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.
Sunday, November 27, 2016
t₀=0, x₀=0, v(x)=√(p+qx). x(t)=?
t₀=0
x₀=0
v(x)=√(p+qx)
x(t)=?
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
v₀=p⁰·⁵
a=½q
Physics-based and algebra-based solution
If a=const then v²=v₀²+2a·Δx, so
v=√(v₀²+2a·Δx)
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
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
then x = √p t + ¼qt²
⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ⁱⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ·ΔΘΣΦΧΨΩαβγδεζηθικλμνξπρςστφχψωϑϓϕ√∛∜∞∠∣∥∫∬∭∮∯≤≥⊥℃ℇ℉ℏ℔№⅀⅏⎛⎞⎜⍊⎝½
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