Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.2 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.) rad/s

Answer 1

Answer:0.2 rad/s

Step-by-step explanation:

Given data

Velocity of the bottom point of the ladder=1.2Ft/s

Length of ladder=10ft

distance of the bottom most point of ladder from origin=8ft

From the data the angle θ with ladder makes with horizontal surface is

Cosθ=
`(8)/(10)`

θ=36.86≈37°

We have to find rate of change of θ

From figure we can say that

`x^(2)`+
`y^(2)`=
`AB^(2)`

Differentiating above equation we get

`(dx)/(dt)`=-
`(dy)/(dt)`

i.e
`{V_A}=-{V_B}=1.2ft/s`

`{at\theta}={37}`

`Y=6ft`

`and\ about\ Instantaneous\ centre\ of\ rotation`

`{\omega r_A}={V_A}`

`{\omega=(1.2)/(6)`

ω=0.2rad/s

i.e.Rate of change of angle=0.2 rad/s