156 N. The reduction in weight is attributed to the thrust generated by the liquid discharge, calculated using the formula T = mdot * V.

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Heya i have this question over here... i have done the first part of it

An open tank weighing 1000 N contains 0.6 m3 of water. Calculate the force required to support the system against gravity.

Answer is 6886 .. the calculation i have done is convert the 0.6 meter cubic into m.liters and then convert it to Kilograms after that i multiply it with Gravity to give me 5886 and then i add the weight of the tank on it to give me the final answer

If a streamlined orifice of 120 mm diameter, discharging a flow of 90.5 kg/s, is fitted in the floor of the tank, what supporting force will now be required? Assume that the orifice discharges to atmosphere, and that the water level in the tank remains steady. Ans:6156

HERE where i am stuck i don't know what to do after this step so please help me thanks
 
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anyhelp guys?
 
weight reduction comes from "thrust" created by the draining of the liquid, i guess.

at anyrate, T = mdot * V. mdot = rho A V. you can solve backwards for V (knowing the orifice diameter, and density of water), then get your V.

you should get: Wnew = Wold - T, hopefully 6886N - T = 6156N
 

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