## Sunday, November 27, 2016

### Find the Water Speed V₂ in the Pipe on the Top Floor

Water at a gauge pressure P₁ at street level (zero, reference height) flows into an office building through a pipe with the radius R₁ of the circular cross-section. The pipe tapers down to the radius R₂ of the circular cross-section by the top floor, located at the height h, where the faucet has been left open. Find the water speed V₂ in the pipe on the top floor in terms of P₁, R₁, R₂, h, water density ρ and g. Assume no branch pipes and ignore viscosity.

### Find the Vertical Support Forces at Ends of the Lower Steel Beam

The lower uniform steel beam has a mass M. Only one-third of an identical steel beam is resting on the lower steel beam, as shown in Fig. Find the vertical support forces at ends of the lower steel beam in terms of M and g

### t₀=0, x₀=0, v(x)=√(p+qx). x(t)=?

t₀=0
x₀=0
v(x)=√(p+qx)
x(t)=?
Calculus-based solution:
dx/dt=q⁰·⁵(p/q+x)⁰·⁵
(p/q+x)⁻⁰·⁵dx = q⁰·⁵dt
(p/q+x)⁻⁰·⁵d(p/q+x) = q⁰·⁵dt
2(p/q+x)⁰·⁵ = q⁰·⁵t + c
2(p/q+x)⁰·⁵ -2(p/q+x₀)⁰·⁵= q⁰·⁵(t-t₀)
2(p/q+x)⁰·⁵ -2(p/q)⁰·⁵= q⁰·⁵t
(p/q+x)⁰·⁵ -(p/q)⁰·⁵= ½q⁰·⁵t
(p/q+x)⁰·⁵ = ½q⁰·⁵t+(p/q)⁰·⁵
p/q+x = ¼qt² + q⁰·⁵t(p/q)⁰·⁵ + p/q
x = ¼qt² + q⁰·⁵t(p/q)⁰·⁵
x = p⁰·⁵t + ¼qt²
x = v₀ t + ½at²
v₀=p⁰·⁵
a=½q
Physics-based and algebra-based solution
If a=const then v²=v₀²+2a·Δx, so
v=√(v₀²+2a·Δx)
As it is given:
v=√(p+qx)
so p=v₀², v₀=√p=p⁰·⁵
as x₀=0 then 2a·Δx=2ax
2ax=qx, 2a=q, a=½q
as x = v₀ t + ½at²
then x = √p t + ¼qt²

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