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

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=$\frac{8}{10}$

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

$\frac{dx}{dt}$=-$\frac{dy}{dt}$

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

${at\theta}={37}$

$Y=6ft$

$and\ about\ Instantaneous\ centre\ of\ rotation$

${\omega r_A}={V_A}$

${\omega=\frac{1.2}{6}$

ω=0.2rad/s

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