# Answer in Mechanics | Relativity for krithi #140169

A jet of water is discharging at a constant rate of 0.06 m 3 /s from the upper tank (Fig. 2). If the jet diameter at

section 1 is 10 cm, what forces will be measured by scales A and B? Assume the empty tank weighs 1335 N, the cross-sectional area of the tank is 0.37 m 2 , h= 30 cm, and H =2.74 m.(density of water at 60°F = 983.2 kg/m)

Q = 0.06 m^{3}/s

d = 0.1 m

“A = frac{u03c0}{4}d^2”

“A = 7.85 times 10^{-3} ;m^2”

“V = frac{Q}{A}”

“V = frac{0.06}{7.85 times 10^{-3}} = 7.64 ;m/s; (horizontal ;velocity)”

When water hits the lower tank, vertical velocity

“V_v = 0 + gt”

“V_v^2 = 2gu03b2”

β = H = 2.74 m

“V_v = sqrt{2 times 9.81 times 2.74}”

“V_v = 7.33 ;m/s”

Force at point A = force due to weight of water (Fw) + Force due to impulse of vertical velocity + weight of empty tank

“F_w = 0.37 times 0.3 times 9.81 = 1.08 ;kN”

“F_{impulse} = u03c1AV_v^2”

“F_{impulse} = 0.9832 times 7.85 times 10^{-3} times 7.33^2 = 0.414 ;kN”

Total force

“F_A = 1.08 + 0.414 + 1.335 = 2.829 ;kN”

Force at B = Force due to impulse of jet dur to horizontal component of velocity.

“F_B = u03c1AV_h^2”

“F_B = 0.9832 times 7.85 times 10^{-3} times 7.64^2 = 0.45 ;kN”