Question

It is desired to design a system for controlling the rate of extension of the fire-truck ladder during elevation of the ladder so that the bucket B will have vertical motion only. Determine if the elongation rate of the hydraulic cylinder is 0.7 ft/sec and θ = 21°, L = 13 ft, h = 3.2 ft, b = 9 ft, and x = 5 ft.

Answer 1

The elongation rate is 1.17 ft/s

Step-by-step explanation:

The expression is:

-(L - x)θsinθ + x*cosθ = 0

x = (L+x)θ * tanθ

CD² = 2b²(1 - cos(θ + δ))

2CC = 2b²θsin(θ + δ)

`O=\frac{CC*C\sqrt{2^(2)(1-cos(O+Y) } }{b^(2)sin(O+Y) }`

Where O = θ and Y = δ

`O=(√(2) )/(b) (\frac{√(1-cos(O+Y)) }{\sqrt{1-cos^(2)(O+Y) } } )C`

`x=(L+x)tanO(√(2) C)/(b√(1+cos(O+Y)) )`

Where

C = 1 ft/s

θ = 28°

L = 13 ft

h = 3.2 ft

b = 9 ft

x = 5 ft

δ = sin⁻¹(h/b) = sin⁻¹(3.2/9) = 20.83°

Replacing:

`x=((13+5)*tan(28)*√(2)*1 )/(9*√(1+cos(28+20.83)) ) =1.17ft/s`