Answer to Question #148530 in Electrical Engineering for Lim

Question #148530
A ladder 15ft long rest against a vertical wall. Its top slides down the wall while its bottom moves away along with the level ground at a speed of 2 ft/s . How fast is the angle between the top of the ladder and the wall changing when the angle is π/3 radians.
1
Expert's answer
2020-12-08T02:41:28-0500

Ladder and wall can be regarded as sides of a right triangle


Att=0,OB=0andOC=BC=15ft.OB=2tftAt t = 0, OB=0 and OC =BC =15 ft. OB =2t ft

sinθ=OBBC    θ=sin1OBBCsin \theta=\frac{OB}{BC} \implies \theta = sin^{-1}\frac{OB}{BC}

θ=sin12t15    dθ=22254t2dt\theta = sin^{-1}\frac{2t}{15} \implies d\theta=\frac{2}{\sqrt{225-4t^2}}dt

dθdt=22254t2\frac{d\theta}{dt}=\frac{2}{\sqrt{225-4t^2}}

Determining t, when θ=π3\theta=\frac{\pi}{3}

sinπ3=2t115    t1=1534sin\frac{\pi}{3}=\frac{2t_1}{15} \implies t_1= \frac{15\sqrt{3}}{4}

dθdtt=t1=22254t2=22254(1534)2\frac{d\theta}{dt}_{t=t_1}=\frac{2}{\sqrt{225-4t^2}}=\frac{2}{\sqrt{225-4(\frac{15 \sqrt{3}}{4})^2}}

=2225225(34)=415=\frac{2}{\sqrt{225-225(\frac{{3}}{4})}}=\frac{4}{15} rad/s


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